{"id":16327,"date":"2019-10-03T03:21:57","date_gmt":"2019-10-03T03:21:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-factor-using-substitution-2\/"},"modified":"2024-04-30T23:21:06","modified_gmt":"2024-04-30T23:21:06","slug":"read-factor-using-substitution-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-factor-using-substitution-2\/","title":{"raw":"More Factoring Methods","rendered":"More Factoring Methods"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor polynomials\u00a0with negative or fractional exponents<\/li>\r\n \t<li>Factor by substitution<\/li>\r\n<\/ul>\r\n<\/div>\r\nExpressions with fractional or negative exponents can be factored using\u00a0the same factoring techniques\u00a0as those with integer exponents. It is important to remember a couple of things first.\r\n<ul>\r\n \t<li>When you multiply two exponentiated terms with the same base, you can add the exponents: [latex]x^{-1}\\cdot{x^{-1}}=x^{-1+(-1)}=x^{-2}[\/latex]<\/li>\r\n \t<li>When you add fractions, you need a common denominator: [latex]\\frac{1}{2}+\\frac{1}{3}=\\frac{3}{3}\\cdot\\frac{1}{2}+\\frac{2}{2}\\cdot\\frac{1}{3}=\\frac{3}{6}+\\frac{2}{6}=\\frac{5}{6}[\/latex]<\/li>\r\n \t<li>Polynomials have positive integer exponents - if it has a fractional or negative exponent it is an expression.<\/li>\r\n<\/ul>\r\nFirst, practice finding a GCF that is\u00a0a negative exponent.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]12y^{-3}-2y^{-2}[\/latex].\r\n[reveal-answer q=\"433582\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"433582\"]\r\n\r\nIf the exponents in this expression\u00a0were positive, we could determine that the GCF is [latex]2y^2[\/latex], but since we have negative exponents, we will need to use [latex]2y^{-3}[\/latex].\r\n\r\nTherefore, [latex]12y^{-3}-2y^{-2}=2y^{-3}(6-y)[\/latex]\r\n\r\nWe can check that we are correct by multiplying:\r\n\r\n[latex]2y^{-3}(6-y)=12y^{-3}-2y^{-3+1}=12y^{-3}-2y^{-2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]93663[\/ohm_question]\r\n\r\n<\/div>\r\nNow\u00a0let us factor a trinomial that has negative exponents.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]x^{-2}+5x^{-1}+6[\/latex].\r\n[reveal-answer q=\"44187\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44187\"]\r\n\r\nIf the exponents on this trinomial were positive, we could factor this as [latex](x+2)(x+3)[\/latex]. \u00a0Note that the exponent on the x's in the factored form is\u00a0[latex]1[\/latex], in other words [latex](x+2)=(x^{1}+2)[\/latex].\r\n\r\nAlso note that [latex]-1+(-1) = -2[\/latex]; therefore, if we factor this trinomial as [latex](x^{-1}+2)(x^{-1}+3)[\/latex], we will get the correct result if we check by multiplying.\r\n\r\n[latex](x^{-1}+2)(x^{-1}+3)=x^{-1+(-1)}+2x^{-1}+3x^{-1}+6=x^{-2}+5x^{-1}+6[\/latex]\r\n\r\nThe factored form is [latex](x^{-1}+2)(x^{-1}+3)[\/latex]\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will see a difference of squares with negative exponents. We can use the same shortcut as we have before, but be careful with the exponent.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]25x^{-4}-36[\/latex].\r\n[reveal-answer q=\"196480\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"196480\"]\r\n\r\nRecall that a difference of squares factors in this way: [latex]a^2-b^2=(a-b)(a+b)[\/latex], and the first thing we did was identify a and b to see whether we could factor this as a difference of squares.\r\n\r\nGiven\u00a0[latex]25x^{-4}-36[\/latex], we can define [latex]a=5x^{-2} \\text{ and }b = 6[\/latex] because [latex]({5x^{-2}})^2=25x^{-4} \\text{ and }6^2=36[\/latex]\r\n\r\nTherefore, the factored form is: [latex](5x^{-2}-6)(5x^{-2}+6)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will see more examples that are similar to the previous three written examples.\r\n\r\nhttps:\/\/youtu.be\/4w99g0GZOCk\r\n<h2>\u00a0Fractional Exponents<\/h2>\r\nAgain, we will first practice finding a GCF that has a fractional exponent.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]x^{\\frac{2}{3}}+3x^{\\frac{1}{3}}[\/latex].\r\n[reveal-answer q=\"533261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"533261\"]\r\n\r\nFirst, look for the term with the lowest value exponent. \u00a0In this case, it is [latex]3x^{\\frac{1}{3}}[\/latex].\r\n\r\nRecall that when you multiply terms with exponents, you add the exponents. To get\u00a0[latex]\\frac{2}{3}[\/latex] you would need to add [latex]\\frac{1}{3}[\/latex] to\u00a0[latex]\\frac{1}{3}[\/latex], so we will need a term whose exponent is\u00a0[latex]\\frac{1}{3}[\/latex].\r\n\r\n[latex]x^{\\frac{1}{3}}\\cdot{x^{\\frac{1}{3}}}=x^{\\frac{2}{3}}[\/latex], therefore:\r\n\r\n[latex]x^{\\frac{2}{3}}+3x^{\\frac{1}{3}}=x^{\\frac{1}{3}}(x^{\\frac{1}{3}}+3)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example, we will factor a perfect square trinomial that has fractional\u00a0exponents.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]25x^{\\frac{1}{2}}+70x^{\\frac{1}{4}}+49[\/latex].\r\n[reveal-answer q=\"703334\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"703334\"]\r\n\r\nRecall that a perfect square trinomial of the form [latex]a^2+2ab+b^2[\/latex] factors as [latex](a+b)^2[\/latex]\r\n\r\nThe first step in factoring a perfect square trinomial \u00a0was to identify a and b.\r\n\r\nTo find a, we ask: \u00a0[latex](?)^2=25x^{\\frac{1}{2}}[\/latex], and recall that [latex](x^a)^b=x^{a\\cdot{b}}[\/latex], therefore we are looking for an exponent for x that when multiplied by\u00a0[latex]2[\/latex], will give [latex]\\frac{1}{2}[\/latex]. You can also think about the fact that the middle term is defined as [latex]2ab[\/latex] so [latex]a[\/latex] will probably have an exponent of [latex]\\frac{1}{4}[\/latex], therefore a choice for [latex]a[\/latex] may be [latex]5x^{\\frac{1}{4}}[\/latex].\r\n\r\nWe can check that this is right by squaring [latex]a[\/latex]: [latex]{(5x^{\\frac{1}{4}})}^{2}=25x^{2\\cdot\\frac{1}{4}}=25x^{\\frac{1}{2}}[\/latex]\r\n\r\n[latex]b = 7\\text{ and }b^2=49[\/latex]\r\n\r\nNow we can check whether [latex]2ab =70x^{\\frac{1}{4}}[\/latex]\r\n\r\n[latex]2ab=2\\cdot{5x^{\\frac{1}{4}}}\\cdot7=70x^{\\frac{1}{4}}[\/latex]\r\n\r\nOur terms work out, so we can use the shortcut to factor:\r\n\r\n[latex]25x^{\\frac{1}{2}}+70x^{\\frac{1}{4}}+49=(5x^{\\frac{1}{4}}+7)^2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]93668[\/ohm_question]\r\n\r\n<\/div>\r\nIn our next video, you will see more examples of how to factor expressions with fractional exponents.\r\n\r\nhttps:\/\/youtu.be\/R6BzjR2O4z8\r\n<h2>Factor Using Substitution<\/h2>\r\nWe are going to move back to factoring polynomials; our exponents will be positive integers. Sometimes we encounter a polynomial that looks similar to something we know how to factor but is not quite the same. Substitution is a useful tool that can be used to \"mask\" a term or expression to make algebraic operations easier.\r\n\r\nYou may recall that substitution can be used to solve systems of linear equations and to check whether a point is a solution to a system of linear equations.\r\n\r\nFor example, consider the following equation:\r\n<div style=\"text-align: center;\">[latex]x+3y=8[\/latex]<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: left;\">To determine whether [latex]x=5[\/latex], and [latex]y=1[\/latex] is a solution to the equation, we can substitute the values\u00a0[latex]x=5[\/latex] and [latex]y=1[\/latex] into the equation.<\/div>\r\n<div><\/div>\r\n<p style=\"text-align: center;\">[latex](5)+3(1)=8[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]8=8[\/latex]\u00a0 \u00a0 \u00a0True<\/p>\r\nWe replaced the variables with numbers and then performed the algebraic operations specified. In the next example, we will see how we can use a similar technique to factor a fourth degree polynomial.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]x^4+3x^2+2[\/latex].\r\n[reveal-answer q=\"98597\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"98597\"]\r\n\r\nThis looks a lot like a trinomial that we know how to factor: [latex]x^2+3x+2=(x+2)(x+1)[\/latex]. The only thing different is the exponents.\r\n\r\nIf we substitute [latex]u=x^2[\/latex] and recognize that [latex]u^2=(x^2)^2=x^4[\/latex], we may be able to factor this beast!\r\n\r\nEverywhere there is a [latex]x^2[\/latex] we will replace it with a [latex]u[\/latex] then factor.\r\n\r\n[latex]u^2+3u+2=(u+1)(u+2)[\/latex]\r\n\r\nWe are not quite done yet. We want to factor the original polynomial which had [latex]x[\/latex] as its variable, so we need to replace [latex]x^2=u[\/latex] now that we are done factoring.\r\n\r\n[latex](u+1)(u+2)=(x^2+1)(x^2+2)[\/latex]\r\n\r\nWe conclude that [latex]x^4+3x^2+2=(x^2+1)(x^2+2)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]1366[\/ohm_question]\r\n\r\n<\/div>\r\nIn the following video, we show two more examples of how to use substitution to factor a fourth degree polynomial and an expression with fractional exponents.\r\n\r\nhttps:\/\/youtu.be\/QUznZt6yrgI\r\n<h2>Factor Completely<\/h2>\r\nSometimes you may encounter a polynomial that takes an extra step to factor. In our next example, we will first find the GCF of a trinomial, and after factoring it out, we will be able to factor again so that we end up with a product of a monomial and two binomials.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]6m^2k-3mk-3k[\/latex] completely.\r\n[reveal-answer q=\"698742\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"698742\"]\r\n\r\nWhenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is\u00a0[latex]3k[\/latex].\r\n\r\nFactor\u00a0[latex]3k[\/latex] from the trinomial:\r\n\r\n[latex]6m^2k-3mk-3k=3k\\left(2m^2-m-1\\right)[\/latex]\r\n\r\nWe are left with a trinomial that can be factored using your choice of factoring methods. We will create a table to find the factors of [latex]2\\cdot{-1}=-2[\/latex] that sum to [latex]-1[\/latex]\r\n<table class=\" aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 and Sum of Factors. The entries in the first column are: 1, 6; -1, -6; 2,3; and -2,-3. The entries in the second column are: 7, -7, 5, and -5.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]2\\cdot-1=-2[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]2,-1[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,1[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nOur factors are [latex]-2,1[\/latex] which will allow us to factor by grouping:\r\n\r\nRewrite the middle term with the\u00a0factors we found:\r\n<p style=\"text-align: center;\">[latex]\\left(2m^2-m-1\\right)=2m^2-2m+m-1[\/latex]<\/p>\r\nRegroup and find the GCF of each group:\r\n<p style=\"text-align: center;\">[latex](2m^2-2m)+(m-1)=2m(m-1)+1(m-1)[\/latex]<\/p>\r\nNow\u00a0factor [latex](m-1)[\/latex] from each term:\r\n<p style=\"text-align: center;\">[latex]2m^2-m-1=(m-1)(2m+1)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Do not forget the original GCF that we factored out! Our final factored form is:<\/p>\r\n<p style=\"text-align: center;\">[latex]6m^2k-3mk-3k=3k (m-1)(2m+1)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last example, we show why it is important to factor out a GCF, if there is one, before you begin using the techniques shown in this module.\r\n\r\nhttps:\/\/youtu.be\/hMAImz2BuPc\r\n<h2>Summary<\/h2>\r\nIn this section, we used factoring with special cases and factoring by grouping to factor expressions with negative and fractional exponents. We also returned to factoring polynomials and used the substitution method to factor a\u00a0[latex]4th[\/latex] degree polynomial. The last topic we covered was what it means to factor completely.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor polynomials\u00a0with negative or fractional exponents<\/li>\n<li>Factor by substitution<\/li>\n<\/ul>\n<\/div>\n<p>Expressions with fractional or negative exponents can be factored using\u00a0the same factoring techniques\u00a0as those with integer exponents. It is important to remember a couple of things first.<\/p>\n<ul>\n<li>When you multiply two exponentiated terms with the same base, you can add the exponents: [latex]x^{-1}\\cdot{x^{-1}}=x^{-1+(-1)}=x^{-2}[\/latex]<\/li>\n<li>When you add fractions, you need a common denominator: [latex]\\frac{1}{2}+\\frac{1}{3}=\\frac{3}{3}\\cdot\\frac{1}{2}+\\frac{2}{2}\\cdot\\frac{1}{3}=\\frac{3}{6}+\\frac{2}{6}=\\frac{5}{6}[\/latex]<\/li>\n<li>Polynomials have positive integer exponents &#8211; if it has a fractional or negative exponent it is an expression.<\/li>\n<\/ul>\n<p>First, practice finding a GCF that is\u00a0a negative exponent.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]12y^{-3}-2y^{-2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q433582\">Show Solution<\/span><\/p>\n<div id=\"q433582\" class=\"hidden-answer\" style=\"display: none\">\n<p>If the exponents in this expression\u00a0were positive, we could determine that the GCF is [latex]2y^2[\/latex], but since we have negative exponents, we will need to use [latex]2y^{-3}[\/latex].<\/p>\n<p>Therefore, [latex]12y^{-3}-2y^{-2}=2y^{-3}(6-y)[\/latex]<\/p>\n<p>We can check that we are correct by multiplying:<\/p>\n<p>[latex]2y^{-3}(6-y)=12y^{-3}-2y^{-3+1}=12y^{-3}-2y^{-2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm93663\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93663&theme=oea&iframe_resize_id=ohm93663&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Now\u00a0let us factor a trinomial that has negative exponents.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]x^{-2}+5x^{-1}+6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44187\">Show Solution<\/span><\/p>\n<div id=\"q44187\" class=\"hidden-answer\" style=\"display: none\">\n<p>If the exponents on this trinomial were positive, we could factor this as [latex](x+2)(x+3)[\/latex]. \u00a0Note that the exponent on the x&#8217;s in the factored form is\u00a0[latex]1[\/latex], in other words [latex](x+2)=(x^{1}+2)[\/latex].<\/p>\n<p>Also note that [latex]-1+(-1) = -2[\/latex]; therefore, if we factor this trinomial as [latex](x^{-1}+2)(x^{-1}+3)[\/latex], we will get the correct result if we check by multiplying.<\/p>\n<p>[latex](x^{-1}+2)(x^{-1}+3)=x^{-1+(-1)}+2x^{-1}+3x^{-1}+6=x^{-2}+5x^{-1}+6[\/latex]<\/p>\n<p>The factored form is [latex](x^{-1}+2)(x^{-1}+3)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will see a difference of squares with negative exponents. We can use the same shortcut as we have before, but be careful with the exponent.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]25x^{-4}-36[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q196480\">Show Solution<\/span><\/p>\n<div id=\"q196480\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recall that a difference of squares factors in this way: [latex]a^2-b^2=(a-b)(a+b)[\/latex], and the first thing we did was identify a and b to see whether we could factor this as a difference of squares.<\/p>\n<p>Given\u00a0[latex]25x^{-4}-36[\/latex], we can define [latex]a=5x^{-2} \\text{ and }b = 6[\/latex] because [latex]({5x^{-2}})^2=25x^{-4} \\text{ and }6^2=36[\/latex]<\/p>\n<p>Therefore, the factored form is: [latex](5x^{-2}-6)(5x^{-2}+6)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will see more examples that are similar to the previous three written examples.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factor Expressions with Negative Exponents\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/4w99g0GZOCk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>\u00a0Fractional Exponents<\/h2>\n<p>Again, we will first practice finding a GCF that has a fractional exponent.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]x^{\\frac{2}{3}}+3x^{\\frac{1}{3}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q533261\">Show Solution<\/span><\/p>\n<div id=\"q533261\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, look for the term with the lowest value exponent. \u00a0In this case, it is [latex]3x^{\\frac{1}{3}}[\/latex].<\/p>\n<p>Recall that when you multiply terms with exponents, you add the exponents. To get\u00a0[latex]\\frac{2}{3}[\/latex] you would need to add [latex]\\frac{1}{3}[\/latex] to\u00a0[latex]\\frac{1}{3}[\/latex], so we will need a term whose exponent is\u00a0[latex]\\frac{1}{3}[\/latex].<\/p>\n<p>[latex]x^{\\frac{1}{3}}\\cdot{x^{\\frac{1}{3}}}=x^{\\frac{2}{3}}[\/latex], therefore:<\/p>\n<p>[latex]x^{\\frac{2}{3}}+3x^{\\frac{1}{3}}=x^{\\frac{1}{3}}(x^{\\frac{1}{3}}+3)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example, we will factor a perfect square trinomial that has fractional\u00a0exponents.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]25x^{\\frac{1}{2}}+70x^{\\frac{1}{4}}+49[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q703334\">Show Solution<\/span><\/p>\n<div id=\"q703334\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recall that a perfect square trinomial of the form [latex]a^2+2ab+b^2[\/latex] factors as [latex](a+b)^2[\/latex]<\/p>\n<p>The first step in factoring a perfect square trinomial \u00a0was to identify a and b.<\/p>\n<p>To find a, we ask: \u00a0[latex](?)^2=25x^{\\frac{1}{2}}[\/latex], and recall that [latex](x^a)^b=x^{a\\cdot{b}}[\/latex], therefore we are looking for an exponent for x that when multiplied by\u00a0[latex]2[\/latex], will give [latex]\\frac{1}{2}[\/latex]. You can also think about the fact that the middle term is defined as [latex]2ab[\/latex] so [latex]a[\/latex] will probably have an exponent of [latex]\\frac{1}{4}[\/latex], therefore a choice for [latex]a[\/latex] may be [latex]5x^{\\frac{1}{4}}[\/latex].<\/p>\n<p>We can check that this is right by squaring [latex]a[\/latex]: [latex]{(5x^{\\frac{1}{4}})}^{2}=25x^{2\\cdot\\frac{1}{4}}=25x^{\\frac{1}{2}}[\/latex]<\/p>\n<p>[latex]b = 7\\text{ and }b^2=49[\/latex]<\/p>\n<p>Now we can check whether [latex]2ab =70x^{\\frac{1}{4}}[\/latex]<\/p>\n<p>[latex]2ab=2\\cdot{5x^{\\frac{1}{4}}}\\cdot7=70x^{\\frac{1}{4}}[\/latex]<\/p>\n<p>Our terms work out, so we can use the shortcut to factor:<\/p>\n<p>[latex]25x^{\\frac{1}{2}}+70x^{\\frac{1}{4}}+49=(5x^{\\frac{1}{4}}+7)^2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm93668\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93668&theme=oea&iframe_resize_id=ohm93668&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In our next video, you will see more examples of how to factor expressions with fractional exponents.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Factor Expressions with Fractional Exponents\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/R6BzjR2O4z8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Factor Using Substitution<\/h2>\n<p>We are going to move back to factoring polynomials; our exponents will be positive integers. Sometimes we encounter a polynomial that looks similar to something we know how to factor but is not quite the same. Substitution is a useful tool that can be used to &#8220;mask&#8221; a term or expression to make algebraic operations easier.<\/p>\n<p>You may recall that substitution can be used to solve systems of linear equations and to check whether a point is a solution to a system of linear equations.<\/p>\n<p>For example, consider the following equation:<\/p>\n<div style=\"text-align: center;\">[latex]x+3y=8[\/latex]<\/div>\n<div><\/div>\n<div style=\"text-align: left;\">To determine whether [latex]x=5[\/latex], and [latex]y=1[\/latex] is a solution to the equation, we can substitute the values\u00a0[latex]x=5[\/latex] and [latex]y=1[\/latex] into the equation.<\/div>\n<div><\/div>\n<p style=\"text-align: center;\">[latex](5)+3(1)=8[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]8=8[\/latex]\u00a0 \u00a0 \u00a0True<\/p>\n<p>We replaced the variables with numbers and then performed the algebraic operations specified. In the next example, we will see how we can use a similar technique to factor a fourth degree polynomial.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]x^4+3x^2+2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q98597\">Show Solution<\/span><\/p>\n<div id=\"q98597\" class=\"hidden-answer\" style=\"display: none\">\n<p>This looks a lot like a trinomial that we know how to factor: [latex]x^2+3x+2=(x+2)(x+1)[\/latex]. The only thing different is the exponents.<\/p>\n<p>If we substitute [latex]u=x^2[\/latex] and recognize that [latex]u^2=(x^2)^2=x^4[\/latex], we may be able to factor this beast!<\/p>\n<p>Everywhere there is a [latex]x^2[\/latex] we will replace it with a [latex]u[\/latex] then factor.<\/p>\n<p>[latex]u^2+3u+2=(u+1)(u+2)[\/latex]<\/p>\n<p>We are not quite done yet. We want to factor the original polynomial which had [latex]x[\/latex] as its variable, so we need to replace [latex]x^2=u[\/latex] now that we are done factoring.<\/p>\n<p>[latex](u+1)(u+2)=(x^2+1)(x^2+2)[\/latex]<\/p>\n<p>We conclude that [latex]x^4+3x^2+2=(x^2+1)(x^2+2)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm1366\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1366&theme=oea&iframe_resize_id=ohm1366&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video, we show two more examples of how to use substitution to factor a fourth degree polynomial and an expression with fractional exponents.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"(New Version Available) Factor Expressions Using Substitution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/QUznZt6yrgI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Factor Completely<\/h2>\n<p>Sometimes you may encounter a polynomial that takes an extra step to factor. In our next example, we will first find the GCF of a trinomial, and after factoring it out, we will be able to factor again so that we end up with a product of a monomial and two binomials.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]6m^2k-3mk-3k[\/latex] completely.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q698742\">Show Solution<\/span><\/p>\n<div id=\"q698742\" class=\"hidden-answer\" style=\"display: none\">\n<p>Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is\u00a0[latex]3k[\/latex].<\/p>\n<p>Factor\u00a0[latex]3k[\/latex] from the trinomial:<\/p>\n<p>[latex]6m^2k-3mk-3k=3k\\left(2m^2-m-1\\right)[\/latex]<\/p>\n<p>We are left with a trinomial that can be factored using your choice of factoring methods. We will create a table to find the factors of [latex]2\\cdot{-1}=-2[\/latex] that sum to [latex]-1[\/latex]<\/p>\n<table class=\"aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 and Sum of Factors. The entries in the first column are: 1, 6; -1, -6; 2,3; and -2,-3. The entries in the second column are: 7, -7, 5, and -5.\">\n<thead>\n<tr>\n<th>Factors of [latex]2\\cdot-1=-2[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]2,-1[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,1[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Our factors are [latex]-2,1[\/latex] which will allow us to factor by grouping:<\/p>\n<p>Rewrite the middle term with the\u00a0factors we found:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(2m^2-m-1\\right)=2m^2-2m+m-1[\/latex]<\/p>\n<p>Regroup and find the GCF of each group:<\/p>\n<p style=\"text-align: center;\">[latex](2m^2-2m)+(m-1)=2m(m-1)+1(m-1)[\/latex]<\/p>\n<p>Now\u00a0factor [latex](m-1)[\/latex] from each term:<\/p>\n<p style=\"text-align: center;\">[latex]2m^2-m-1=(m-1)(2m+1)[\/latex]<\/p>\n<p style=\"text-align: left;\">Do not forget the original GCF that we factored out! Our final factored form is:<\/p>\n<p style=\"text-align: center;\">[latex]6m^2k-3mk-3k=3k (m-1)(2m+1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our last example, we show why it is important to factor out a GCF, if there is one, before you begin using the techniques shown in this module.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  Factoring Polynomials with Common Factors\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/hMAImz2BuPc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>In this section, we used factoring with special cases and factoring by grouping to factor expressions with negative and fractional exponents. We also returned to factoring polynomials and used the substitution method to factor a\u00a0[latex]4th[\/latex] degree polynomial. The last topic we covered was what it means to factor completely.<\/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-16327\">\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>Factor Expressions with Fractional Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/R6BzjR2O4z8\">https:\/\/youtu.be\/R6BzjR2O4z8<\/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>Factor Expressions Using Substitution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/QUznZt6yrgI\">https:\/\/youtu.be\/QUznZt6yrgI<\/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>Revision and Adaptation. <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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Factor Expressions with Negative Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/4w99g0GZOCk\">https:\/\/youtu.be\/4w99g0GZOCk<\/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: Factoring Polynomials with Common Factors. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/hMAImz2BuPc\">https:\/\/youtu.be\/hMAImz2BuPc<\/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>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/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 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