{"id":149,"date":"2023-11-08T16:10:06","date_gmt":"2023-11-08T16:10:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-divide-polynomials-part-i\/"},"modified":"2024-11-01T19:30:05","modified_gmt":"2024-11-01T19:30:05","slug":"4-8-polynomial-division","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/4-8-polynomial-division\/","title":{"raw":"4.8 Polynomial Division","rendered":"4.8 Polynomial Division"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning OutcomeS<\/h3>\r\n<ul>\r\n \t<li>Divide a polynomial by a monomial.<\/li>\r\n \t<li>Divide a polynomial by a polynomial.<\/li>\r\n \t<li>Divide polynomial functions given in function notation.<\/li>\r\n<\/ul>\r\n<\/div>\r\nLet's review the terminology related to the division of whole numbers. If we divide [latex]\\require{color}7[\/latex] by [latex]3[\/latex], [latex]7[\/latex] is called the <strong>dividend<\/strong>, and [latex]3[\/latex] is called the <strong>divisor<\/strong>. The division result can be written as a mixed number [latex]2 \\dfrac{1}{3}[\/latex] that means [latex]2+\\dfrac{1}{3}[\/latex]. The whole part of the division result, [latex]2[\/latex], is called the <strong>quotient<\/strong>, and the numerator of the \u201cfractional part\u201d, [latex]1[\/latex], is called the <strong>remainder<\/strong>. The entire division statement can be written mathematically in two ways, using a fraction for the division, or using the division symbol.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{7}{3}=2+\\dfrac{1}{3}[\/latex]\u00a0 \u00a0or\u00a0 \u00a0[latex]7\\div3=2+\\dfrac{1}{3}[\/latex]<\/p>\r\nIn general,\r\n<p style=\"text-align: center;\">[latex]\\dfrac{{\\color{Green}{\\textsf{dividend}}}}{{\\color{blue}{\\textsf{divisor}}}}={\\color{Orange}{\\textsf{quotient}}}+\\dfrac{{\\color{Plum}{\\textsf{remainder}}}}{{\\color{blue}{\\textsf{divisor}}}}[\/latex]\u00a0 \u00a0or\u00a0 \u00a0[latex]{\\color{Green}{\\textsf{dividend}}}\\div {\\color{blue}{\\textsf{divisor}}}={\\color{Orange}{\\textsf{quotient}}}+\\dfrac{{\\color{Plum}{\\textsf{remainder}}}}{{\\color{blue}{\\textsf{divisor}}}}[\/latex]<\/p>\r\nWe say that the division is performed on <strong>improper fractions<\/strong> which means that the divisor is not greater than the dividend. The quotient is never greater than the dividend, and the remainder is nonnegative and always less than the divisor. When the remainder is zero, we say that the dividend is <strong>divisible<\/strong> by the divisor. In such case, the quotient itself is the result of the division. A way of verifying correctness of a division statement is based on the fact that the division statement can be presented in equivalent form as\r\n<p style=\"text-align: center;\">[latex]{\\color{Green}{\\textsf{dividend}}} = {\\color{blue}{\\textsf{divisor}}}\\cdot{\\color{Orange}{\\textsf{quotient}}}+{\\color{Plum}{\\textsf{remainder}}}[\/latex]<\/p>\r\nUsing our example of the division of [latex]{\\color{Green}{7}}[\/latex] by [latex]{\\color{blue}{3}}[\/latex], the statement [latex]{\\color{Green}{7}}\\div{\\color{blue}{3}}={\\color{Orange}{2}}+\\dfrac{{\\color{Plum}{1}}}{{\\color{blue}{3}}}[\/latex] is true only if the statement [latex]{\\color{Green}{7}}={\\color{blue}{3}}\\cdot {\\color{Orange}{2}}+{\\color{Plum}{1}}[\/latex] is true. In such verification, we usually start with the RHS (right hand side) of the latter statement and simplify to check if it matches the LHS (left hand side). Since\u00a0[latex]3\\cdot 2+1[\/latex] is indeed equal to [latex]7[\/latex], both statements are true.\u00a0Notice that the quotient ([latex]\\color{Orange}{2}[\/latex]) is not greater than the dividend ([latex]\\color{Green}{7}[\/latex]). Also, the remainder ([latex]\\color{Plum}{1}[\/latex]) is not negative.\r\n<h2>Dividing a Polynomial by a Monomial<\/h2>\r\nDividing a polynomial by a monomial can be handled by dividing each term in the dividend polynomial separately by the divisor monomial. This is based on the fact that adding or subtracting rational expressions with like denominators is equivalent to operating on the numerators keeping the common denominator the same: [latex]\\displaystyle \\frac{a}{c}\\pm\\frac{b}{c}=\\frac{a\\pm b}{c}[\/latex]. We just use that equivalency\u00a0\u201cbackwards.\u201d We will refer to this fact as \u201cusing linearity of operations on rational expressions.\u201d\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nDivide [latex]12x^3+2x^2-8x+1[\/latex] by [latex]4x[\/latex].\r\n\r\n[reveal-answer q=\"38801\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"38801\"]\r\n\r\nWe first rewrite the division in fractional form and use linearity of operations on rational expressions. We then simplify each rational expression if possible.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{12x^3+2x^2-8x+1}{4x} &amp;= \\frac{12x^3}{4x} + \\frac{2x^2}{4x} - \\frac{8x}{4x} + \\frac{1}{4x}\\\\[5pt]\r\n&amp;= \\frac{3\\cdot {\\color{red}\\cancel{\\color{black}{4\\cdot x}}}\\cdot x^2}{{\\color{red}\\cancel{\\color{black}{4\\cdot x}}} \\cdot 1} + \\frac{{\\color{red}\\cancel{\\color{black}{2\\cdot x}}} \\cdot x}{2\\cdot {\\color{red}\\cancel{\\color{black}{2\\cdot x}}}} - \\frac{2\\cdot {\\color{red}\\cancel{\\color{black}{4\\cdot x}}}}{{\\color{red}\\cancel{\\color{black}{4\\cdot x}}}\\cdot 1} + \\frac{1}{4x}\\\\[5pt]\r\n&amp;= 3x^2 + \\frac{x}{2} - 2 + \\frac{1}{4x}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOf course, when we divide by a variable expression, we have to determine restricted values of the variable that result in division by zero. In the example above, [latex]x\\ne 0[\/latex].\r\n\r\nWe usually refrain from using the notions of quotient and remainder when dividing by a monomial.\r\n<h2>Divide a Polynomial by a Polynomial<\/h2>\r\nThe easiest case of dividing a polynomial by a general polynomial is when the dividend contains the divisor as a factor. For example,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{x^2-4x-12}{x+2} &amp;= \\frac{{\\color{red}\\cancel{\\color{black}{(x+2)}}}\\cdot (x-6)}{{\\color{red}\\cancel{\\color{black}{(x+2)}}}\\cdot 1}\\\\[5pt]\r\n&amp;= x-6\r\n\\end{align}\r\n[\/latex]<\/p>\r\nThat is of course dependent on the ability to factor that might prove useful if you encounter a division example in which the divisor is a factor of the dividend. As always, remember about the restricted values of the variable (the real zeros of the divisor). In the example above, [latex]x\\ne -2[\/latex].\r\n\r\nIt is also possible that\u00a0the dividend and divisor contain\u00a0common factors that can be divided out. For example,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{x^2-x-2}{x^2+3x+2} &amp;= \\frac{{\\color{red}\\cancel{\\color{black}{(x+1)}}}\\cdot (x-2)}{{\\color{red}\\cancel{\\color{black}{(x+1)}}}\\cdot (x+2)}\\\\[5pt]\r\n&amp;= \\frac{x-2}{x+2}\r\n\\end{align}\r\n[\/latex]<\/p>\r\nThe final rational expression is still improper and requires further division.\r\n\r\nIn general, when dividing a polynomial by a polynomial, we will use long division, a process similar to long division of whole numbers, or synthetic division (discussed in the next section). The notions of the dividend, divisor, quotient, and reminder extend to division of polynomials with some adjustments:\r\n<ul>\r\n \t<li>In an <strong>improper<\/strong> rational expression, the <strong>degree<\/strong> of the divisor is not greater than the <strong>degree<\/strong> of the dividend.<\/li>\r\n \t<li>The <strong>degree<\/strong> of the quotient is never greater than the <strong>degree<\/strong> of the dividend. Moreover, the degree of the quotient is the difference of the degrees of the dividend and the divisor.<\/li>\r\n \t<li>The <strong>degree<\/strong> of the remainder is always less than the <strong>degree<\/strong> of the divisor or the remainder is [latex]0[\/latex].<\/li>\r\n<\/ul>\r\nLet's discuss the setup of the long division diagram. The basic setup is similar to long division of whole numbers:\r\n<p style=\"text-align: center;\">[latex]\r\n\\require{enclose}\r\n{\\color{blue}{\\textsf{divisor}}} \\enclose{longdiv}{{\\color{Green}{\\textsf{dividend}}}}\r\n[\/latex]<\/p>\r\nThe place above the line over the dividend is designated for the <strong><span style=\"color: #ff6600;\">quotient<\/span><\/strong>. The dividend and divisor should be written in descending powers of the variable. To avoid mistakes, the dividend should have terms of all degrees with a coefficient of zero for all missing degrees of terms. There is no need to include terms of missing degrees in the divisor. We suggest that you include vertical lines separating columns of terms of specific degree. Since the constant term might be any real number and only nonzero real numbers perceived as a polynomial have the degree of zero, we will label the last column as\u00a0\u201cc.t.\u201d For example, for the division of [latex]{\\color{Green}{x^4-2x^2+3x-4}}[\/latex] by [latex]{\\color{blue}{x^2+1}}[\/latex], the long division diagram should look like this:\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{10pt}4\\hspace{22pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{x^2+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{{} - 2x^2} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{-4}\\\\[-10pt]\r\n{\\color{blue}{x^2+1}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{x^4}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+0x^3}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-2x^2}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+3x}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-4}}}\\\\[-10pt]\r\n\\phantom{x^2+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\vdots}} \\phantom{{} + 0x^3} {\\color{Gray}{\\vdots}} \\phantom{{} - 2x^2} {\\color{Gray}{\\vdots}} \\phantom{{} + 3x} {\\color{Gray}{\\vdots}} \\phantom{-4}\\\\[-10pt]\r\n\\end{array}\r\n[\/latex]<\/p>\r\nLet's now describe a <strong>long division step<\/strong>. In long division of whole numbers, the question we'd ask is \u201chow many times does the divisor go into the number made of the first sufficient number of digits of the dividend?\u201d In long division of polynomials, we simply divide the first (leading) term of the dividend (actually, what we will refer to as the <em>current dividend<\/em>, which we will explain in a bit) by the first (leading) term of the divisor and place the simplified result of that division in the proper column of the quotient space in the diagram. In our example, we divide [latex]\\dfrac{\\;{\\color{Green}\\enclose{box}{x^4}}\\;}{{\\color{blue}{\\enclose{box}{x^2}}}}[\/latex]. We write the simplified result, [latex]{\\color{Orange}{\\enclose{box}{x^2}}}[\/latex], in the [latex]2[\/latex]<sup>nd<\/sup> degree column of the quotient:\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{14pt}\\textsf{deg.}\\hspace{12pt}4\\hspace{24pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{x^2}+1-\\enclose{box}{x^4}}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{{}+0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{-2}{\\color{Orange}{\\enclose{box}{x^2}}} {\\color{Gray}{\\Bigl|}} \\phantom{{}+3x} {\\color{Gray}{\\Bigl|}} \\phantom{{}-4}\\\\[-10pt]\r\n{\\color{blue}{\\enclose{box}{x^2}+1}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{\\enclose{box}{x^4}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+0x^3}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-2x^2}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+3x}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-4}}}\\\\[-10pt]\r\n\\end{array}\r\n[\/latex]<\/p>\r\nNext we multiply the newly obtained term of the quotient by the (entire) divisor and subtract the simplified result from the (current) dividend. Since addition is easier than subtraction and to subtract means to add the opposite, we write the opposite of that simplified result under the (current) dividend placing the terms of specific degree in the proper column, and add. In our example, we multiply [latex]{\\color{Orange}{x^2}}\\cdot\\left({\\color{blue}{x^2+1}}\\right)[\/latex], simplify the result to [latex]x^4+x^2[\/latex], write the opposite of that simplified result, [latex]{\\color{red}{-x^4-x^2}}[\/latex], under the (current) dividend with proper placement of terms, and add:\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{10pt}4\\hspace{22pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{x^2+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{{} - 2}{\\color{Orange}{x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{{ }- 4}\\\\[-10pt]\r\n{\\color{blue}{x^2+1}} \\enclose{longdiv}{\\phantom{-} x^4 {\\color{Gray}{\\Bigl|}} +0x^3 {\\color{Gray}{\\Bigl|}} -2x^2 {\\color{Gray}{\\Bigl|}} +3x {\\color{Gray}{\\Bigl|}} -4}\\\\[-10pt]\r\n\\phantom{x^2+1\\hspace{3pt}} \\enclose{bottom}{{\\color{red}{- x^4}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}}\\hspace{5pt} {\\color{red}{{} - x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{{} -4}}\\\\[-10pt]\r\n\\phantom{x^2+1\u00a0-x^4 {\\color{Gray}{\\Bigl|}} + 0x^3 {\\color{Gray}{\\Bigl|}}}\\hspace{-9pt} -3x^2 {\\color{Gray}{\\Bigl|}} + 3x {\\color{Gray}{\\Bigl|}} -4\r\n\\end{array}\r\n[\/latex]<\/p>\r\nWe ask now if the <strong>degree<\/strong> of the polynomial resulting from the addition is <strong>less than<\/strong> the <strong>degree<\/strong> of the divisor or if that resulting polynomial is [latex]0[\/latex]. If <strong>yes<\/strong>, we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. That resulting polynomial is the <strong>remainder<\/strong>. If <strong>no<\/strong>, the resulting polynomial becomes the new (<strong>current<\/strong>) <strong>dividend<\/strong> and we <span style=\"color: #ff0000;\"><strong>continue<\/strong><\/span> with another long division step. In our example, the degree of [latex]-3x^2+3x-4[\/latex] is [latex]2[\/latex]. Since [latex]2[\/latex] is not less than [latex]2[\/latex], the degree of the divisor, [latex]{\\color{blue}{x^2+1}}[\/latex],\u00a0[latex]{\\color{Green}{-3x^2+3x-4}}[\/latex] becomes the current dividend and we continue with another long division step. We divide the leading term of the current dividend, [latex]{\\color{Green}{\\enclose{box}{-3x^2}}}[\/latex] by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{x^2}}}[\/latex], and write the simplified result, [latex]{\\color{Orange}{\\enclose{box}{-3}}}[\/latex] in the [latex]\\textsf{c.t.}[\/latex]\u00a0column of the quotient:\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{14pt}\\textsf{deg.}\\hspace{10pt}4\\hspace{22pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{x^2}+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{{} - 2}{\\color{Orange}{x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} {\\color{Orange}{\\enclose{box}{-3}}}\\\\[-10pt]\r\n{\\color{blue}{\\enclose{box}{x^2}+1}} \\enclose{longdiv}{\\phantom{-} x^4 {\\color{Gray}{\\Bigl|}} +0x^3 {\\color{Gray}{\\Bigl|}} -2x^2 {\\color{Gray}{\\Bigl|}} +3x {\\color{Gray}{\\Bigl|}} -4}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{x^2}+1\\hspace{3pt}} \\enclose{bottom}{- x^4 {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}}\\hspace{5pt} - x^2 {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{-4}}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{x^2}+1\\hspace{1pt} -x^4 \\color{Gray}{\\Bigl|} + 0x^3 {\\color{Gray}{\\Bigl|}}} {\\color{Green}{\\enclose{box}{-3x^2}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+ 3x}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-4}}\\\\[-10pt]\r\n\\end{array}\r\n[\/latex]<\/p>\r\nWe follow by\u00a0multiplying [latex]{\\color{Orange}{-3}}\\cdot\\left({\\color{blue}{x^2+1}}\\right)[\/latex], simplify the result to [latex]-3x^2-3[\/latex], write the opposite of that simplified result, [latex]{\\color{red}{+3x^2+3}}[\/latex], under the current dividend with proper placement of terms, and add:\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{10pt}4\\hspace{22pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{x^2+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{{} - 2}{\\color{Orange}{x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} {\\color{Orange}{-3}}\\\\[-10pt]\r\n{\\color{blue}{x^2+1}} \\enclose{longdiv}{\\phantom{-} x^4 {\\color{Gray}{\\Bigl|}} +0x^3 {\\color{Gray}{\\Bigl|}} -2x^2 {\\color{Gray}{\\Bigl|}} +3x {\\color{Gray}{\\Bigl|}} -4}\\\\[-10pt]\r\n\\phantom{x^2+1\\hspace{3pt}} \\enclose{bottom}{- x^4 {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}}\\hspace{5pt} - x^2 {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{{} -4}}\\\\[-10pt]\r\n\\phantom{x^2+1\u00a0-x^4 {\\color{Gray}{\\Bigl|}} + 0x^3 {\\color{Gray}{\\Bigl|}}}\\hspace{-9pt} -3x^2 {\\color{Gray}{\\Bigl|}} + 3x {\\color{Gray}{\\Bigl|}} -4\\\\[-10pt]\r\n\\phantom{x^2+1 -x^4 {\\color{Gray}{\\Bigl|}} + 0x^3 {\\color{Gray}{\\Bigl|}}}\\hspace{-1pt} \\enclose{bottom}{{\\color{red}{{} + 3x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} {\\color{red}{{} + 3}}}\\\\[-10pt]\r\n\\phantom{x^2+1 -x^4 {\\color{Gray}{\\Bigl|}} + 0x^3 {\\color{Gray}{\\Bigl|}}\u00a0- 3x^2 {\\color{Gray}{\\Bigl|}}} \\hspace{3.5pt} 3x {\\color{Gray}{\\Bigl|}} -1\r\n\\end{array}\r\n[\/latex]<\/p>\r\nSince the degree of [latex]3x-1[\/latex], [latex]1[\/latex], is less than [latex]2[\/latex], the degree of the divisor, we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. [latex]{\\color{Plum}{3x-1}}[\/latex] is the remainder and [latex]{\\color{Orange}{x^2-3}}[\/latex] is the complete quotient. We can write the division statement in the form\r\n<p style=\"text-align: center;\">[latex]\r\n\\displaystyle\\frac{{\\color{Green}{x^4-2x^2+3x-4}}}{{\\color{blue}{x^2+1}}} = {\\color{Orange}{x^2-3}} + \\frac{{\\color{Plum}{3x-1}}}{{\\color{blue}{x^2+1}}}\r\n[\/latex]<\/p>\r\nand verify the correctness of this statement by simplifying [latex]\\left({\\color{blue}{x^2+1}}\\right)\\cdot \\left({\\color{Orange}{x^2-3}}\\right)+{\\color{Plum}{3x-1}}[\/latex] to verify if it matches [latex]{\\color{Green}{x^4-2x^2+3x-4}}[\/latex]. We have\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\left(x^2+1\\right)\\cdot \\left(x^2-3\\right)+3x-1 &amp;= x^4-3x^2+x^2-3+3x-1\\\\\r\n&amp;= x^4-2x^2+3x-4\\ \\large\\checkmark\r\n\\end{align}\r\n[\/latex]<\/p>\r\nNote, that since [latex]x^2+1\\ne 0[\/latex] for all real numbers [latex]x[\/latex], there are no restricted values of the variable in the example above.\r\n\r\nIn the video that follows, we show another example of dividing a degree three trinomial by a binomial.\r\n\r\nhttps:\/\/youtu.be\/Rxds7Q_UTeo\r\n\r\nLet's attempt long division on an earlier example where the divisor was a factor of the dividend.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse long division to divide [latex]x^2-4x-12[\/latex] by [latex]x+2[\/latex].\r\n\r\n[reveal-answer q=\"455187\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455187\"]\r\n\r\nSetup the long division diagram, divide the leading term of the dividend, [latex]{\\color{Green}{\\enclose{box}{x^2}}}[\/latex], by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{x}}}[\/latex], and place the simplified result, [latex]{\\color{Orange}{\\enclose{box}{x}}}[\/latex], in the proper column of the quotient.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{12pt}2\\hspace{22pt}1\\hspace{17pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{x}+2-\\enclose{box}{x^2}}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{-4}{\\color{Orange}{\\enclose{box}{x}}} {\\color{Gray}{\\Bigl|}} \\phantom{-12}\\\\[-10pt]\r\n{\\color{blue}{\\enclose{box}{x}+2}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{\\enclose{box}{x^2}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-4x}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-12}}}\r\n\\end{array}\r\n[\/latex]<\/p>\r\nMultiply [latex]{\\color{Orange}{x}}[\/latex] by [latex]{\\color{blue}{x+2}}[\/latex], simplify the result to [latex]x^2+2x[\/latex], write the opposite of it, [latex]{\\color{red}{-x^2-2x}}[\/latex], under the (current) dividend, and add.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{7pt}\\textsf{deg.}\\hspace{10pt}2\\hspace{20pt}1\\hspace{17pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{x+2 -x^2}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{{}-4}{\\color{Orange}{x}} {\\color{Gray}{\\Bigl|}} \\phantom{-12}\\\\[-10pt]\r\n{\\color{blue}{x+2}} \\enclose{longdiv}{\\phantom{-}x^2 {\\color{Gray}{\\Bigl|}} -4x {\\color{Gray}{\\Bigl|}} -12}\\\\[-10pt]\r\n\\phantom{x+2\\hspace{3pt}} \\enclose{bottom}{{\\color{red}{- x^2}} {\\color{Gray}{\\Bigl|}} {\\color{red}{{} - 2x}} {\\color{Gray}{\\Bigl|}} \\phantom{{} -12}}\\\\[-10pt]\r\n\\phantom{x+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-9pt}} -6x {\\color{Gray}{\\Bigl|}} -12\r\n\\end{array}\r\n[\/latex]<\/p>\r\nSince the degree of [latex]-6x-12[\/latex], [latex]1[\/latex], is not less than [latex]1[\/latex], the degree of the divisor, [latex]{\\color{blue}{x+2}}[\/latex],\u00a0[latex]{\\color{Green}{-6x-12}}[\/latex] becomes the current dividend and we continue with another long division step. We divide the leading term of the current dividend, [latex]{\\color{Green}{\\enclose{box}{-6x}}}[\/latex], by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{x}}}[\/latex], and write the simplified result, [latex]{\\color{Orange}{\\enclose{box}{-6}}}[\/latex], in the proper\u00a0column of the quotient.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{10pt}2\\hspace{20pt}1\\hspace{17pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{x}+2 -x^2}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{{}-4}{\\color{Orange}{x}} {\\color{Gray}{\\Bigl|}} {\\color{Orange}{\\enclose{box}{-6}}}\\\\[-10pt]\r\n{\\color{blue}{\\enclose{box}{x}+2}} \\enclose{longdiv}{\\phantom{-}x^2 {\\color{Gray}{\\Bigl|}} -4x {\\color{Gray}{\\Bigl|}} -12}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{x}+2\\hspace{3pt}} \\enclose{bottom}{- x^2 {\\color{Gray}{\\Bigl|}} - 2x {\\color{Gray}{\\Bigl|}} \\phantom{{} -12}}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{x}+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-8.5pt}} {\\color{Green}{\\enclose{box}{-6x}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-12}}\r\n\\end{array}\r\n[\/latex]<\/p>\r\nWe follow by\u00a0multiplying [latex]{\\color{Orange}{-6}}\\cdot({\\color{blue}{x+2}})[\/latex], simplify the result to [latex]-6x-12[\/latex], write the opposite of it, [latex]{\\color{red}{+6x+12}}[\/latex], under the current dividend with proper placement of terms, and add.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{7pt}\\textsf{deg.}\\hspace{10pt}2\\hspace{20pt}1\\hspace{17pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{x+2 -x^2}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{{}-4}{\\color{Orange}{x}} {\\color{Gray}{\\Bigl|}} {\\color{Orange}{-6}}\\\\[-10pt]\r\n{\\color{blue}{x+2}} \\enclose{longdiv}{\\phantom{-}x^2 {\\color{Gray}{\\Bigl|}} -4x {\\color{Gray}{\\Bigl|}} -12}\\\\[-10pt]\r\n\\phantom{x+2\\hspace{3pt}} \\enclose{bottom}{- x^2 {\\color{Gray}{\\Bigl|}} - 2x {\\color{Gray}{\\Bigl|}} \\phantom{{} -12}}\\\\[-10pt]\r\n\\phantom{x+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-9pt}} -6x {\\color{Gray}{\\Bigl|}} -12\\\\[-10pt]\r\n\\phantom{x+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-10.5pt}} \\enclose{bottom}{{\\color{red}{{}+6x}} {\\color{Gray}{\\Bigl|}} {\\color{red}{{}+12}}}\\\\[-10pt]\r\n\\phantom{x+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-8.5pt} {\\color{red}{{}+6x}} {\\color{Gray}{\\Bigl|}} {\\color{red}{{}+{}}}}0\r\n\\end{array}\r\n[\/latex]<\/p>\r\nSince the resulting polynomial is [latex]0[\/latex], we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. The remainder is [latex]{\\color{Plum}{0}}[\/latex] and [latex]{\\color{Orange}{x-6}}[\/latex] is the complete quotient. We can write the division statement in the form\r\n<p style=\"text-align: center;\">[latex]\r\n\\displaystyle\\frac{{\\color{Green}{x^2-4x-12}}}{{\\color{blue}{x+2}}} = {\\color{Orange}{x-6}} + \\frac{{\\color{Plum}{0}}}{{\\color{blue}{x+2}}}\r\n[\/latex]<\/p>\r\nor, in simplified form, as\r\n<p style=\"text-align: center;\">[latex]\r\n\\displaystyle\\frac{{\\color{Green}{x^2-4x-12}}}{{\\color{blue}{x+2}}} = {\\color{Orange}{x-6}}\r\n[\/latex]<\/p>\r\nand verify the correctness of this statement by simplifying [latex]({\\color{blue}{x+2}})\\cdot ({\\color{Orange}{x-6}}) + {\\color{Plum}{0}}[\/latex] to verify if it matches [latex]{\\color{Green}{x^2-4x-12}}[\/latex]. We have\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n(x+2)\\cdot (x-6) + 0 &amp;= x^2-6x+2x-12\\\\\r\n&amp;= x^2-4x-12\\ \\large\\checkmark\r\n\\end{align}\r\n[\/latex]<\/p>\r\nWe also note that [latex]x\\ne -2[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show a couple more examples of dividing a degree two trinomial by a degree one binomial.\r\n\r\nhttps:\/\/youtu.be\/KUPFg__Djzw\r\n\r\nIn the next video, we present one more example of polynomial long division.\r\n\r\nhttps:\/\/youtu.be\/P6OTbUf8f60\r\n<h2>Polynomial Division and Function Notation<\/h2>\r\nIn <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/3-2-the-algebra-of-functions\/\" target=\"_blank\" rel=\"noopener\">Sec. 3.2<\/a>, we discussed quotient functions of the form [latex]\\dfrac{f}{g}[\/latex]. If both [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are polynomial functions, the quotient functions\u00a0[latex]\\dfrac{f}{g}(x)=\\dfrac{f(x)}{g(x)}[\/latex] and\/or [latex]\\dfrac{g}{f}(x)=\\dfrac{g(x)}{f(x)}[\/latex] can possibly be simplified using polynomial division. Recall that we only perform division on improper rational expressions. Hence, if the degrees of both [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are the same, both quotient functions can be simplified using polynomial division. Always make sure to identify the domain of the quotient function by excluding the real zeros of the <em>divisor<\/em> function.\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nLet [latex]f(x)=2x+1[\/latex] and [latex]g(x)=3x-2[\/latex]. Find\u00a0[latex]\\dfrac{f}{g}(x)[\/latex] and [latex]\\dfrac{g}{f}(x)[\/latex], identify their domains, and simplify both using polynomial division.\r\n\r\n[reveal-answer q=\"284757\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"284757\"]\r\n\r\nLet's start with [latex]\\dfrac{f}{g}(x)=\\dfrac{f(x)}{g(x)}=\\dfrac{{\\color{Green}{2x+1}}}{{\\color{blue}{3x-2}}}[\/latex]. Since [latex]3x-2=0[\/latex] only if [latex]x=\\dfrac{2}{3}[\/latex], the domain of\u00a0[latex]\\dfrac{f}{g}(x)[\/latex] consists of all real numbers [latex]x[\/latex] except [latex]\\dfrac{2}{3}[\/latex].\u00a0Set up the long division diagram, divide the leading term of the dividend, [latex]{\\color{Green}{\\enclose{box}{2x}}}[\/latex], by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{3x}}}[\/latex], and place the simplified result, [latex]{\\color{Orange}{\\enclose{box}{\\dfrac{2}{3}}}}[\/latex], in the proper column of the quotient.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{14pt}\\textsf{deg.}\\hspace{12pt}1\\hspace{15.5pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{3x}-2-\\enclose{box}{2x}}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}\\ {\\color{Orange}{\\enclose{box}{\\dfrac{2}{3}}}}\\\\[-10pt]\r\n{\\color{blue}{\\enclose{box}{3x}-2}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{\\enclose{box}{2x}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+1}}}\r\n\\end{array}\r\n[\/latex]<\/p>\r\nMultiply [latex]{\\color{Orange}{\\dfrac{2}{3}}}[\/latex] by [latex]{\\color{blue}{3x-2}}[\/latex], simplify the result to [latex]2x-\\dfrac{4}{3}[\/latex], write the opposite of it, [latex]{\\color{red}{-2x+\\dfrac{4}{3}}}[\/latex], under the (current) dividend, and add.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{11pt}\\textsf{deg.}\\hspace{12pt}1\\hspace{11.5pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{3x-2-2x}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}\\hspace{5.5pt} {\\color{Orange}{\\dfrac{2}{3}}}\\\\[-10pt]\r\n{\\color{blue}{3x-2}} \\enclose{longdiv}{\\phantom{-}2x {\\color{Gray}{\\Bigl|}} +1}\\\\[-10pt]\r\n\\phantom{3x-2\\hspace{2.5pt}}\\enclose{bottom}{{\\color{red}{-2x}}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}{\\color{red}{{}+ \\dfrac{4}{3}}}}\\\\[-10pt]\r\n\\phantom{3x-2-2x\\hspace{5.5pt} {\\color{Gray}{\\Biggl|}}}\\ \\dfrac{7}{3}\r\n\\end{array}\r\n[\/latex]<\/p>\r\nSince the degree of [latex]\\dfrac{7}{3}[\/latex] as a polynomial, [latex]0[\/latex], is less than [latex]1[\/latex], the degree of the divisor, we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. The remainder is [latex]{\\color{Plum}{\\dfrac{7}{3}}}[\/latex] and [latex]{\\color{Orange}{\\dfrac{2}{3}}}[\/latex] is the complete quotient. We can write the division statement in the form\r\n<p style=\"text-align: center;\">[latex]\r\n\\displaystyle\\frac{{\\color{Green}{2x+1}}}{{\\color{blue}{3x-2}}} = {\\color{Orange}{\\frac{2}{3}}} + \\frac{{\\color{Plum}{\\dfrac{7}{3}}}}{\\;{\\color{blue}{3x-2}}\\;}\r\n[\/latex]<\/p>\r\nand verify the correctness of this statement by simplifying [latex]({\\color{blue}{3x-2}})\\cdot {\\color{Orange}{\\dfrac{2}{3}}} + {\\color{Plum}{\\dfrac{7}{3}}}[\/latex] to verify if it matches [latex]{\\color{Green}{2x+1}}[\/latex]. We have\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n(3x-2)\\cdot \\frac{2}{3} + \\frac{7}{3} &amp;= 2x-\\dfrac{4}{3} + \\dfrac{7}{3}\\\\[5pt]\r\n&amp;= 2x+\\dfrac{3}{3}\\\\[5pt]\r\n&amp;= 2x+1\\ \\large\\checkmark\r\n\\end{align}\r\n[\/latex]<\/p>\r\n\r\n\r\n<hr style=\"width: 80%;\" \/>\r\n\r\nFor [latex]\\dfrac{g}{f}(x)=\\dfrac{g(x)}{f(x)}=\\dfrac{{\\color{Green}{3x-2}}}{{\\color{blue}{2x+1}}}[\/latex], since [latex]2x+1=0[\/latex] only if [latex]x=-\\dfrac{1}{2}[\/latex], the domain of\u00a0[latex]\\dfrac{g}{f}(x)[\/latex] consists of all real numbers [latex]x[\/latex] except [latex]-\\dfrac{1}{2}[\/latex].\u00a0Setup the long division diagram, divide the leading term of the dividend, [latex]{\\color{Green}{\\enclose{box}{3x}}}[\/latex], by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{2x}}}[\/latex], and place the simplified result, [latex]{\\color{Orange}{\\enclose{box}{\\dfrac{3}{2}}}}[\/latex], in the proper column of the quotient.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{14pt}\\textsf{deg.}\\hspace{12pt}1\\hspace{15.5pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{\\enclose{box}{2x}+1-\\enclose{box}{3x}}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}\\ {\\color{Orange}{\\enclose{box}{\\dfrac{3}{2}}}}\\\\[-10pt]\r\n{\\color{blue}{\\enclose{box}{2x}+1}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{\\enclose{box}{3x}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-2}}}\r\n\\end{array}\r\n[\/latex]<\/p>\r\nMultiply [latex]{\\color{Orange}{\\dfrac{3}{2}}}[\/latex] by [latex]{\\color{blue}{2x+1}}[\/latex], simplify the result to [latex]3x+\\dfrac{3}{2}[\/latex], write the opposite of it, [latex]{\\color{red}{-3x-\\dfrac{3}{2}}}[\/latex], under the (current) dividend, and add.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{array}{l}\r\n\\color{Gray}{\\hspace{11pt}\\textsf{deg.}\\hspace{12pt}1\\hspace{11.5pt}\\textsf{c.t.}}\\\\[-10pt]\r\n\\phantom{2x+1-3x}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}\\hspace{5.5pt} {\\color{Orange}{\\dfrac{3}{2}}}\\\\[-10pt]\r\n{\\color{blue}{2x+1}} \\enclose{longdiv}{\\phantom{-}3x {\\color{Gray}{\\Bigl|}} -2}\\\\[-10pt]\r\n\\phantom{2x+1\\hspace{2.5pt}}\\enclose{bottom}{{\\color{red}{-3x}}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}{\\color{red}{{}- \\dfrac{3}{2}}}}\\\\[-10pt]\r\n\\phantom{2x+1-3x\\hspace{1pt} {\\color{Gray}{\\Biggl|}}}- \\dfrac{7}{2}\r\n\\end{array}\r\n[\/latex]<\/p>\r\nSince the degree of [latex]-\\dfrac{7}{2}[\/latex] as a polynomial, [latex]0[\/latex], is less than [latex]1[\/latex], the degree of the divisor, we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. The remainder is [latex]{\\color{Plum}{-\\dfrac{7}{2}}}[\/latex] and [latex]{\\color{Orange}{\\dfrac{3}{2}}}[\/latex] is the complete quotient. We can write the division statement in the form\r\n<p style=\"text-align: center;\">[latex]\r\n\\displaystyle\\frac{{\\color{Green}{3x-2}}}{{\\color{blue}{2x+1}}} = {\\color{Orange}{\\frac{3}{2}}} + \\frac{{\\color{Plum}{-\\dfrac{7}{2}}}}{\\;{\\color{blue}{2x+1}}\\;}\r\n[\/latex]<\/p>\r\nor, in equivalent form, as\r\n<p style=\"text-align: center;\">[latex]\r\n\\displaystyle\\frac{{\\color{Green}{3x-2}}}{{\\color{blue}{2x+1}}} = {\\color{Orange}{\\frac{3}{2}}} {\\color{Plum}{{}-{}}} \\frac{{\\color{Plum}{\\dfrac{7}{2}}}}{\\;{\\color{blue}{2x+1}}\\;}\r\n[\/latex]<\/p>\r\nand verify the correctness of this statement by simplifying [latex]({\\color{blue}{2x+1}})\\cdot {\\color{Orange}{\\dfrac{3}{2}}} + \\left({\\color{Plum}{-\\dfrac{7}{2}}}\\right)[\/latex] to verify if it matches [latex]{\\color{Green}{3x-2}}[\/latex]. We have\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n(2x+1)\\cdot \\dfrac{3}{2} + \\left(-\\dfrac{7}{2}\\right) &amp;= 3x+\\dfrac{3}{2} - \\dfrac{7}{2}\\\\[5pt]\r\n&amp;= 3x-\\dfrac{4}{2}\\\\[5pt]\r\n&amp;= 3x-2\\ \\large\\checkmark\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Summary<\/h2>\r\nDividing polynomials by polynomials of more than one term can be done using a process very much like long division of whole numbers. You must be careful to subtract entire expressions, not just the first term. Stop when the degree of the remainder is less than the degree of the divisor. The division result can be written as a fraction added to the quotient with the remainder in the numerator and the divisor in the denominator.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning OutcomeS<\/h3>\n<ul>\n<li>Divide a polynomial by a monomial.<\/li>\n<li>Divide a polynomial by a polynomial.<\/li>\n<li>Divide polynomial functions given in function notation.<\/li>\n<\/ul>\n<\/div>\n<p>Let&#8217;s review the terminology related to the division of whole numbers. If we divide [latex]\\require{color}7[\/latex] by [latex]3[\/latex], [latex]7[\/latex] is called the <strong>dividend<\/strong>, and [latex]3[\/latex] is called the <strong>divisor<\/strong>. The division result can be written as a mixed number [latex]2 \\dfrac{1}{3}[\/latex] that means [latex]2+\\dfrac{1}{3}[\/latex]. The whole part of the division result, [latex]2[\/latex], is called the <strong>quotient<\/strong>, and the numerator of the \u201cfractional part\u201d, [latex]1[\/latex], is called the <strong>remainder<\/strong>. The entire division statement can be written mathematically in two ways, using a fraction for the division, or using the division symbol.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{7}{3}=2+\\dfrac{1}{3}[\/latex]\u00a0 \u00a0or\u00a0 \u00a0[latex]7\\div3=2+\\dfrac{1}{3}[\/latex]<\/p>\n<p>In general,<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{{\\color{Green}{\\textsf{dividend}}}}{{\\color{blue}{\\textsf{divisor}}}}={\\color{Orange}{\\textsf{quotient}}}+\\dfrac{{\\color{Plum}{\\textsf{remainder}}}}{{\\color{blue}{\\textsf{divisor}}}}[\/latex]\u00a0 \u00a0or\u00a0 \u00a0[latex]{\\color{Green}{\\textsf{dividend}}}\\div {\\color{blue}{\\textsf{divisor}}}={\\color{Orange}{\\textsf{quotient}}}+\\dfrac{{\\color{Plum}{\\textsf{remainder}}}}{{\\color{blue}{\\textsf{divisor}}}}[\/latex]<\/p>\n<p>We say that the division is performed on <strong>improper fractions<\/strong> which means that the divisor is not greater than the dividend. The quotient is never greater than the dividend, and the remainder is nonnegative and always less than the divisor. When the remainder is zero, we say that the dividend is <strong>divisible<\/strong> by the divisor. In such case, the quotient itself is the result of the division. A way of verifying correctness of a division statement is based on the fact that the division statement can be presented in equivalent form as<\/p>\n<p style=\"text-align: center;\">[latex]{\\color{Green}{\\textsf{dividend}}} = {\\color{blue}{\\textsf{divisor}}}\\cdot{\\color{Orange}{\\textsf{quotient}}}+{\\color{Plum}{\\textsf{remainder}}}[\/latex]<\/p>\n<p>Using our example of the division of [latex]{\\color{Green}{7}}[\/latex] by [latex]{\\color{blue}{3}}[\/latex], the statement [latex]{\\color{Green}{7}}\\div{\\color{blue}{3}}={\\color{Orange}{2}}+\\dfrac{{\\color{Plum}{1}}}{{\\color{blue}{3}}}[\/latex] is true only if the statement [latex]{\\color{Green}{7}}={\\color{blue}{3}}\\cdot {\\color{Orange}{2}}+{\\color{Plum}{1}}[\/latex] is true. In such verification, we usually start with the RHS (right hand side) of the latter statement and simplify to check if it matches the LHS (left hand side). Since\u00a0[latex]3\\cdot 2+1[\/latex] is indeed equal to [latex]7[\/latex], both statements are true.\u00a0Notice that the quotient ([latex]\\color{Orange}{2}[\/latex]) is not greater than the dividend ([latex]\\color{Green}{7}[\/latex]). Also, the remainder ([latex]\\color{Plum}{1}[\/latex]) is not negative.<\/p>\n<h2>Dividing a Polynomial by a Monomial<\/h2>\n<p>Dividing a polynomial by a monomial can be handled by dividing each term in the dividend polynomial separately by the divisor monomial. This is based on the fact that adding or subtracting rational expressions with like denominators is equivalent to operating on the numerators keeping the common denominator the same: [latex]\\displaystyle \\frac{a}{c}\\pm\\frac{b}{c}=\\frac{a\\pm b}{c}[\/latex]. We just use that equivalency\u00a0\u201cbackwards.\u201d We will refer to this fact as \u201cusing linearity of operations on rational expressions.\u201d<\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Divide [latex]12x^3+2x^2-8x+1[\/latex] by [latex]4x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q38801\">Show Solution<\/span><\/p>\n<div id=\"q38801\" class=\"hidden-answer\" style=\"display: none\">\n<p>We first rewrite the division in fractional form and use linearity of operations on rational expressions. We then simplify each rational expression if possible.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{12x^3+2x^2-8x+1}{4x} &= \\frac{12x^3}{4x} + \\frac{2x^2}{4x} - \\frac{8x}{4x} + \\frac{1}{4x}\\\\[5pt]  &= \\frac{3\\cdot {\\color{red}\\cancel{\\color{black}{4\\cdot x}}}\\cdot x^2}{{\\color{red}\\cancel{\\color{black}{4\\cdot x}}} \\cdot 1} + \\frac{{\\color{red}\\cancel{\\color{black}{2\\cdot x}}} \\cdot x}{2\\cdot {\\color{red}\\cancel{\\color{black}{2\\cdot x}}}} - \\frac{2\\cdot {\\color{red}\\cancel{\\color{black}{4\\cdot x}}}}{{\\color{red}\\cancel{\\color{black}{4\\cdot x}}}\\cdot 1} + \\frac{1}{4x}\\\\[5pt]  &= 3x^2 + \\frac{x}{2} - 2 + \\frac{1}{4x}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Of course, when we divide by a variable expression, we have to determine restricted values of the variable that result in division by zero. In the example above, [latex]x\\ne 0[\/latex].<\/p>\n<p>We usually refrain from using the notions of quotient and remainder when dividing by a monomial.<\/p>\n<h2>Divide a Polynomial by a Polynomial<\/h2>\n<p>The easiest case of dividing a polynomial by a general polynomial is when the dividend contains the divisor as a factor. For example,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{x^2-4x-12}{x+2} &= \\frac{{\\color{red}\\cancel{\\color{black}{(x+2)}}}\\cdot (x-6)}{{\\color{red}\\cancel{\\color{black}{(x+2)}}}\\cdot 1}\\\\[5pt]  &= x-6  \\end{align}[\/latex]<\/p>\n<p>That is of course dependent on the ability to factor that might prove useful if you encounter a division example in which the divisor is a factor of the dividend. As always, remember about the restricted values of the variable (the real zeros of the divisor). In the example above, [latex]x\\ne -2[\/latex].<\/p>\n<p>It is also possible that\u00a0the dividend and divisor contain\u00a0common factors that can be divided out. For example,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{x^2-x-2}{x^2+3x+2} &= \\frac{{\\color{red}\\cancel{\\color{black}{(x+1)}}}\\cdot (x-2)}{{\\color{red}\\cancel{\\color{black}{(x+1)}}}\\cdot (x+2)}\\\\[5pt]  &= \\frac{x-2}{x+2}  \\end{align}[\/latex]<\/p>\n<p>The final rational expression is still improper and requires further division.<\/p>\n<p>In general, when dividing a polynomial by a polynomial, we will use long division, a process similar to long division of whole numbers, or synthetic division (discussed in the next section). The notions of the dividend, divisor, quotient, and reminder extend to division of polynomials with some adjustments:<\/p>\n<ul>\n<li>In an <strong>improper<\/strong> rational expression, the <strong>degree<\/strong> of the divisor is not greater than the <strong>degree<\/strong> of the dividend.<\/li>\n<li>The <strong>degree<\/strong> of the quotient is never greater than the <strong>degree<\/strong> of the dividend. Moreover, the degree of the quotient is the difference of the degrees of the dividend and the divisor.<\/li>\n<li>The <strong>degree<\/strong> of the remainder is always less than the <strong>degree<\/strong> of the divisor or the remainder is [latex]0[\/latex].<\/li>\n<\/ul>\n<p>Let&#8217;s discuss the setup of the long division diagram. The basic setup is similar to long division of whole numbers:<\/p>\n<p style=\"text-align: center;\">[latex]\\require{enclose}  {\\color{blue}{\\textsf{divisor}}} \\enclose{longdiv}{{\\color{Green}{\\textsf{dividend}}}}[\/latex]<\/p>\n<p>The place above the line over the dividend is designated for the <strong><span style=\"color: #ff6600;\">quotient<\/span><\/strong>. The dividend and divisor should be written in descending powers of the variable. To avoid mistakes, the dividend should have terms of all degrees with a coefficient of zero for all missing degrees of terms. There is no need to include terms of missing degrees in the divisor. We suggest that you include vertical lines separating columns of terms of specific degree. Since the constant term might be any real number and only nonzero real numbers perceived as a polynomial have the degree of zero, we will label the last column as\u00a0\u201cc.t.\u201d For example, for the division of [latex]{\\color{Green}{x^4-2x^2+3x-4}}[\/latex] by [latex]{\\color{blue}{x^2+1}}[\/latex], the long division diagram should look like this:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{10pt}4\\hspace{22pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{x^2+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{{} - 2x^2} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{-4}\\\\[-10pt]  {\\color{blue}{x^2+1}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{x^4}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+0x^3}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-2x^2}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+3x}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-4}}}\\\\[-10pt]  \\phantom{x^2+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\vdots}} \\phantom{{} + 0x^3} {\\color{Gray}{\\vdots}} \\phantom{{} - 2x^2} {\\color{Gray}{\\vdots}} \\phantom{{} + 3x} {\\color{Gray}{\\vdots}} \\phantom{-4}\\\\[-10pt]  \\end{array}[\/latex]<\/p>\n<p>Let&#8217;s now describe a <strong>long division step<\/strong>. In long division of whole numbers, the question we&#8217;d ask is \u201chow many times does the divisor go into the number made of the first sufficient number of digits of the dividend?\u201d In long division of polynomials, we simply divide the first (leading) term of the dividend (actually, what we will refer to as the <em>current dividend<\/em>, which we will explain in a bit) by the first (leading) term of the divisor and place the simplified result of that division in the proper column of the quotient space in the diagram. In our example, we divide [latex]\\dfrac{\\;{\\color{Green}\\enclose{box}{x^4}}\\;}{{\\color{blue}{\\enclose{box}{x^2}}}}[\/latex]. We write the simplified result, [latex]{\\color{Orange}{\\enclose{box}{x^2}}}[\/latex], in the [latex]2[\/latex]<sup>nd<\/sup> degree column of the quotient:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{14pt}\\textsf{deg.}\\hspace{12pt}4\\hspace{24pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{\\enclose{box}{x^2}+1-\\enclose{box}{x^4}}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{{}+0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{-2}{\\color{Orange}{\\enclose{box}{x^2}}} {\\color{Gray}{\\Bigl|}} \\phantom{{}+3x} {\\color{Gray}{\\Bigl|}} \\phantom{{}-4}\\\\[-10pt]  {\\color{blue}{\\enclose{box}{x^2}+1}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{\\enclose{box}{x^4}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+0x^3}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-2x^2}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+3x}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-4}}}\\\\[-10pt]  \\end{array}[\/latex]<\/p>\n<p>Next we multiply the newly obtained term of the quotient by the (entire) divisor and subtract the simplified result from the (current) dividend. Since addition is easier than subtraction and to subtract means to add the opposite, we write the opposite of that simplified result under the (current) dividend placing the terms of specific degree in the proper column, and add. In our example, we multiply [latex]{\\color{Orange}{x^2}}\\cdot\\left({\\color{blue}{x^2+1}}\\right)[\/latex], simplify the result to [latex]x^4+x^2[\/latex], write the opposite of that simplified result, [latex]{\\color{red}{-x^4-x^2}}[\/latex], under the (current) dividend with proper placement of terms, and add:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{10pt}4\\hspace{22pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{x^2+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{{} - 2}{\\color{Orange}{x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{{ }- 4}\\\\[-10pt]  {\\color{blue}{x^2+1}} \\enclose{longdiv}{\\phantom{-} x^4 {\\color{Gray}{\\Bigl|}} +0x^3 {\\color{Gray}{\\Bigl|}} -2x^2 {\\color{Gray}{\\Bigl|}} +3x {\\color{Gray}{\\Bigl|}} -4}\\\\[-10pt]  \\phantom{x^2+1\\hspace{3pt}} \\enclose{bottom}{{\\color{red}{- x^4}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}}\\hspace{5pt} {\\color{red}{{} - x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{{} -4}}\\\\[-10pt]  \\phantom{x^2+1\u00a0-x^4 {\\color{Gray}{\\Bigl|}} + 0x^3 {\\color{Gray}{\\Bigl|}}}\\hspace{-9pt} -3x^2 {\\color{Gray}{\\Bigl|}} + 3x {\\color{Gray}{\\Bigl|}} -4  \\end{array}[\/latex]<\/p>\n<p>We ask now if the <strong>degree<\/strong> of the polynomial resulting from the addition is <strong>less than<\/strong> the <strong>degree<\/strong> of the divisor or if that resulting polynomial is [latex]0[\/latex]. If <strong>yes<\/strong>, we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. That resulting polynomial is the <strong>remainder<\/strong>. If <strong>no<\/strong>, the resulting polynomial becomes the new (<strong>current<\/strong>) <strong>dividend<\/strong> and we <span style=\"color: #ff0000;\"><strong>continue<\/strong><\/span> with another long division step. In our example, the degree of [latex]-3x^2+3x-4[\/latex] is [latex]2[\/latex]. Since [latex]2[\/latex] is not less than [latex]2[\/latex], the degree of the divisor, [latex]{\\color{blue}{x^2+1}}[\/latex],\u00a0[latex]{\\color{Green}{-3x^2+3x-4}}[\/latex] becomes the current dividend and we continue with another long division step. We divide the leading term of the current dividend, [latex]{\\color{Green}{\\enclose{box}{-3x^2}}}[\/latex] by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{x^2}}}[\/latex], and write the simplified result, [latex]{\\color{Orange}{\\enclose{box}{-3}}}[\/latex] in the [latex]\\textsf{c.t.}[\/latex]\u00a0column of the quotient:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{14pt}\\textsf{deg.}\\hspace{10pt}4\\hspace{22pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{\\enclose{box}{x^2}+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{{} - 2}{\\color{Orange}{x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} {\\color{Orange}{\\enclose{box}{-3}}}\\\\[-10pt]  {\\color{blue}{\\enclose{box}{x^2}+1}} \\enclose{longdiv}{\\phantom{-} x^4 {\\color{Gray}{\\Bigl|}} +0x^3 {\\color{Gray}{\\Bigl|}} -2x^2 {\\color{Gray}{\\Bigl|}} +3x {\\color{Gray}{\\Bigl|}} -4}\\\\[-10pt]  \\phantom{\\enclose{box}{x^2}+1\\hspace{3pt}} \\enclose{bottom}{- x^4 {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}}\\hspace{5pt} - x^2 {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{-4}}\\\\[-10pt]  \\phantom{\\enclose{box}{x^2}+1\\hspace{1pt} -x^4 \\color{Gray}{\\Bigl|} + 0x^3 {\\color{Gray}{\\Bigl|}}} {\\color{Green}{\\enclose{box}{-3x^2}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+ 3x}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-4}}\\\\[-10pt]  \\end{array}[\/latex]<\/p>\n<p>We follow by\u00a0multiplying [latex]{\\color{Orange}{-3}}\\cdot\\left({\\color{blue}{x^2+1}}\\right)[\/latex], simplify the result to [latex]-3x^2-3[\/latex], write the opposite of that simplified result, [latex]{\\color{red}{+3x^2+3}}[\/latex], under the current dividend with proper placement of terms, and add:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{10pt}4\\hspace{22pt}3\\hspace{25pt}2\\hspace{22pt}1\\hspace{14pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{x^2+1\\ \\ } \\phantom{-x^4} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}} \\phantom{{} - 2}{\\color{Orange}{x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} {\\color{Orange}{-3}}\\\\[-10pt]  {\\color{blue}{x^2+1}} \\enclose{longdiv}{\\phantom{-} x^4 {\\color{Gray}{\\Bigl|}} +0x^3 {\\color{Gray}{\\Bigl|}} -2x^2 {\\color{Gray}{\\Bigl|}} +3x {\\color{Gray}{\\Bigl|}} -4}\\\\[-10pt]  \\phantom{x^2+1\\hspace{3pt}} \\enclose{bottom}{- x^4 {\\color{Gray}{\\Bigl|}} \\phantom{{} + 0x^3} {\\color{Gray}{\\Bigl|}}\\hspace{5pt} - x^2 {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} \\phantom{{} -4}}\\\\[-10pt]  \\phantom{x^2+1\u00a0-x^4 {\\color{Gray}{\\Bigl|}} + 0x^3 {\\color{Gray}{\\Bigl|}}}\\hspace{-9pt} -3x^2 {\\color{Gray}{\\Bigl|}} + 3x {\\color{Gray}{\\Bigl|}} -4\\\\[-10pt]  \\phantom{x^2+1 -x^4 {\\color{Gray}{\\Bigl|}} + 0x^3 {\\color{Gray}{\\Bigl|}}}\\hspace{-1pt} \\enclose{bottom}{{\\color{red}{{} + 3x^2}} {\\color{Gray}{\\Bigl|}} \\phantom{{} + 3x} {\\color{Gray}{\\Bigl|}} {\\color{red}{{} + 3}}}\\\\[-10pt]  \\phantom{x^2+1 -x^4 {\\color{Gray}{\\Bigl|}} + 0x^3 {\\color{Gray}{\\Bigl|}}\u00a0- 3x^2 {\\color{Gray}{\\Bigl|}}} \\hspace{3.5pt} 3x {\\color{Gray}{\\Bigl|}} -1  \\end{array}[\/latex]<\/p>\n<p>Since the degree of [latex]3x-1[\/latex], [latex]1[\/latex], is less than [latex]2[\/latex], the degree of the divisor, we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. [latex]{\\color{Plum}{3x-1}}[\/latex] is the remainder and [latex]{\\color{Orange}{x^2-3}}[\/latex] is the complete quotient. We can write the division statement in the form<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{{\\color{Green}{x^4-2x^2+3x-4}}}{{\\color{blue}{x^2+1}}} = {\\color{Orange}{x^2-3}} + \\frac{{\\color{Plum}{3x-1}}}{{\\color{blue}{x^2+1}}}[\/latex]<\/p>\n<p>and verify the correctness of this statement by simplifying [latex]\\left({\\color{blue}{x^2+1}}\\right)\\cdot \\left({\\color{Orange}{x^2-3}}\\right)+{\\color{Plum}{3x-1}}[\/latex] to verify if it matches [latex]{\\color{Green}{x^4-2x^2+3x-4}}[\/latex]. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\left(x^2+1\\right)\\cdot \\left(x^2-3\\right)+3x-1 &= x^4-3x^2+x^2-3+3x-1\\\\  &= x^4-2x^2+3x-4\\ \\large\\checkmark  \\end{align}[\/latex]<\/p>\n<p>Note, that since [latex]x^2+1\\ne 0[\/latex] for all real numbers [latex]x[\/latex], there are no restricted values of the variable in the example above.<\/p>\n<p>In the video that follows, we show another example of dividing a degree three trinomial by a binomial.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Divide a Degree 3 Polynomial by a Degree 1 Polynomial (Long Division with Missing Term)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Rxds7Q_UTeo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Let&#8217;s attempt long division on an earlier example where the divisor was a factor of the dividend.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use long division to divide [latex]x^2-4x-12[\/latex] by [latex]x+2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455187\">Show Solution<\/span><\/p>\n<div id=\"q455187\" class=\"hidden-answer\" style=\"display: none\">\n<p>Setup the long division diagram, divide the leading term of the dividend, [latex]{\\color{Green}{\\enclose{box}{x^2}}}[\/latex], by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{x}}}[\/latex], and place the simplified result, [latex]{\\color{Orange}{\\enclose{box}{x}}}[\/latex], in the proper column of the quotient.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{12pt}2\\hspace{22pt}1\\hspace{17pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{\\enclose{box}{x}+2-\\enclose{box}{x^2}}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{-4}{\\color{Orange}{\\enclose{box}{x}}} {\\color{Gray}{\\Bigl|}} \\phantom{-12}\\\\[-10pt]  {\\color{blue}{\\enclose{box}{x}+2}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{\\enclose{box}{x^2}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-4x}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-12}}}  \\end{array}[\/latex]<\/p>\n<p>Multiply [latex]{\\color{Orange}{x}}[\/latex] by [latex]{\\color{blue}{x+2}}[\/latex], simplify the result to [latex]x^2+2x[\/latex], write the opposite of it, [latex]{\\color{red}{-x^2-2x}}[\/latex], under the (current) dividend, and add.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{7pt}\\textsf{deg.}\\hspace{10pt}2\\hspace{20pt}1\\hspace{17pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{x+2 -x^2}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{{}-4}{\\color{Orange}{x}} {\\color{Gray}{\\Bigl|}} \\phantom{-12}\\\\[-10pt]  {\\color{blue}{x+2}} \\enclose{longdiv}{\\phantom{-}x^2 {\\color{Gray}{\\Bigl|}} -4x {\\color{Gray}{\\Bigl|}} -12}\\\\[-10pt]  \\phantom{x+2\\hspace{3pt}} \\enclose{bottom}{{\\color{red}{- x^2}} {\\color{Gray}{\\Bigl|}} {\\color{red}{{} - 2x}} {\\color{Gray}{\\Bigl|}} \\phantom{{} -12}}\\\\[-10pt]  \\phantom{x+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-9pt}} -6x {\\color{Gray}{\\Bigl|}} -12  \\end{array}[\/latex]<\/p>\n<p>Since the degree of [latex]-6x-12[\/latex], [latex]1[\/latex], is not less than [latex]1[\/latex], the degree of the divisor, [latex]{\\color{blue}{x+2}}[\/latex],\u00a0[latex]{\\color{Green}{-6x-12}}[\/latex] becomes the current dividend and we continue with another long division step. We divide the leading term of the current dividend, [latex]{\\color{Green}{\\enclose{box}{-6x}}}[\/latex], by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{x}}}[\/latex], and write the simplified result, [latex]{\\color{Orange}{\\enclose{box}{-6}}}[\/latex], in the proper\u00a0column of the quotient.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{10pt}\\textsf{deg.}\\hspace{10pt}2\\hspace{20pt}1\\hspace{17pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{\\enclose{box}{x}+2 -x^2}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{{}-4}{\\color{Orange}{x}} {\\color{Gray}{\\Bigl|}} {\\color{Orange}{\\enclose{box}{-6}}}\\\\[-10pt]  {\\color{blue}{\\enclose{box}{x}+2}} \\enclose{longdiv}{\\phantom{-}x^2 {\\color{Gray}{\\Bigl|}} -4x {\\color{Gray}{\\Bigl|}} -12}\\\\[-10pt]  \\phantom{\\enclose{box}{x}+2\\hspace{3pt}} \\enclose{bottom}{- x^2 {\\color{Gray}{\\Bigl|}} - 2x {\\color{Gray}{\\Bigl|}} \\phantom{{} -12}}\\\\[-10pt]  \\phantom{\\enclose{box}{x}+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-8.5pt}} {\\color{Green}{\\enclose{box}{-6x}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-12}}  \\end{array}[\/latex]<\/p>\n<p>We follow by\u00a0multiplying [latex]{\\color{Orange}{-6}}\\cdot({\\color{blue}{x+2}})[\/latex], simplify the result to [latex]-6x-12[\/latex], write the opposite of it, [latex]{\\color{red}{+6x+12}}[\/latex], under the current dividend with proper placement of terms, and add.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{7pt}\\textsf{deg.}\\hspace{10pt}2\\hspace{20pt}1\\hspace{17pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{x+2 -x^2}\\hspace{.5pt} {\\color{Gray}{\\Bigl|}} \\phantom{{}-4}{\\color{Orange}{x}} {\\color{Gray}{\\Bigl|}} {\\color{Orange}{-6}}\\\\[-10pt]  {\\color{blue}{x+2}} \\enclose{longdiv}{\\phantom{-}x^2 {\\color{Gray}{\\Bigl|}} -4x {\\color{Gray}{\\Bigl|}} -12}\\\\[-10pt]  \\phantom{x+2\\hspace{3pt}} \\enclose{bottom}{- x^2 {\\color{Gray}{\\Bigl|}} - 2x {\\color{Gray}{\\Bigl|}} \\phantom{{} -12}}\\\\[-10pt]  \\phantom{x+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-9pt}} -6x {\\color{Gray}{\\Bigl|}} -12\\\\[-10pt]  \\phantom{x+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-10.5pt}} \\enclose{bottom}{{\\color{red}{{}+6x}} {\\color{Gray}{\\Bigl|}} {\\color{red}{{}+12}}}\\\\[-10pt]  \\phantom{x+2\u00a0-x^2 {\\color{Gray}{\\Bigl|}} \\hspace{-8.5pt} {\\color{red}{{}+6x}} {\\color{Gray}{\\Bigl|}} {\\color{red}{{}+{}}}}0  \\end{array}[\/latex]<\/p>\n<p>Since the resulting polynomial is [latex]0[\/latex], we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. The remainder is [latex]{\\color{Plum}{0}}[\/latex] and [latex]{\\color{Orange}{x-6}}[\/latex] is the complete quotient. We can write the division statement in the form<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{{\\color{Green}{x^2-4x-12}}}{{\\color{blue}{x+2}}} = {\\color{Orange}{x-6}} + \\frac{{\\color{Plum}{0}}}{{\\color{blue}{x+2}}}[\/latex]<\/p>\n<p>or, in simplified form, as<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{{\\color{Green}{x^2-4x-12}}}{{\\color{blue}{x+2}}} = {\\color{Orange}{x-6}}[\/latex]<\/p>\n<p>and verify the correctness of this statement by simplifying [latex]({\\color{blue}{x+2}})\\cdot ({\\color{Orange}{x-6}}) + {\\color{Plum}{0}}[\/latex] to verify if it matches [latex]{\\color{Green}{x^2-4x-12}}[\/latex]. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  (x+2)\\cdot (x-6) + 0 &= x^2-6x+2x-12\\\\  &= x^2-4x-12\\ \\large\\checkmark  \\end{align}[\/latex]<\/p>\n<p>We also note that [latex]x\\ne -2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show a couple more examples of dividing a degree two trinomial by a degree one binomial.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 1:  Divide a Trinomial by a Binomial Using Long Division\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/KUPFg__Djzw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next video, we present one more example of polynomial long division.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 6:  Divide a Polynomial by a Degree Two Binomial Using Long Division\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/P6OTbUf8f60?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Polynomial Division and Function Notation<\/h2>\n<p>In <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/3-2-the-algebra-of-functions\/\" target=\"_blank\" rel=\"noopener\">Sec. 3.2<\/a>, we discussed quotient functions of the form [latex]\\dfrac{f}{g}[\/latex]. If both [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are polynomial functions, the quotient functions\u00a0[latex]\\dfrac{f}{g}(x)=\\dfrac{f(x)}{g(x)}[\/latex] and\/or [latex]\\dfrac{g}{f}(x)=\\dfrac{g(x)}{f(x)}[\/latex] can possibly be simplified using polynomial division. Recall that we only perform division on improper rational expressions. Hence, if the degrees of both [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are the same, both quotient functions can be simplified using polynomial division. Always make sure to identify the domain of the quotient function by excluding the real zeros of the <em>divisor<\/em> function.<\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Let [latex]f(x)=2x+1[\/latex] and [latex]g(x)=3x-2[\/latex]. Find\u00a0[latex]\\dfrac{f}{g}(x)[\/latex] and [latex]\\dfrac{g}{f}(x)[\/latex], identify their domains, and simplify both using polynomial division.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q284757\">Show Solution<\/span><\/p>\n<div id=\"q284757\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let&#8217;s start with [latex]\\dfrac{f}{g}(x)=\\dfrac{f(x)}{g(x)}=\\dfrac{{\\color{Green}{2x+1}}}{{\\color{blue}{3x-2}}}[\/latex]. Since [latex]3x-2=0[\/latex] only if [latex]x=\\dfrac{2}{3}[\/latex], the domain of\u00a0[latex]\\dfrac{f}{g}(x)[\/latex] consists of all real numbers [latex]x[\/latex] except [latex]\\dfrac{2}{3}[\/latex].\u00a0Set up the long division diagram, divide the leading term of the dividend, [latex]{\\color{Green}{\\enclose{box}{2x}}}[\/latex], by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{3x}}}[\/latex], and place the simplified result, [latex]{\\color{Orange}{\\enclose{box}{\\dfrac{2}{3}}}}[\/latex], in the proper column of the quotient.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{14pt}\\textsf{deg.}\\hspace{12pt}1\\hspace{15.5pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{\\enclose{box}{3x}-2-\\enclose{box}{2x}}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}\\ {\\color{Orange}{\\enclose{box}{\\dfrac{2}{3}}}}\\\\[-10pt]  {\\color{blue}{\\enclose{box}{3x}-2}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{\\enclose{box}{2x}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}+1}}}  \\end{array}[\/latex]<\/p>\n<p>Multiply [latex]{\\color{Orange}{\\dfrac{2}{3}}}[\/latex] by [latex]{\\color{blue}{3x-2}}[\/latex], simplify the result to [latex]2x-\\dfrac{4}{3}[\/latex], write the opposite of it, [latex]{\\color{red}{-2x+\\dfrac{4}{3}}}[\/latex], under the (current) dividend, and add.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{11pt}\\textsf{deg.}\\hspace{12pt}1\\hspace{11.5pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{3x-2-2x}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}\\hspace{5.5pt} {\\color{Orange}{\\dfrac{2}{3}}}\\\\[-10pt]  {\\color{blue}{3x-2}} \\enclose{longdiv}{\\phantom{-}2x {\\color{Gray}{\\Bigl|}} +1}\\\\[-10pt]  \\phantom{3x-2\\hspace{2.5pt}}\\enclose{bottom}{{\\color{red}{-2x}}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}{\\color{red}{{}+ \\dfrac{4}{3}}}}\\\\[-10pt]  \\phantom{3x-2-2x\\hspace{5.5pt} {\\color{Gray}{\\Biggl|}}}\\ \\dfrac{7}{3}  \\end{array}[\/latex]<\/p>\n<p>Since the degree of [latex]\\dfrac{7}{3}[\/latex] as a polynomial, [latex]0[\/latex], is less than [latex]1[\/latex], the degree of the divisor, we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. The remainder is [latex]{\\color{Plum}{\\dfrac{7}{3}}}[\/latex] and [latex]{\\color{Orange}{\\dfrac{2}{3}}}[\/latex] is the complete quotient. We can write the division statement in the form<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{{\\color{Green}{2x+1}}}{{\\color{blue}{3x-2}}} = {\\color{Orange}{\\frac{2}{3}}} + \\frac{{\\color{Plum}{\\dfrac{7}{3}}}}{\\;{\\color{blue}{3x-2}}\\;}[\/latex]<\/p>\n<p>and verify the correctness of this statement by simplifying [latex]({\\color{blue}{3x-2}})\\cdot {\\color{Orange}{\\dfrac{2}{3}}} + {\\color{Plum}{\\dfrac{7}{3}}}[\/latex] to verify if it matches [latex]{\\color{Green}{2x+1}}[\/latex]. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  (3x-2)\\cdot \\frac{2}{3} + \\frac{7}{3} &= 2x-\\dfrac{4}{3} + \\dfrac{7}{3}\\\\[5pt]  &= 2x+\\dfrac{3}{3}\\\\[5pt]  &= 2x+1\\ \\large\\checkmark  \\end{align}[\/latex]<\/p>\n<hr style=\"width: 80%;\" \/>\n<p>For [latex]\\dfrac{g}{f}(x)=\\dfrac{g(x)}{f(x)}=\\dfrac{{\\color{Green}{3x-2}}}{{\\color{blue}{2x+1}}}[\/latex], since [latex]2x+1=0[\/latex] only if [latex]x=-\\dfrac{1}{2}[\/latex], the domain of\u00a0[latex]\\dfrac{g}{f}(x)[\/latex] consists of all real numbers [latex]x[\/latex] except [latex]-\\dfrac{1}{2}[\/latex].\u00a0Setup the long division diagram, divide the leading term of the dividend, [latex]{\\color{Green}{\\enclose{box}{3x}}}[\/latex], by the leading term of the divisor, [latex]{\\color{blue}{\\enclose{box}{2x}}}[\/latex], and place the simplified result, [latex]{\\color{Orange}{\\enclose{box}{\\dfrac{3}{2}}}}[\/latex], in the proper column of the quotient.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{14pt}\\textsf{deg.}\\hspace{12pt}1\\hspace{15.5pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{\\enclose{box}{2x}+1-\\enclose{box}{3x}}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}\\ {\\color{Orange}{\\enclose{box}{\\dfrac{3}{2}}}}\\\\[-10pt]  {\\color{blue}{\\enclose{box}{2x}+1}} \\enclose{longdiv}{\\phantom{-}{\\color{Green}{\\enclose{box}{3x}}} {\\color{Gray}{\\Bigl|}} {\\color{Green}{{}-2}}}  \\end{array}[\/latex]<\/p>\n<p>Multiply [latex]{\\color{Orange}{\\dfrac{3}{2}}}[\/latex] by [latex]{\\color{blue}{2x+1}}[\/latex], simplify the result to [latex]3x+\\dfrac{3}{2}[\/latex], write the opposite of it, [latex]{\\color{red}{-3x-\\dfrac{3}{2}}}[\/latex], under the (current) dividend, and add.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}  \\color{Gray}{\\hspace{11pt}\\textsf{deg.}\\hspace{12pt}1\\hspace{11.5pt}\\textsf{c.t.}}\\\\[-10pt]  \\phantom{2x+1-3x}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}\\hspace{5.5pt} {\\color{Orange}{\\dfrac{3}{2}}}\\\\[-10pt]  {\\color{blue}{2x+1}} \\enclose{longdiv}{\\phantom{-}3x {\\color{Gray}{\\Bigl|}} -2}\\\\[-10pt]  \\phantom{2x+1\\hspace{2.5pt}}\\enclose{bottom}{{\\color{red}{-3x}}\\hspace{.5pt} {\\color{Gray}{\\Biggl|}}{\\color{red}{{}- \\dfrac{3}{2}}}}\\\\[-10pt]  \\phantom{2x+1-3x\\hspace{1pt} {\\color{Gray}{\\Biggl|}}}- \\dfrac{7}{2}  \\end{array}[\/latex]<\/p>\n<p>Since the degree of [latex]-\\dfrac{7}{2}[\/latex] as a polynomial, [latex]0[\/latex], is less than [latex]1[\/latex], the degree of the divisor, we <span style=\"color: #ff0000;\"><strong>stop<\/strong><\/span>. The remainder is [latex]{\\color{Plum}{-\\dfrac{7}{2}}}[\/latex] and [latex]{\\color{Orange}{\\dfrac{3}{2}}}[\/latex] is the complete quotient. We can write the division statement in the form<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{{\\color{Green}{3x-2}}}{{\\color{blue}{2x+1}}} = {\\color{Orange}{\\frac{3}{2}}} + \\frac{{\\color{Plum}{-\\dfrac{7}{2}}}}{\\;{\\color{blue}{2x+1}}\\;}[\/latex]<\/p>\n<p>or, in equivalent form, as<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{{\\color{Green}{3x-2}}}{{\\color{blue}{2x+1}}} = {\\color{Orange}{\\frac{3}{2}}} {\\color{Plum}{{}-{}}} \\frac{{\\color{Plum}{\\dfrac{7}{2}}}}{\\;{\\color{blue}{2x+1}}\\;}[\/latex]<\/p>\n<p>and verify the correctness of this statement by simplifying [latex]({\\color{blue}{2x+1}})\\cdot {\\color{Orange}{\\dfrac{3}{2}}} + \\left({\\color{Plum}{-\\dfrac{7}{2}}}\\right)[\/latex] to verify if it matches [latex]{\\color{Green}{3x-2}}[\/latex]. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  (2x+1)\\cdot \\dfrac{3}{2} + \\left(-\\dfrac{7}{2}\\right) &= 3x+\\dfrac{3}{2} - \\dfrac{7}{2}\\\\[5pt]  &= 3x-\\dfrac{4}{2}\\\\[5pt]  &= 3x-2\\ \\large\\checkmark  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Summary<\/h2>\n<p>Dividing polynomials by polynomials of more than one term can be done using a process very much like long division of whole numbers. You must be careful to subtract entire expressions, not just the first term. Stop when the degree of the remainder is less than the degree of the divisor. The division result can be written as a fraction added to the quotient with the remainder in the numerator and the divisor in the denominator.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-149\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Divide a Degree 3 Polynomial by a Degree 1 Polynomial (Long Division with Missing Term). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Rxds7Q_UTeo\">https:\/\/youtu.be\/Rxds7Q_UTeo<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 1: Divide a Trinomial by a Binomial Using Long Division. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/KUPFg__Djzw\">https:\/\/youtu.be\/KUPFg__Djzw<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 6: Divide a Polynomial by a Degree Two Binomial Using Long Division. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/P6OTbUf8f60\">https:\/\/youtu.be\/P6OTbUf8f60<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Divide a Trinomial by a Binomial Using Long Division\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/KUPFg__Djzw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Divide a Degree 3 Polynomial by a Degree 1 Polynomial (Long Division with Missing Term)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Rxds7Q_UTeo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 6: Divide a Polynomial by a Degree Two Binomial Using Long Division\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/P6OTbUf8f60\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"fb2fd6ff-a26d-48d9-9ee4-bcbaeea28ee3","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-149","chapter","type-chapter","status-publish","hentry"],"part":171,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/149","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":17,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/149\/revisions"}],"predecessor-version":[{"id":1897,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/149\/revisions\/1897"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/171"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/149\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=149"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=149"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=149"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=149"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}