{"id":2726,"date":"2016-07-19T20:25:57","date_gmt":"2016-07-19T20:25:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2726"},"modified":"2018-05-16T23:45:38","modified_gmt":"2018-05-16T23:45:38","slug":"read-special-cases-squares","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/read-special-cases-squares\/","title":{"raw":"Factoring Special Cases","rendered":"Factoring Special Cases"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Special Cases - Squares\r\n<ul>\r\n \t<li>Factor a polynomial of the form:\u00a0[latex]{a}^{2}+2ab+{b}^{2}[\/latex]<\/li>\r\n \t<li>Factor a polynomial of the form:\u00a0[latex]{a}^{2}-{b}^{2}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Special Cases - Cubes\r\n<ul>\r\n \t<li>Factor the sum of cubes.<\/li>\r\n \t<li>Factor the difference of cubes<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>More Factoring Methods\r\n<ul>\r\n \t<li>Factor by substitution<\/li>\r\n \t<li>Factor completely<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>General Factoring Strategy<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h3>Why learn how to factor special cases?<\/h3>\r\n<img class=\"size-medium wp-image-2786 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/19191115\/Screen-Shot-2016-07-19-at-12.10.30-PM-300x153.png\" alt=\"Repeated pattern of interlocking plus signs, each row a different color following the rainbow spectrum.\" width=\"300\" height=\"153\" \/>\r\n\r\n&nbsp;\r\n\r\nSome people like to find patterns in the world around them, like a game. \u00a0There are some polynomials that, when factored, follow a specific pattern.\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<p style=\"text-align: center\"><span style=\"text-decoration: underline\">These include:<\/span><\/p>\r\n<p style=\"text-align: center\">Perfect square trinomials of the form:\u00a0[latex]{a}^{2}+2ab+{b}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: center\">A difference of squares:\u00a0[latex]{a}^{2}-{b}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: center\">A sum of cubes:\u00a0[latex]{a}^{3}+{b}^{3}[\/latex]<\/p>\r\n<p style=\"text-align: center\">A difference of cubes:\u00a0[latex]{a}^{3}-{b}^{3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn this lesson you will see\u00a0you can factor each of these types of polynomials following a specific pattern. \u00a0You will also learn how to factor polynomials that have negative or fractional exponents.\r\n\r\n<img class=\"wp-image-4884 alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/18145450\/Screen-Shot-2016-06-14-at-10.43.58-AM-300x167.png\" alt=\"Picture of a sidewalk leading to a parking lot. There is a path through the grass to teh right of the sidewalk through the trees that has been made by people walking on the grass. The shortcut to the parking lot is the preferred way.\" width=\"581\" height=\"324\" \/>\r\n\r\nSome people find it helpful to know when they can take a shortcut to avoid doing extra work. \u00a0There are some polynomials that will always factor a certain way, and for those we offer a shortcut. \u00a0Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares. \u00a0The most important skill you will use in this section will be recognizing when you can use the shortcuts.\r\n<h2>Factoring a Perfect Square Trinomial<\/h2>\r\nA perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}&amp; =&amp; {\\left(a+b\\right)}^{2}\\hfill \\\\ &amp; \\text{and}&amp; \\\\ \\hfill {a}^{2}-2ab+{b}^{2}&amp; =&amp; {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\nWe can use this equation to factor any perfect square trinomial.\r\n<h3>A General Note: Perfect Square Trinomials<\/h3>\r\nA perfect square trinomial can be written as the square of a binomial:\r\n<div style=\"text-align: center\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\r\nIn the following example we will show you how to define a, and b so you can use the shortcut.\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nFactor [latex]25{x}^{2}+20x+4[\/latex].\r\n[reveal-answer q=\"119279\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"119279\"]\r\n\r\nFirst, notice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex].\r\n\r\nThis means that [latex]a=5x\\text{ and }b=2[\/latex]\r\n\r\nNext, check to see if the middle term is equal to [latex]2ab[\/latex], which it is:\r\n<p style=\"text-align: center\">[latex]2ab = 2\\left(5x\\right)\\left(2\\right)=20x[\/latex].<\/p>\r\n&nbsp;\r\n\r\nTherefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(a+b\\right)}^{2}={\\left(5x+2\\right)}^{2}[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]25{x}^{2}+20x+4={\\left(5x+2\\right)}^{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will show that we can use [latex]1 = 1^2[\/latex] to factor a polynomial with a term equal to 1.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]49{x}^{2}-14x+1[\/latex].\r\n[reveal-answer q=\"865849\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"865849\"]\r\n\r\nFirst, notice that [latex]49{x}^{2}[\/latex] and [latex]1[\/latex] are perfect squares because [latex]49{x}^{2}={\\left(7x\\right)}^{2}[\/latex] and [latex]1={1}^{2}[\/latex].\r\n\r\nThis means that [latex]a=7x[\/latex], we could say that [latex]b=1[\/latex], but would that give a middle term of [latex]-14x[\/latex]? We will need to choose [latex]b = -1[\/latex] to get the results we want:\r\n<p style=\"text-align: center\">[latex]2ab = 2\\left(7x\\right)\\left(-1\\right)=-14x[\/latex].<\/p>\r\nTherefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(a-b\\right)}^{2}={\\left(7x-1\\right)}^{2}[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]49{x}^{2}-14x+1={\\left(7x-1\\right)}^{2}[\/latex]\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we provide another short description of what a perfect square trinomial is, and show how to factor them using a the formula.\r\n\r\nhttps:\/\/youtu.be\/UMCVGDTxxTI\r\n\r\nWe can summarize our process in the following way:\r\n<h3>Given a perfect square trinomial, factor it into the square of a binomial.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\r\n \t<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex], or[latex]{\\left(a-b\\right)}^{2}[\/latex].<\/li>\r\n<\/ol>\r\n<h2>Factoring a Difference of Squares<\/h2>\r\nA difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.\r\n<div style=\"text-align: center\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\nWe can use this equation to factor any differences of squares.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Differences of Squares<\/h3>\r\nA difference of squares can be rewritten as two factors containing the same terms but opposite signs.\r\n<div style=\"text-align: center\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]9{x}^{2}-25[\/latex].\r\n[reveal-answer q=\"960938\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"960938\"]\r\n\r\nNotice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex].\r\n\r\nThis means that [latex]a=3x,\\text{ and }b=5[\/latex]\r\n\r\nThe polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].\r\n\r\nCheck that you are correct by multiplying.\r\n\r\n[latex]\\left(3x+5\\right)\\left(3x - 5\\right)=9x^2-15x+15x-25=9x^2-25[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex]9{x}^{2}-25=\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe most helpful\u00a0thing for recognizing a difference of squares that can be factored with the shortcut is knowing which numbers are perfect squares, as you will see in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]81{y}^{2}-144[\/latex].\r\n[reveal-answer q=\"193159\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"193159\"]\r\n\r\nNotice that [latex]81{y}^{2}[\/latex] and [latex]144[\/latex] are perfect squares because [latex]81{y}^{2}={\\left(9x\\right)}^{2}[\/latex] and [latex]144={12}^{2}[\/latex].\r\n\r\nThis means that [latex]a=9x,\\text{ and }b=12[\/latex]\r\n\r\nThe polynomial represents a difference of squares and can be rewritten as [latex]\\left(9x+12\\right)\\left(9x - 12\\right)[\/latex].\r\n\r\nCheck that you are correct by multiplying.\r\n\r\n[latex]\\left(9x+12\\right)\\left(9x - 12\\right)=81x^2-108x+108x-144=81x^2-144[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex]81{y}^{2}-144=\\left(9x+12\\right)\\left(9x - 12\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\nIn the following video we show another example of how to use the formula for fact a difference of squares.\r\n\r\nhttps:\/\/youtu.be\/Li9IBp5HrFA\r\n\r\nWe can summarize the process for factoring a difference of squares with the shortcut this way:\r\n<div class=\"textbox\">\r\n<h3>How To: Given a difference of squares, factor it into binomials.<\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nIs there a formula to factor the sum of squares, [latex]a^2+b^2[\/latex], into a product of two binomials?\r\n\r\nWrite down some ideas for how you would answer this in the box below before you look at the answer.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"121734\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"121734\"]\r\n\r\nThere is no way to factor a sum of squares into a product of two binomials, this is because of addition - the middle term needs to \"disappear\" and the only way to do that is with opposite signs. \u00a0to get a positive result, you must multiply two numbers with the same signs.\r\n\r\nThe only time a sum of squares can be factored is if they share any common factors, as in the following case:\r\n\r\n[latex]9x^2+36[\/latex]\r\n\r\n[latex]9x^2={(3x)}^2, \\text{ and }36 = 6^2[\/latex]\r\n\r\nThe only way to factor this expression is by pulling out the GCF which is 9.\r\n\r\n[latex]9x^2+36=9(x^2+4)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Cubes<\/h2>\r\nSome interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider: [latex]a^{3}+b^{3}[\/latex] and [latex]a^{3}-b^{3}[\/latex].\r\n\r\nLet\u2019s take a look at how to factor sums and differences of cubes.\r\n<h2>Sum of Cubes<\/h2>\r\nThe term \u201ccubed\u201d is used to describe a number raised to the third power. In geometry, a cube is a six-sided shape with equal width, length, and height; since all these measures are equal, the volume of a cube with width <i>x<\/i> can be represented by [latex]x^{3}[\/latex].\u00a0(Notice the exponent!)\r\n\r\nCubed numbers get large very quickly. [latex]1^{3}=1[\/latex], [latex]2^{3}=8[\/latex], [latex]3^{3}=27[\/latex], [latex]4^{3}=64[\/latex], and [latex]5^{3}=125[\/latex].\r\n\r\nBefore looking at factoring a sum of two cubes, let\u2019s look at the possible factors.\r\n\r\nIt turns out that [latex]a^{3}+b^{3}[\/latex] can actually be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]. Let\u2019s check these factors by multiplying.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nDoes [latex](a+b)(a^{2}\u2013ab+b^{2})=a^{3}+b^{3}[\/latex]?\r\n[reveal-answer q=\"386615\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"386615\"]\r\n\r\nApply the distributive property.\r\n<p style=\"text-align: center\">[latex]\\left(a\\right)\\left(a^{2}\u2013ab+b^{2}\\right)+\\left(b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]<\/p>\r\nMultiply by <i>a<\/i>.\r\n<p style=\"text-align: center\">[latex]\\left(a^{3}\u2013a^{2}b+ab^{2}\\right)+\\left(b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex]<\/p>\r\nMultiply by <i>b<\/i>.\r\n<p style=\"text-align: center\">[latex]\\left(a^{3}\u2013a^{2}b+ab^{2}\\right)+\\left(a^{2}b\u2013ab^{2}+b^{3}\\right)[\/latex]<\/p>\r\nRearrange terms in order to combine the like terms.\r\n<p style=\"text-align: center\">[latex]a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}[\/latex]<\/p>\r\nSimplify.\r\n<h4>Answer<\/h4>\r\n<p style=\"text-align: right\">[latex]a^{3}+b^{3}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nDid you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial [latex]a^{3}+b^{3}[\/latex].\u00a0So, the factors are correct.\r\n\r\nYou can use this pattern to factor binomials in the form [latex]a^{3}+b^{3}[\/latex], otherwise known as \u201cthe sum of cubes.\u201d\r\n<div class=\"textbox shaded\">\r\n<h3>The Sum of Cubes<\/h3>\r\nA binomial in the form [latex]a^{3}+b^{3}[\/latex] can be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex].\r\n<h3>Examples:<\/h3>\r\nThe factored form of [latex]x^{3}+64[\/latex]\u00a0is [latex]\\left(x+4\\right)\\left(x^{2}\u20134x+16\\right)[\/latex].\r\n\r\nThe factored form of [latex]8x^{3}+y^{3}[\/latex] is [latex]\\left(2x+y\\right)\\left(4x^{2}\u20132xy+y^{2}\\right)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]x^{3}+8y^{3}[\/latex].\r\n[reveal-answer q=\"354149\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"354149\"]\r\n\r\nIdentify that this binomial fits the sum of cubes pattern:\u00a0[latex]a^{3}+b^{3}[\/latex].\r\n\r\n[latex]a=x[\/latex], and [latex]b=2y[\/latex] (since [latex]2y\\cdot2y\\cdot2y=8y^{3}[\/latex]).\r\n<p style=\"text-align: center\">[latex]x^{3}+8y^{3}[\/latex]<\/p>\r\nFactor the binomial as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex], substituting [latex]a=x[\/latex] and [latex]b=2y[\/latex] into the expression.\r\n<p style=\"text-align: center\">[latex]\\left(x+2y\\right)\\left(x^{2}-x\\left(2y\\right)+\\left(2y\\right)^{2}\\right)[\/latex]<\/p>\r\nSquare\u00a0[latex](2y)^{2}=4y^{2}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\left(x+2y\\right)\\left(x^{2}-x\\left(2y\\right)+4y^{2}\\right)[\/latex]<\/p>\r\nMultiply [latex]\u2212x\\left(2y\\right)=\u22122xy[\/latex] (writing the coefficient first).\r\n<h4>Answer<\/h4>\r\n<p style=\"text-align: right\">[latex]\\left(x+2y\\right)\\left(x^{2}-2xy+4y^{2}\\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nAnd that\u2019s it. The binomial [latex]x^{3}+8y^{3}[\/latex]\u00a0can be factored as [latex]\\left(x+2y\\right)\\left(x^{2}\u20132xy+4y^{2}\\right)[\/latex]! Let\u2019s try another one.\r\n\r\nYou should always look for a common factor before you follow any of the patterns for factoring.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]16m^{3}+54n^{3}[\/latex].\r\n[reveal-answer q=\"254227\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"254227\"]\r\n\r\nFactor out the common factor 2.\r\n<p style=\"text-align: center\">[latex]16m^{3}+54n^{3}[\/latex]<\/p>\r\n[latex]8m^{3}[\/latex] and [latex]27n^{3}[\/latex]\u00a0are cubes, so you can factor [latex]8m^{3}+27n^{3}[\/latex]\u00a0as the sum of two cubes: [latex]a=2m[\/latex],\u00a0and\u00a0[latex]b=3n[\/latex].\r\n<p style=\"text-align: center\">[latex]2\\left(8m^{3}+27^n{3}\\right)[\/latex]<\/p>\r\nFactor the binomial [latex]8m^{3}+27n^{3}[\/latex] substituting [latex]a=2m[\/latex]\u00a0and [latex]b=3n[\/latex] into the expression [latex]\\left(a+b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex].\r\n<p style=\"text-align: center\">[latex]2\\left(2m+3n\\right)\\left[\\left(2m\\right)^{2}-\\left(2m\\right)\\left(3n\\right)+\\left(3n\\right)^{2}\\right][\/latex]<\/p>\r\nSquare: [latex](2m)^{2}=4m^{2}[\/latex] and [latex](3n)^{2}=9n^{2}[\/latex].\r\n<p style=\"text-align: center\">[latex]2\\left(2m+3n\\right)\\left[4m^{2}-\\left(2m\\right)\\left(3n\\right)+9n^{2}\\right][\/latex]<\/p>\r\nMultiply [latex]-\\left(2m\\right)\\left(3n\\right)=-6mn[\/latex].\r\n<h4>Answer<\/h4>\r\n<p style=\"text-align: right\">[latex]2\\left(2m+3n\\right)\\left(4m^{2}-6mn+9n^{2}\\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<h2>Difference of Cubes<\/h2>\r\nHaving seen how binomials in the form [latex]a^{3}+b^{3}[\/latex]\u00a0can be factored, it should not come as a surprise that binomials in the form [latex]a^{3}-b^{3}[\/latex]\u00a0can be factored in a similar way.\r\n<div class=\"textbox shaded\">\r\n<h3>The Difference of Cubes<\/h3>\r\nA binomial in the form [latex]a^{3}\u2013b^{3}[\/latex] can be factored as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex].\r\n<h4>Examples<\/h4>\r\nThe factored form of [latex]x^{3}\u201364[\/latex]\u00a0is [latex]\\left(x\u20134\\right)\\left(x^{2}+4x+16\\right)[\/latex].\r\n\r\nThe factored form of [latex]27x^{3}\u20138y^{3}[\/latex] is [latex]\\left(3x\u20132y\\left)\\right(9x^{2}+6xy+4y^{2}\\right)[\/latex].\r\n\r\n<\/div>\r\nNotice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the [latex]+[\/latex] and [latex]\u2013[\/latex] signs. Take a moment to compare the factored form of [latex]a^{3}+b^{3}[\/latex]\u00a0with the factored form of [latex]a^{3}-b^{3}[\/latex].\r\n\r\nFactored form of\u00a0[latex]a^{3}+b^{3}[\/latex]:\u00a0[latex]\\left(a+b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex]\r\n\r\nFactored form of [latex]a^{3}-b^{3}[\/latex]: [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex]\r\n\r\nThis can be tricky to remember because of the different signs\u2014the factored form of [latex]a^{3}+b^{3}[\/latex]\u00a0contains a negative, and the factored form of [latex]a^{3}-b^{3}[\/latex]<i> <\/i>contains a positive! Some people remember the different forms like this:\r\n\r\n\u201cRemember one sequence of variables: [latex]a^{3}b^{3}=\\left(a\\,b\\right)\\left(a^{2}ab\\,b^{2}\\right)[\/latex].\u00a0There are 4 missing signs. Whatever the first sign is, it is also the second sign. The third sign is the opposite, and the fourth sign is always [latex]+[\/latex].\u201d\r\n\r\nTry this for yourself. If the first sign is [latex]+[\/latex], as in [latex]a^{3}+b^{3}[\/latex], according to this strategy how do you fill in the rest: [latex]\\left(a\\,b\\right)\\left(a^{2}ab\\,b^{2}\\right)[\/latex]? Does this method help you remember the factored form of\u00a0[latex]a^{3}+b^{3}[\/latex]\u00a0and\u00a0[latex]a^{3}\u2013b^{3}[\/latex]?\r\n\r\nLet\u2019s go ahead and look at a couple of examples. Remember to factor out all common factors first.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]8x^{3}\u20131,000[\/latex].\r\n[reveal-answer q=\"809204\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"809204\"]\r\n\r\nFactor out 8.\r\n<p style=\"text-align: center\">[latex]8(x^{3}\u2013125)[\/latex]<\/p>\r\nIdentify that the binomial fits the pattern\u00a0[latex]a^{3}-b^{3}:a=x[\/latex], and [latex]b=5[\/latex] (since [latex]5^{3}=125[\/latex]).\r\n<p style=\"text-align: center\">[latex]8\\left(x^{3}\u2013125\\right)[\/latex]<\/p>\r\nFactor [latex]x^{3}\u2013125[\/latex]\u00a0as [latex]\\left(a\u2013b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex], substituting [latex]a=x[\/latex] and [latex]b=5[\/latex] into the expression.\r\n<p style=\"text-align: center\">[latex]8\\left(x-5\\right)\\left[x^{2}+\\left(x\\right)\\left(5\\right)+5^{2}\\right][\/latex]<\/p>\r\nSquare the first and last terms, and rewrite\u00a0[latex]\\left(x\\right)\\left(5\\right)[\/latex] as [latex]5x[\/latex].\r\n<p style=\"text-align: center\">[latex]8\\left(x\u20135\\right)\\left(x^{2}+5x+25\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n<p style=\"text-align: right\">[latex]8\\left(x\u20135\\right)\\left(x^{2}+5x+25\\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nLet\u2019s see what happens if you don\u2019t factor out the common factor first. In this example, it can still be factored as the difference of two cubes. However, the factored form still has common factors, which need to be factored out.\r\n\r\nAs you can see, this last example still worked, but required a couple of extra steps. It is always a good idea to factor out all common factors first. In some cases, the only efficient way to factor the binomial is to factor out the common factors first.\r\n\r\nHere is one more example. Note that [latex]r^{9}=\\left(r^{3}\\right)^{3}[\/latex]\u00a0and that [latex]8s^{6}=\\left(2s^{2}\\right)^{3}[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]r^{9}-8s^{6}[\/latex].\r\n[reveal-answer q=\"609560\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"609560\"]Identify this binomial as the difference of two cubes. As shown above, it is. Using the laws of exponents, rewrite [latex]r^{9}[\/latex] as [latex]\\left(r^{3}\\right)^{3}[\/latex].\r\n<p style=\"text-align: center\">[latex]r^{9}-8s^{6}[\/latex]<\/p>\r\nRewrite [latex]r^{9}[\/latex] as [latex]\\left(r^{3}\\right)^{3}[\/latex] and rewrite [latex]8s^{6}[\/latex] as [latex]\\left(2s^{2}\\right)^{3}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\left(r^{3}\\right)^{3}-\\left(2s^{2}\\right)^{3}[\/latex]<\/p>\r\nNow the binomial is written in terms of cubed quantities. Thinking of [latex]a^{3}-b^{3}[\/latex], [latex]a=r^{3}[\/latex] and [latex]b=2s^{2}[\/latex].\r\n\r\nFactor the binomial as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex], substituting [latex]a=r^{3}[\/latex] and [latex]b=2s^{2}[\/latex] into the expression.\r\n<p style=\"text-align: center\">[latex]\\left(r^{3}-2s^{2}\\right)\\left[\\left(r^{3}\\right)^{2}+\\left(r^{3}\\right)\\left(2s^{2}\\right)+\\left(2s^{2}\\right)^{2}\\right][\/latex]<\/p>\r\nMultiply and square the terms.\r\n<p style=\"text-align: center\">[latex]\\left(r^{3}-2s^{2}\\right)\\left(r^{6}+2r^{3}s^{2}+4s^{4}\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n<p style=\"text-align: right\">[latex]\\left(r^{3}-2s^{2}\\right)\\left(r^{6}+2r^{3}s^{2}+4s^{4}\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following two video examples we show more binomials that can be factored as a sum or difference of cubes.\r\n\r\nhttps:\/\/youtu.be\/tFSEpOB262M\r\n\r\nhttps:\/\/youtu.be\/J_0ctMrl5_0\r\n\r\nYou encounter some interesting patterns when factoring. Two special cases\u2014the sum of cubes and the difference of cubes\u2014can help you factor some binomials that have a degree of three (or higher, in some cases). The special cases are:\r\n<ul>\r\n \t<li>A\u00a0binomial in the form [latex]a^{3}+b^{3}[\/latex]\u00a0can be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]<\/li>\r\n \t<li>A binomial in the form [latex]a^{3}-b^{3}[\/latex]\u00a0can be factored as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex]<\/li>\r\n<\/ul>\r\nAlways remember to factor out any common factors first.\r\n<h2>More Factoring Methods<\/h2>\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 isn't 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. for example:\r\n\r\nTo determine whether the ordered pair [latex]\\left(5,1\\right)[\/latex] is a solution to the given system of equations.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{l}x+3y=8\\hfill \\\\ 2x - 9=y\\hfill \\end{array}[\/latex]<\/div>\r\nWe can substitute the ordered pair [latex]\\left(5,1\\right)[\/latex] into both equations.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{ll}\\left(5\\right)+3\\left(1\\right)=8\\hfill &amp; \\hfill \\\\ \\text{ }8=8\\hfill &amp; \\text{True}\\hfill \\\\ 2\\left(5\\right)-9=\\left(1\\right)\\hfill &amp; \\hfill \\\\ \\text{ }\\text{1=1}\\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]<\/p>\r\nWe replaced the variable with a number and then performed the algebraic operations specified. \u00a0In 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 Answer[\/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] except for 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 an [latex]x^2[\/latex] we will replace it with a u, then factor.\r\n\r\n[latex]u^2+3u+2=(u+1)(u+2)[\/latex]\r\n\r\nWe aren't quite done yet, we want to factor the original polynomial which had x as it's 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<h4>Answer<\/h4>\r\n[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\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 completely [latex]6m^2k-3mk-3k[\/latex].\r\n[reveal-answer q=\"698742\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"698742\"]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 3k.\r\n\r\nFactor 3k 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 method. 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], so we can factor by grouping:\r\n\r\nRewrite the middle term with the\u00a0factors we found with the table:\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\">Don't 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 shoe that it is important to factor out a GCF if there is one before you being using the techniques shown in this module.\r\n\r\nhttps:\/\/youtu.be\/hMAImz2BuPc\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]40018[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]124794[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]97334[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>General Factoring Strategy<\/h2>\r\nWith so many different tools used to factor, it is easy to get lost as to\u00a0which tool to use when. Here we will attempt to organize all the different\u00a0factoring types we have seen. A large part of deciding how to solve a problem\u00a0is based on how many terms are in the problem. For all problem types we will\u00a0always try to factor out the GCF first.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Factoring Strategy<\/h3>\r\n1) Factor out the GCF!\r\n\r\n2) <strong>2 terms<\/strong>: sum or difference of squares or cubes:\r\n<ul>\r\n \t<li>[latex]a^2 - b^2 = (a + b) (a - b)[\/latex]<\/li>\r\n \t<li>[latex]a^2 + b^2 = \\text{ Prime}[\/latex]<\/li>\r\n \t<li>[latex]a^3 + b^3 = (a + b) (a^2 - a b + b^2)[\/latex]<\/li>\r\n \t<li>[latex]a^3 - b^3 = (a - b) (a^2 + \\text{ ab} + b^2)[\/latex]<\/li>\r\n<\/ul>\r\n3) <strong>3 terms:<\/strong> watch for perfect square!\r\n<ul>\r\n \t<li>[latex]a^2 + 2 a b + b^2 = (a + b)^2[\/latex]<\/li>\r\n \t<li>[latex]a^2 - 2ab +b^2 = (a-b)^2[\/latex]<\/li>\r\n \t<li>grouping<\/li>\r\n \t<li>factoring by trial and error<\/li>\r\n<\/ul>\r\n4) <strong>4 terms<\/strong>:\r\n<ul>\r\n \t<li>grouping<\/li>\r\n<\/ul>\r\n&nbsp;\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Special Cases &#8211; Squares\n<ul>\n<li>Factor a polynomial of the form:\u00a0[latex]{a}^{2}+2ab+{b}^{2}[\/latex]<\/li>\n<li>Factor a polynomial of the form:\u00a0[latex]{a}^{2}-{b}^{2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>Special Cases &#8211; Cubes\n<ul>\n<li>Factor the sum of cubes.<\/li>\n<li>Factor the difference of cubes<\/li>\n<\/ul>\n<\/li>\n<li>More Factoring Methods\n<ul>\n<li>Factor by substitution<\/li>\n<li>Factor completely<\/li>\n<\/ul>\n<\/li>\n<li>General Factoring Strategy<\/li>\n<\/ul>\n<\/div>\n<h3>Why learn how to factor special cases?<\/h3>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-2786 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/19191115\/Screen-Shot-2016-07-19-at-12.10.30-PM-300x153.png\" alt=\"Repeated pattern of interlocking plus signs, each row a different color following the rainbow spectrum.\" width=\"300\" height=\"153\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Some people like to find patterns in the world around them, like a game. \u00a0There are some polynomials that, when factored, follow a specific pattern.<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<p style=\"text-align: center\"><span style=\"text-decoration: underline\">These include:<\/span><\/p>\n<p style=\"text-align: center\">Perfect square trinomials of the form:\u00a0[latex]{a}^{2}+2ab+{b}^{2}[\/latex]<\/p>\n<p style=\"text-align: center\">A difference of squares:\u00a0[latex]{a}^{2}-{b}^{2}[\/latex]<\/p>\n<p style=\"text-align: center\">A sum of cubes:\u00a0[latex]{a}^{3}+{b}^{3}[\/latex]<\/p>\n<p style=\"text-align: center\">A difference of cubes:\u00a0[latex]{a}^{3}-{b}^{3}[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In this lesson you will see\u00a0you can factor each of these types of polynomials following a specific pattern. \u00a0You will also learn how to factor polynomials that have negative or fractional exponents.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4884 alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/18145450\/Screen-Shot-2016-06-14-at-10.43.58-AM-300x167.png\" alt=\"Picture of a sidewalk leading to a parking lot. There is a path through the grass to teh right of the sidewalk through the trees that has been made by people walking on the grass. The shortcut to the parking lot is the preferred way.\" width=\"581\" height=\"324\" \/><\/p>\n<p>Some people find it helpful to know when they can take a shortcut to avoid doing extra work. \u00a0There are some polynomials that will always factor a certain way, and for those we offer a shortcut. \u00a0Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares. \u00a0The most important skill you will use in this section will be recognizing when you can use the shortcuts.<\/p>\n<h2>Factoring a Perfect Square Trinomial<\/h2>\n<p>A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}& =& {\\left(a+b\\right)}^{2}\\hfill \\\\ & \\text{and}& \\\\ \\hfill {a}^{2}-2ab+{b}^{2}& =& {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>We can use this equation to factor any perfect square trinomial.<\/p>\n<h3>A General Note: Perfect Square Trinomials<\/h3>\n<p>A perfect square trinomial can be written as the square of a binomial:<\/p>\n<div style=\"text-align: center\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\n<p>In the following example we will show you how to define a, and b so you can use the shortcut.<\/p>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>Factor [latex]25{x}^{2}+20x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q119279\">Show Answer<\/span><\/p>\n<div id=\"q119279\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, notice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex].<\/p>\n<p>This means that [latex]a=5x\\text{ and }b=2[\/latex]<\/p>\n<p>Next, check to see if the middle term is equal to [latex]2ab[\/latex], which it is:<\/p>\n<p style=\"text-align: center\">[latex]2ab = 2\\left(5x\\right)\\left(2\\right)=20x[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p>Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(a+b\\right)}^{2}={\\left(5x+2\\right)}^{2}[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]25{x}^{2}+20x+4={\\left(5x+2\\right)}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will show that we can use [latex]1 = 1^2[\/latex] to factor a polynomial with a term equal to 1.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]49{x}^{2}-14x+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q865849\">Show Answer<\/span><\/p>\n<div id=\"q865849\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, notice that [latex]49{x}^{2}[\/latex] and [latex]1[\/latex] are perfect squares because [latex]49{x}^{2}={\\left(7x\\right)}^{2}[\/latex] and [latex]1={1}^{2}[\/latex].<\/p>\n<p>This means that [latex]a=7x[\/latex], we could say that [latex]b=1[\/latex], but would that give a middle term of [latex]-14x[\/latex]? We will need to choose [latex]b = -1[\/latex] to get the results we want:<\/p>\n<p style=\"text-align: center\">[latex]2ab = 2\\left(7x\\right)\\left(-1\\right)=-14x[\/latex].<\/p>\n<p>Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(a-b\\right)}^{2}={\\left(7x-1\\right)}^{2}[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]49{x}^{2}-14x+1={\\left(7x-1\\right)}^{2}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we provide another short description of what a perfect square trinomial is, and show how to factor them using a the formula.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factor Perfect Square Trinomials Using a Formula\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/UMCVGDTxxTI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We can summarize our process in the following way:<\/p>\n<h3>Given a perfect square trinomial, factor it into the square of a binomial.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\n<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex], or[latex]{\\left(a-b\\right)}^{2}[\/latex].<\/li>\n<\/ol>\n<h2>Factoring a Difference of Squares<\/h2>\n<p>A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.<\/p>\n<div style=\"text-align: center\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<p>We can use this equation to factor any differences of squares.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Differences of Squares<\/h3>\n<p>A difference of squares can be rewritten as two factors containing the same terms but opposite signs.<\/p>\n<div style=\"text-align: center\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]9{x}^{2}-25[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q960938\">Show Answer<\/span><\/p>\n<div id=\"q960938\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex].<\/p>\n<p>This means that [latex]a=3x,\\text{ and }b=5[\/latex]<\/p>\n<p>The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].<\/p>\n<p>Check that you are correct by multiplying.<\/p>\n<p>[latex]\\left(3x+5\\right)\\left(3x - 5\\right)=9x^2-15x+15x-25=9x^2-25[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]9{x}^{2}-25=\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The most helpful\u00a0thing for recognizing a difference of squares that can be factored with the shortcut is knowing which numbers are perfect squares, as you will see in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]81{y}^{2}-144[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q193159\">Show Answer<\/span><\/p>\n<div id=\"q193159\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]81{y}^{2}[\/latex] and [latex]144[\/latex] are perfect squares because [latex]81{y}^{2}={\\left(9x\\right)}^{2}[\/latex] and [latex]144={12}^{2}[\/latex].<\/p>\n<p>This means that [latex]a=9x,\\text{ and }b=12[\/latex]<\/p>\n<p>The polynomial represents a difference of squares and can be rewritten as [latex]\\left(9x+12\\right)\\left(9x - 12\\right)[\/latex].<\/p>\n<p>Check that you are correct by multiplying.<\/p>\n<p>[latex]\\left(9x+12\\right)\\left(9x - 12\\right)=81x^2-108x+108x-144=81x^2-144[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]81{y}^{2}-144=\\left(9x+12\\right)\\left(9x - 12\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>In the following video we show another example of how to use the formula for fact a difference of squares.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Factor a Difference of Squares\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Li9IBp5HrFA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We can summarize the process for factoring a difference of squares with the shortcut this way:<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a difference of squares, factor it into binomials.<\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Is there a formula to factor the sum of squares, [latex]a^2+b^2[\/latex], into a product of two binomials?<\/p>\n<p>Write down some ideas for how you would answer this in the box below before you look at the answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q121734\">Show Answer<\/span><\/p>\n<div id=\"q121734\" class=\"hidden-answer\" style=\"display: none\">\n<p>There is no way to factor a sum of squares into a product of two binomials, this is because of addition &#8211; the middle term needs to &#8220;disappear&#8221; and the only way to do that is with opposite signs. \u00a0to get a positive result, you must multiply two numbers with the same signs.<\/p>\n<p>The only time a sum of squares can be factored is if they share any common factors, as in the following case:<\/p>\n<p>[latex]9x^2+36[\/latex]<\/p>\n<p>[latex]9x^2={(3x)}^2, \\text{ and }36 = 6^2[\/latex]<\/p>\n<p>The only way to factor this expression is by pulling out the GCF which is 9.<\/p>\n<p>[latex]9x^2+36=9(x^2+4)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Cubes<\/h2>\n<p>Some interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider: [latex]a^{3}+b^{3}[\/latex] and [latex]a^{3}-b^{3}[\/latex].<\/p>\n<p>Let\u2019s take a look at how to factor sums and differences of cubes.<\/p>\n<h2>Sum of Cubes<\/h2>\n<p>The term \u201ccubed\u201d is used to describe a number raised to the third power. In geometry, a cube is a six-sided shape with equal width, length, and height; since all these measures are equal, the volume of a cube with width <i>x<\/i> can be represented by [latex]x^{3}[\/latex].\u00a0(Notice the exponent!)<\/p>\n<p>Cubed numbers get large very quickly. [latex]1^{3}=1[\/latex], [latex]2^{3}=8[\/latex], [latex]3^{3}=27[\/latex], [latex]4^{3}=64[\/latex], and [latex]5^{3}=125[\/latex].<\/p>\n<p>Before looking at factoring a sum of two cubes, let\u2019s look at the possible factors.<\/p>\n<p>It turns out that [latex]a^{3}+b^{3}[\/latex] can actually be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]. Let\u2019s check these factors by multiplying.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Does [latex](a+b)(a^{2}\u2013ab+b^{2})=a^{3}+b^{3}[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q386615\">Show Answer<\/span><\/p>\n<div id=\"q386615\" class=\"hidden-answer\" style=\"display: none\">\n<p>Apply the distributive property.<\/p>\n<p style=\"text-align: center\">[latex]\\left(a\\right)\\left(a^{2}\u2013ab+b^{2}\\right)+\\left(b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]<\/p>\n<p>Multiply by <i>a<\/i>.<\/p>\n<p style=\"text-align: center\">[latex]\\left(a^{3}\u2013a^{2}b+ab^{2}\\right)+\\left(b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex]<\/p>\n<p>Multiply by <i>b<\/i>.<\/p>\n<p style=\"text-align: center\">[latex]\\left(a^{3}\u2013a^{2}b+ab^{2}\\right)+\\left(a^{2}b\u2013ab^{2}+b^{3}\\right)[\/latex]<\/p>\n<p>Rearrange terms in order to combine the like terms.<\/p>\n<p style=\"text-align: center\">[latex]a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<h4>Answer<\/h4>\n<p style=\"text-align: right\">[latex]a^{3}+b^{3}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial [latex]a^{3}+b^{3}[\/latex].\u00a0So, the factors are correct.<\/p>\n<p>You can use this pattern to factor binomials in the form [latex]a^{3}+b^{3}[\/latex], otherwise known as \u201cthe sum of cubes.\u201d<\/p>\n<div class=\"textbox shaded\">\n<h3>The Sum of Cubes<\/h3>\n<p>A binomial in the form [latex]a^{3}+b^{3}[\/latex] can be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex].<\/p>\n<h3>Examples:<\/h3>\n<p>The factored form of [latex]x^{3}+64[\/latex]\u00a0is [latex]\\left(x+4\\right)\\left(x^{2}\u20134x+16\\right)[\/latex].<\/p>\n<p>The factored form of [latex]8x^{3}+y^{3}[\/latex] is [latex]\\left(2x+y\\right)\\left(4x^{2}\u20132xy+y^{2}\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]x^{3}+8y^{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q354149\">Show Answer<\/span><\/p>\n<div id=\"q354149\" class=\"hidden-answer\" style=\"display: none\">\n<p>Identify that this binomial fits the sum of cubes pattern:\u00a0[latex]a^{3}+b^{3}[\/latex].<\/p>\n<p>[latex]a=x[\/latex], and [latex]b=2y[\/latex] (since [latex]2y\\cdot2y\\cdot2y=8y^{3}[\/latex]).<\/p>\n<p style=\"text-align: center\">[latex]x^{3}+8y^{3}[\/latex]<\/p>\n<p>Factor the binomial as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex], substituting [latex]a=x[\/latex] and [latex]b=2y[\/latex] into the expression.<\/p>\n<p style=\"text-align: center\">[latex]\\left(x+2y\\right)\\left(x^{2}-x\\left(2y\\right)+\\left(2y\\right)^{2}\\right)[\/latex]<\/p>\n<p>Square\u00a0[latex](2y)^{2}=4y^{2}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\left(x+2y\\right)\\left(x^{2}-x\\left(2y\\right)+4y^{2}\\right)[\/latex]<\/p>\n<p>Multiply [latex]\u2212x\\left(2y\\right)=\u22122xy[\/latex] (writing the coefficient first).<\/p>\n<h4>Answer<\/h4>\n<p style=\"text-align: right\">[latex]\\left(x+2y\\right)\\left(x^{2}-2xy+4y^{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>And that\u2019s it. The binomial [latex]x^{3}+8y^{3}[\/latex]\u00a0can be factored as [latex]\\left(x+2y\\right)\\left(x^{2}\u20132xy+4y^{2}\\right)[\/latex]! Let\u2019s try another one.<\/p>\n<p>You should always look for a common factor before you follow any of the patterns for factoring.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]16m^{3}+54n^{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q254227\">Show Answer<\/span><\/p>\n<div id=\"q254227\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factor out the common factor 2.<\/p>\n<p style=\"text-align: center\">[latex]16m^{3}+54n^{3}[\/latex]<\/p>\n<p>[latex]8m^{3}[\/latex] and [latex]27n^{3}[\/latex]\u00a0are cubes, so you can factor [latex]8m^{3}+27n^{3}[\/latex]\u00a0as the sum of two cubes: [latex]a=2m[\/latex],\u00a0and\u00a0[latex]b=3n[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]2\\left(8m^{3}+27^n{3}\\right)[\/latex]<\/p>\n<p>Factor the binomial [latex]8m^{3}+27n^{3}[\/latex] substituting [latex]a=2m[\/latex]\u00a0and [latex]b=3n[\/latex] into the expression [latex]\\left(a+b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]2\\left(2m+3n\\right)\\left[\\left(2m\\right)^{2}-\\left(2m\\right)\\left(3n\\right)+\\left(3n\\right)^{2}\\right][\/latex]<\/p>\n<p>Square: [latex](2m)^{2}=4m^{2}[\/latex] and [latex](3n)^{2}=9n^{2}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]2\\left(2m+3n\\right)\\left[4m^{2}-\\left(2m\\right)\\left(3n\\right)+9n^{2}\\right][\/latex]<\/p>\n<p>Multiply [latex]-\\left(2m\\right)\\left(3n\\right)=-6mn[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p style=\"text-align: right\">[latex]2\\left(2m+3n\\right)\\left(4m^{2}-6mn+9n^{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<h2>Difference of Cubes<\/h2>\n<p>Having seen how binomials in the form [latex]a^{3}+b^{3}[\/latex]\u00a0can be factored, it should not come as a surprise that binomials in the form [latex]a^{3}-b^{3}[\/latex]\u00a0can be factored in a similar way.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Difference of Cubes<\/h3>\n<p>A binomial in the form [latex]a^{3}\u2013b^{3}[\/latex] can be factored as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex].<\/p>\n<h4>Examples<\/h4>\n<p>The factored form of [latex]x^{3}\u201364[\/latex]\u00a0is [latex]\\left(x\u20134\\right)\\left(x^{2}+4x+16\\right)[\/latex].<\/p>\n<p>The factored form of [latex]27x^{3}\u20138y^{3}[\/latex] is [latex]\\left(3x\u20132y\\left)\\right(9x^{2}+6xy+4y^{2}\\right)[\/latex].<\/p>\n<\/div>\n<p>Notice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the [latex]+[\/latex] and [latex]\u2013[\/latex] signs. Take a moment to compare the factored form of [latex]a^{3}+b^{3}[\/latex]\u00a0with the factored form of [latex]a^{3}-b^{3}[\/latex].<\/p>\n<p>Factored form of\u00a0[latex]a^{3}+b^{3}[\/latex]:\u00a0[latex]\\left(a+b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex]<\/p>\n<p>Factored form of [latex]a^{3}-b^{3}[\/latex]: [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex]<\/p>\n<p>This can be tricky to remember because of the different signs\u2014the factored form of [latex]a^{3}+b^{3}[\/latex]\u00a0contains a negative, and the factored form of [latex]a^{3}-b^{3}[\/latex]<i> <\/i>contains a positive! Some people remember the different forms like this:<\/p>\n<p>\u201cRemember one sequence of variables: [latex]a^{3}b^{3}=\\left(a\\,b\\right)\\left(a^{2}ab\\,b^{2}\\right)[\/latex].\u00a0There are 4 missing signs. Whatever the first sign is, it is also the second sign. The third sign is the opposite, and the fourth sign is always [latex]+[\/latex].\u201d<\/p>\n<p>Try this for yourself. If the first sign is [latex]+[\/latex], as in [latex]a^{3}+b^{3}[\/latex], according to this strategy how do you fill in the rest: [latex]\\left(a\\,b\\right)\\left(a^{2}ab\\,b^{2}\\right)[\/latex]? Does this method help you remember the factored form of\u00a0[latex]a^{3}+b^{3}[\/latex]\u00a0and\u00a0[latex]a^{3}\u2013b^{3}[\/latex]?<\/p>\n<p>Let\u2019s go ahead and look at a couple of examples. Remember to factor out all common factors first.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]8x^{3}\u20131,000[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q809204\">Show Answer<\/span><\/p>\n<div id=\"q809204\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factor out 8.<\/p>\n<p style=\"text-align: center\">[latex]8(x^{3}\u2013125)[\/latex]<\/p>\n<p>Identify that the binomial fits the pattern\u00a0[latex]a^{3}-b^{3}:a=x[\/latex], and [latex]b=5[\/latex] (since [latex]5^{3}=125[\/latex]).<\/p>\n<p style=\"text-align: center\">[latex]8\\left(x^{3}\u2013125\\right)[\/latex]<\/p>\n<p>Factor [latex]x^{3}\u2013125[\/latex]\u00a0as [latex]\\left(a\u2013b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex], substituting [latex]a=x[\/latex] and [latex]b=5[\/latex] into the expression.<\/p>\n<p style=\"text-align: center\">[latex]8\\left(x-5\\right)\\left[x^{2}+\\left(x\\right)\\left(5\\right)+5^{2}\\right][\/latex]<\/p>\n<p>Square the first and last terms, and rewrite\u00a0[latex]\\left(x\\right)\\left(5\\right)[\/latex] as [latex]5x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]8\\left(x\u20135\\right)\\left(x^{2}+5x+25\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p style=\"text-align: right\">[latex]8\\left(x\u20135\\right)\\left(x^{2}+5x+25\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Let\u2019s see what happens if you don\u2019t factor out the common factor first. In this example, it can still be factored as the difference of two cubes. However, the factored form still has common factors, which need to be factored out.<\/p>\n<p>As you can see, this last example still worked, but required a couple of extra steps. It is always a good idea to factor out all common factors first. In some cases, the only efficient way to factor the binomial is to factor out the common factors first.<\/p>\n<p>Here is one more example. Note that [latex]r^{9}=\\left(r^{3}\\right)^{3}[\/latex]\u00a0and that [latex]8s^{6}=\\left(2s^{2}\\right)^{3}[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]r^{9}-8s^{6}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q609560\">Show Answer<\/span><\/p>\n<div id=\"q609560\" class=\"hidden-answer\" style=\"display: none\">Identify this binomial as the difference of two cubes. As shown above, it is. Using the laws of exponents, rewrite [latex]r^{9}[\/latex] as [latex]\\left(r^{3}\\right)^{3}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]r^{9}-8s^{6}[\/latex]<\/p>\n<p>Rewrite [latex]r^{9}[\/latex] as [latex]\\left(r^{3}\\right)^{3}[\/latex] and rewrite [latex]8s^{6}[\/latex] as [latex]\\left(2s^{2}\\right)^{3}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\left(r^{3}\\right)^{3}-\\left(2s^{2}\\right)^{3}[\/latex]<\/p>\n<p>Now the binomial is written in terms of cubed quantities. Thinking of [latex]a^{3}-b^{3}[\/latex], [latex]a=r^{3}[\/latex] and [latex]b=2s^{2}[\/latex].<\/p>\n<p>Factor the binomial as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex], substituting [latex]a=r^{3}[\/latex] and [latex]b=2s^{2}[\/latex] into the expression.<\/p>\n<p style=\"text-align: center\">[latex]\\left(r^{3}-2s^{2}\\right)\\left[\\left(r^{3}\\right)^{2}+\\left(r^{3}\\right)\\left(2s^{2}\\right)+\\left(2s^{2}\\right)^{2}\\right][\/latex]<\/p>\n<p>Multiply and square the terms.<\/p>\n<p style=\"text-align: center\">[latex]\\left(r^{3}-2s^{2}\\right)\\left(r^{6}+2r^{3}s^{2}+4s^{4}\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p style=\"text-align: right\">[latex]\\left(r^{3}-2s^{2}\\right)\\left(r^{6}+2r^{3}s^{2}+4s^{4}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following two video examples we show more binomials that can be factored as a sum or difference of cubes.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 1:  Factor a Sum or Difference of Cubes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tFSEpOB262M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 3:  Factor a Sum or Difference of Cubes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/J_0ctMrl5_0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You encounter some interesting patterns when factoring. Two special cases\u2014the sum of cubes and the difference of cubes\u2014can help you factor some binomials that have a degree of three (or higher, in some cases). The special cases are:<\/p>\n<ul>\n<li>A\u00a0binomial in the form [latex]a^{3}+b^{3}[\/latex]\u00a0can be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]<\/li>\n<li>A binomial in the form [latex]a^{3}-b^{3}[\/latex]\u00a0can be factored as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex]<\/li>\n<\/ul>\n<p>Always remember to factor out any common factors first.<\/p>\n<h2>More Factoring Methods<\/h2>\n<h2>Factor Using Substitution<\/h2>\n<p>We are going to move back to factoring polynomials &#8211; our exponents will be positive integers. Sometimes we encounter a polynomial that looks similar to something we know how to factor, but isn&#8217;t 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. for example:<\/p>\n<p>To determine whether the ordered pair [latex]\\left(5,1\\right)[\/latex] is a solution to the given system of equations.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{l}x+3y=8\\hfill \\\\ 2x - 9=y\\hfill \\end{array}[\/latex]<\/div>\n<p>We can substitute the ordered pair [latex]\\left(5,1\\right)[\/latex] into both equations.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{ll}\\left(5\\right)+3\\left(1\\right)=8\\hfill & \\hfill \\\\ \\text{ }8=8\\hfill & \\text{True}\\hfill \\\\ 2\\left(5\\right)-9=\\left(1\\right)\\hfill & \\hfill \\\\ \\text{ }\\text{1=1}\\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/p>\n<p>We replaced the variable with a number and then performed the algebraic operations specified. \u00a0In 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 Answer<\/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 &#8211; [latex]x^2+3x+2=(x+2)(x+1)[\/latex] except for 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 an [latex]x^2[\/latex] we will replace it with a u, then factor.<\/p>\n<p>[latex]u^2+3u+2=(u+1)(u+2)[\/latex]<\/p>\n<p>We aren&#8217;t quite done yet, we want to factor the original polynomial which had x as it&#8217;s 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<h4>Answer<\/h4>\n<p>[latex]x^4+3x^2+2=(x^2+1)(x^2+2)[\/latex]<\/p>\n<\/div>\n<\/div>\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-5\" 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 completely [latex]6m^2k-3mk-3k[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q698742\">Show Answer<\/span><\/p>\n<div id=\"q698742\" class=\"hidden-answer\" style=\"display: none\">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 3k.<\/p>\n<p>Factor 3k 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 method. 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], so we can factor by grouping:<\/p>\n<p>Rewrite the middle term with the\u00a0factors we found with the table:<\/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\">Don&#8217;t 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 shoe that it is important to factor out a GCF if there is one before you being using the techniques shown in this module.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" 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<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm40018\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=40018&theme=oea&iframe_resize_id=ohm40018&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm124794\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=124794&theme=oea&iframe_resize_id=ohm124794&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm97334\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=97334&theme=oea&iframe_resize_id=ohm97334&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>General Factoring Strategy<\/h2>\n<p>With so many different tools used to factor, it is easy to get lost as to\u00a0which tool to use when. Here we will attempt to organize all the different\u00a0factoring types we have seen. A large part of deciding how to solve a problem\u00a0is based on how many terms are in the problem. For all problem types we will\u00a0always try to factor out the GCF first.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Factoring Strategy<\/h3>\n<p>1) Factor out the GCF!<\/p>\n<p>2) <strong>2 terms<\/strong>: sum or difference of squares or cubes:<\/p>\n<ul>\n<li>[latex]a^2 - b^2 = (a + b) (a - b)[\/latex]<\/li>\n<li>[latex]a^2 + b^2 = \\text{ Prime}[\/latex]<\/li>\n<li>[latex]a^3 + b^3 = (a + b) (a^2 - a b + b^2)[\/latex]<\/li>\n<li>[latex]a^3 - b^3 = (a - b) (a^2 + \\text{ ab} + b^2)[\/latex]<\/li>\n<\/ul>\n<p>3) <strong>3 terms:<\/strong> watch for perfect square!<\/p>\n<ul>\n<li>[latex]a^2 + 2 a b + b^2 = (a + b)^2[\/latex]<\/li>\n<li>[latex]a^2 - 2ab +b^2 = (a-b)^2[\/latex]<\/li>\n<li>grouping<\/li>\n<li>factoring by trial and error<\/li>\n<\/ul>\n<p>4) <strong>4 terms<\/strong>:<\/p>\n<ul>\n<li>grouping<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/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-2726\">\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>Screenshot: Method to the Madness. <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>Image: Shortcut This Way. <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>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 Perfect Square Trinomials Using a Formula. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/UMCVGDTxxTI\">https:\/\/youtu.be\/UMCVGDTxxTI<\/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: Factor a Difference of Squares. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Li9IBp5HrFA\">https:\/\/youtu.be\/Li9IBp5HrFA<\/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><li><strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tFSEpOB262M\">https:\/\/youtu.be\/tFSEpOB262M<\/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 3: Factor a Sum or Difference of Cubes. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/J_0ctMrl5_0\">https:\/\/youtu.be\/J_0ctMrl5_0<\/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 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>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>Ex: Factoring Polynomials with Common Factors. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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>Question ID# 97334. <strong>Authored by<\/strong>: Lippman,David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID # 40018. <strong>Authored by<\/strong>: Jenck,Michael, mb Lewis,Matthew. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID#124794. <strong>Authored by<\/strong>: James Sousa . <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Factoring - Factoring Strategy. <strong>Authored by<\/strong>: Tyler Wallacy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.wallace.ccfaculty.org\/book\/book.html\">http:\/\/www.wallace.ccfaculty.org\/book\/book.html<\/a>. <strong>Project<\/strong>: Beginning and Intermediate Algebra. <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":21,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Factor Perfect Square Trinomials Using a Formula\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/UMCVGDTxxTI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Factor a Difference of Squares\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Li9IBp5HrFA\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot: Method to the Madness\",\"author\":\"\",\"organization\":\"Lumen 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