{"id":4838,"date":"2025-10-31T12:40:36","date_gmt":"2025-10-31T12:40:36","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=4838"},"modified":"2025-10-31T12:42:47","modified_gmt":"2025-10-31T12:42:47","slug":"introduction-to-derivatives-background-youll-need-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/introduction-to-derivatives-background-youll-need-3\/","title":{"raw":"Introduction to Derivatives: Background You'll Need 3","rendered":"Introduction to Derivatives: Background You&#8217;ll Need 3"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Break down different types of polynomial expressions into simpler parts using several factoring methods<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Factor Polynomials<\/h2>\r\n<p>Factoring is central to simplifying expressions, solving equations, and understanding polynomial behavior. Factoring involves breaking down expressions into simpler, constituent parts. A key step in this process is identifying the greatest common factor (GCF), which simplifies polynomials by dividing out commonalities and reducing complexity.<\/p>\r\n<h3>Greatest Common Factor<\/h3>\r\n<p>The <strong>greatest common factor<\/strong> (GCF) of two numbers is the largest number that divides evenly into both numbers.\u00a0<\/p>\r\n<section class=\"textbox example\">\r\n<p>[latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest number that divides evenly into both [latex]16[\/latex] and [latex]20[\/latex].<\/p>\r\n<\/section>\r\n<p>The GCF of polynomials works the same way. The <strong>greatest common factor<\/strong> (GCF) of a polynomial is the largest polynomial that divides evenly into each term of the polynomial.<\/p>\r\n<section class=\"textbox example\">\r\n<p>[latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest polynomial that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].<\/p>\r\n<\/section>\r\n<p>When factoring a polynomial expression, our first step is to check to see if each term contains a common factor. If so, we factor out the greatest amount we can from each term.\u00a0<\/p>\r\n<section class=\"textbox proTip\">\r\n<p>To make it less challenging to find this GCF of the polynomial terms, first look for the GCF of the coefficients, and then look for the GCF of the variables.<\/p>\r\n<\/section>\r\n<p>To factor out a GCF from a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property \"backwards\" to rewrite the polynomial in a factored form.<\/p>\r\n<section class=\"textbox recall\">\r\n<p>The distributive property allows us to multiply a number by a sum or difference inside parentheses and add or subtract the results. Conversely, when we see a common factor shared by all terms, we can factor it out, effectively reversing the distributive process.<\/p>\r\n<ul>\r\n\t<li>Using the distributive property: [latex]a\\left(b+c\\right)=ab+ac[\/latex].<\/li>\r\n\t<li>Factoring out a common factor: [latex]ab+ac=a\\left(b+c\\right)[\/latex].<\/li>\r\n<\/ul>\r\n<p>This principle shows us that multiplication distributed across a sum can be \"undone\" through factoring, revealing the GCF and the remaining terms of the polynomial.<\/p>\r\n<\/section>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How To: Given a Polynomial Expression, Factor Out the Greatest Common Factor<\/strong><\/p>\r\n<ol>\r\n\t<li>Identify the GCF of the coefficients.<\/li>\r\n\t<li>Identify the GCF of the variables.<\/li>\r\n\t<li>Combine to find the GCF of the expression.<\/li>\r\n\t<li>Determine what the GCF needs to be multiplied by to obtain each term in the expression.<\/li>\r\n\t<li>Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Factor [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].<\/p>\r\n<p>[reveal-answer q=\"113189\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"113189\"]<\/p>\r\n<p>First find the GCF of the expression. <br \/>\r\n<br \/>\r\nThe GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. <br \/>\r\n<br \/>\r\nThe GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions of the form [latex]{x}^{n}[\/latex] will always be the lowest exponent.) <br \/>\r\n<br \/>\r\nThe GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. <br \/>\r\n<br \/>\r\nCombine these to find the GCF of the polynomial, [latex]3xy[\/latex].<\/p>\r\n<p>Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. <br \/>\r\n<br \/>\r\nWe find that:<\/p>\r\n<center>[latex]\\begin{array}{c} 3xy(2x^2y^2) = 6x^3y^3, \\\\ 3xy(15xy) = 45x^2y^2, \\\\ 3xy(7) = 21xy \\end{array}[\/latex]<\/center>\r\n<p>Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.<\/p>\r\n<div style=\"text-align: center;\">[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]<\/div>\r\n<p>After factoring, we can check our work by multiplying. Use the distributive property to confirm that<\/p>\r\n<center>[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex]<\/center>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h3><strong>Factoring Quadratic Trinomials with a Leading Coefficient of [latex]1[\/latex]<\/strong><\/h3>\r\n<p>When factoring polynomials, starting with the greatest common factor (GCF) is standard. However, the GCF is not always the key to simplification, particularly for polynomials without a common factor. Let's look at some examples.<\/p>\r\n<section class=\"textbox example\">\r\n<p>The quadratic [pb_glossary id=\"1526\"]trinomial[\/pb_glossary] [latex]{x}^{2}+5x+6[\/latex] has a GCF of [latex]1[\/latex], but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].<br \/>\r\n<br \/>\r\nThe trinomial [latex]{x}^{2}+10x+16[\/latex] can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex], because [latex]2 \\times 8 =16[\/latex] and [latex]2 + 8 = 10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].<\/p>\r\n<\/section>\r\n<p>To factor trinomials like [latex]{x}^{2}+bx+c[\/latex], find two numbers that multiply to [latex]c[\/latex] and add up to [latex]b[\/latex].<\/p>\r\n<section class=\"textbox proTip\">\r\n<p>It's a common misconception that all trinomials can be broken down into binomial factors, but this isn't always the case. While many polynomials can be factored in this way, revealing a product of simpler binomials, there are instances where a trinomial is prime and cannot be factored further using real numbers<\/p>\r\n<\/section>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How To: Factoring a Trinomial of the Form [latex]{x}^{2}+bx+c[\/latex]<\/strong><\/p>\r\n<ol>\r\n\t<li>Identify all the numbers that multiple together to get [latex]c[\/latex].<\/li>\r\n\t<li>Of these numbers, find the pair of numbers where the sum equals [latex]b[\/latex].<\/li>\r\n\t<li>Write the trinomial as the product of two binomials, [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Factor [latex]{x}^{2}+2x - 15[\/latex]. <br \/>\r\n<br \/>\r\n[reveal-answer q=\"88306\"]Show Solution[\/reveal-answer] [hidden-answer a=\"88306\"] We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. <br \/>\r\n<br \/>\r\nWe need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. <br \/>\r\n<br \/>\r\nIn the table, we list factors until we find a pair with the desired sum.<\/p>\r\n<table summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]-15[\/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]1,-15[\/latex]<\/td>\r\n<td>[latex]-14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,15[\/latex]<\/td>\r\n<td>[latex]14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-5[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,5[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].<br \/>\r\n<br \/>\r\nWe can check our work by multiplying. <br \/>\r\n<br \/>\r\nUse FOIL to confirm that [latex]\\left(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex]. [\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288370[\/ohm_question]<\/p>\r\n<\/section>\r\n<h2>Factoring by Grouping<\/h2>\r\n<p>When we have trinomials with leading coefficients other than 1, they can often be <strong>factored by grouping<\/strong>. This method involves breaking the middle term into two terms that can be factored separately, and then extracting the greatest common factor (GCF).\u00a0<\/p>\r\n<section class=\"textbox example\">\r\n<p>The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be approached by writing it as [latex](2x^2+2x)+(3x+3)[\/latex] and then factoring each group separately. This gives us [latex]2x(x+1)+3(x+1)[\/latex]. We then factor out the common binomial [latex]\\left(x+1\\right)[\/latex] to get the final factored form [latex](2x+3)(x+1)[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How To: Given a Trinomial in the Form [latex]a{x}^{2}+bx+c[\/latex], Factor by Grouping<\/strong><\/p>\r\n<ol>\r\n\t<li>Multiply [latex]a[\/latex] and [latex]c[\/latex] to find the key number.<\/li>\r\n\t<li>Find two numbers that multiply to the key number and add to [latex]b[\/latex].<\/li>\r\n\t<li>Split the middle term, [latex]bx[\/latex], using these two numbers and rewrite the trinomial.<\/li>\r\n\t<li>Group the terms into pairs and factor out the common factor from each group.<\/li>\r\n\t<li>Extract the common binomial factor from the groups.<\/li>\r\n\t<li>Write the original expression as the product of two binomials.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Factor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.<\/p>\r\n<p>[reveal-answer q=\"806328\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"806328\"]<\/p>\r\n<p>We have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\r\n<table summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]-30[\/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]1,-30[\/latex]<\/td>\r\n<td>[latex]-29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,30[\/latex]<\/td>\r\n<td>[latex]29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2,-15[\/latex]<\/td>\r\n<td>[latex]-13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,15[\/latex]<\/td>\r\n<td>[latex]13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-10[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,10[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>So [latex]p=-3[\/latex] and [latex]q=10[\/latex].<\/p>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill &amp; \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill &amp; \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill &amp; \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n<p>We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].<\/p>\r\n<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288372[\/ohm_question]<\/p>\r\n<\/section>\r\n<h3>Factoring a Perfect Square Trinomial<\/h3>\r\n<p>A perfect square trinomial is one that can be expressed as a binomial squared. It occurs when you square a binomial, resulting in a trinomial where the first and last terms are perfect squares and the middle term is twice the product of the terms being squared.<\/p>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}&amp; =&amp; {\\left(a+b\\right)}^{2} \\text{\u00a0 \u00a0 \u00a0 \u00a0 \u00a0When adding.} \\hfill \\\\ &amp; \\text{and}&amp; \\\\ \\hfill {a}^{2}-2ab+{b}^{2}&amp; =&amp; {\\left(a-b\\right)}^{2} \\text{\u00a0 \u00a0 \u00a0 \u00a0 When subtracting.}\\hfill \\end{array}[\/latex]<\/div>\r\n<div style=\"text-align: left;\">These formulas allow us to rewrite any perfect square trinomial in its factored form.<\/div>\r\n<section class=\"textbox example\">\r\n<p>The trinomial [latex]49{x}^{2}-14x+1[\/latex] factors into a binomial squared.<\/p>\r\n<p>The first term [latex]49x^2[\/latex] is the square of [latex]7x[\/latex], and the last term [latex]1[\/latex] is the square of [latex]1[\/latex]. The middle term, [latex]\u221214x[\/latex], is equal to twice the product of [latex]7x[\/latex] and [latex]\u22121[\/latex].<\/p>\r\n<p>Therefore, the trinomial is a perfect square and its factored form is [latex]{\\left(7x - 1\\right)}^{2}[\/latex].<\/p>\r\n<\/section>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How to: Factor a Perfect Square Trinomial:<\/strong><\/p>\r\n<ol>\r\n\t<li>Check that both the first and last terms are perfect squares.<\/li>\r\n\t<li>Verify that the middle term is double the product of the square roots of the first and last terms.<\/li>\r\n\t<li>Express the trinomial as a squared binomial, [latex]{\\left(a+b\\right)}^{2}[\/latex] or [latex]{\\left(a-b\\right)}^{2}[\/latex], based on the sign of the middle term.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Factor [latex]25{x}^{2}+20x+4[\/latex].<\/p>\r\n<p>[reveal-answer q=\"114092\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"114092\"]<\/p>\r\n<p>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]. Then check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(5x+2\\right)}^{2}[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288373[\/ohm_question]<\/p>\r\n<\/section>\r\n<h3>Factoring a Difference of Squares<\/h3>\r\n<p>A <strong>difference of squares<\/strong> occurs when you subtract one perfect square from another. It's a special pattern in algebra where two square terms are separated by a minus sign, and it can be factored into two binomials with opposite signs.<\/p>\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n<p>This equation represents the factored form of a difference of squares.<\/p>\r\n<section class=\"textbox example\">\r\n<p>Take the expression [latex]81y^2-100[\/latex], for example.<\/p>\r\n<p>Both [latex]81y^2[\/latex] and [latex]100[\/latex] are perfect squares, with [latex]81y^2[\/latex] being [latex](9y)^2[\/latex] and [latex]100[\/latex] being [latex]10^2[\/latex].<\/p>\r\n<p>This expression can be factored into binomials as follows:<\/p>\r\n<center>The factored form of [latex]81{y}^{2}-100[\/latex] is [latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex].<\/center><\/section>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How To: Factor a Difference of Squares<\/strong><\/p>\r\n<ol>\r\n\t<li><strong>Identify the Squares:<\/strong> Start by ensuring both terms are perfect squares. In other words, each term can be written as some expression squared, such as [latex]a^2[\/latex] and [latex]b^2[\/latex].<\/li>\r\n\t<li><strong>Determine the Roots:<\/strong> Find the square root of each term. The square root of [latex]a^2[\/latex] is [latex]a[\/latex], and the square root of [latex]b^2[\/latex] is [latex]b[\/latex].<\/li>\r\n\t<li><strong>Set Up Binomials:<\/strong> Create two binomial expressions. One binomial will have a plus sign, and the other will have a minus sign between the terms.<\/li>\r\n\t<li><strong>Write the Factored Form:<\/strong> Combine the binomials to form the factored expression: [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Factor [latex]9{x}^{2}-25[\/latex]. [reveal-answer q=\"830417\"]Show Solution[\/reveal-answer] [hidden-answer a=\"830417\"] 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]. The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex]. [\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288374[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Break down different types of polynomial expressions into simpler parts using several factoring methods<\/li>\n<\/ul>\n<\/section>\n<h2>Factor Polynomials<\/h2>\n<p>Factoring is central to simplifying expressions, solving equations, and understanding polynomial behavior. Factoring involves breaking down expressions into simpler, constituent parts. A key step in this process is identifying the greatest common factor (GCF), which simplifies polynomials by dividing out commonalities and reducing complexity.<\/p>\n<h3>Greatest Common Factor<\/h3>\n<p>The <strong>greatest common factor<\/strong> (GCF) of two numbers is the largest number that divides evenly into both numbers.\u00a0<\/p>\n<section class=\"textbox example\">\n<p>[latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest number that divides evenly into both [latex]16[\/latex] and [latex]20[\/latex].<\/p>\n<\/section>\n<p>The GCF of polynomials works the same way. The <strong>greatest common factor<\/strong> (GCF) of a polynomial is the largest polynomial that divides evenly into each term of the polynomial.<\/p>\n<section class=\"textbox example\">\n<p>[latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest polynomial that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].<\/p>\n<\/section>\n<p>When factoring a polynomial expression, our first step is to check to see if each term contains a common factor. If so, we factor out the greatest amount we can from each term.\u00a0<\/p>\n<section class=\"textbox proTip\">\n<p>To make it less challenging to find this GCF of the polynomial terms, first look for the GCF of the coefficients, and then look for the GCF of the variables.<\/p>\n<\/section>\n<p>To factor out a GCF from a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property &#8220;backwards&#8221; to rewrite the polynomial in a factored form.<\/p>\n<section class=\"textbox recall\">\n<p>The distributive property allows us to multiply a number by a sum or difference inside parentheses and add or subtract the results. Conversely, when we see a common factor shared by all terms, we can factor it out, effectively reversing the distributive process.<\/p>\n<ul>\n<li>Using the distributive property: [latex]a\\left(b+c\\right)=ab+ac[\/latex].<\/li>\n<li>Factoring out a common factor: [latex]ab+ac=a\\left(b+c\\right)[\/latex].<\/li>\n<\/ul>\n<p>This principle shows us that multiplication distributed across a sum can be &#8220;undone&#8221; through factoring, revealing the GCF and the remaining terms of the polynomial.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Given a Polynomial Expression, Factor Out the Greatest Common Factor<\/strong><\/p>\n<ol>\n<li>Identify the GCF of the coefficients.<\/li>\n<li>Identify the GCF of the variables.<\/li>\n<li>Combine to find the GCF of the expression.<\/li>\n<li>Determine what the GCF needs to be multiplied by to obtain each term in the expression.<\/li>\n<li>Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Factor [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q113189\">Show Solution<\/button><\/p>\n<div id=\"q113189\" class=\"hidden-answer\" style=\"display: none\">\n<p>First find the GCF of the expression. <\/p>\n<p>The GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. <\/p>\n<p>The GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions of the form [latex]{x}^{n}[\/latex] will always be the lowest exponent.) <\/p>\n<p>The GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. <\/p>\n<p>Combine these to find the GCF of the polynomial, [latex]3xy[\/latex].<\/p>\n<p>Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. <\/p>\n<p>We find that:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c} 3xy(2x^2y^2) = 6x^3y^3, \\\\ 3xy(15xy) = 45x^2y^2, \\\\ 3xy(7) = 21xy \\end{array}[\/latex]<\/div>\n<p>Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.<\/p>\n<div style=\"text-align: center;\">[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]<\/div>\n<p>After factoring, we can check our work by multiplying. Use the distributive property to confirm that<\/p>\n<div style=\"text-align: center;\">[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h3><strong>Factoring Quadratic Trinomials with a Leading Coefficient of [latex]1[\/latex]<\/strong><\/h3>\n<p>When factoring polynomials, starting with the greatest common factor (GCF) is standard. However, the GCF is not always the key to simplification, particularly for polynomials without a common factor. Let&#8217;s look at some examples.<\/p>\n<section class=\"textbox example\">\n<p>The quadratic <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_4838_1526\">trinomial<\/a> [latex]{x}^{2}+5x+6[\/latex] has a GCF of [latex]1[\/latex], but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].<\/p>\n<p>The trinomial [latex]{x}^{2}+10x+16[\/latex] can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex], because [latex]2 \\times 8 =16[\/latex] and [latex]2 + 8 = 10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].<\/p>\n<\/section>\n<p>To factor trinomials like [latex]{x}^{2}+bx+c[\/latex], find two numbers that multiply to [latex]c[\/latex] and add up to [latex]b[\/latex].<\/p>\n<section class=\"textbox proTip\">\n<p>It&#8217;s a common misconception that all trinomials can be broken down into binomial factors, but this isn&#8217;t always the case. While many polynomials can be factored in this way, revealing a product of simpler binomials, there are instances where a trinomial is prime and cannot be factored further using real numbers<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Factoring a Trinomial of the Form [latex]{x}^{2}+bx+c[\/latex]<\/strong><\/p>\n<ol>\n<li>Identify all the numbers that multiple together to get [latex]c[\/latex].<\/li>\n<li>Of these numbers, find the pair of numbers where the sum equals [latex]b[\/latex].<\/li>\n<li>Write the trinomial as the product of two binomials, [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Factor [latex]{x}^{2}+2x - 15[\/latex]. <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q88306\">Show Solution<\/button> <\/p>\n<div id=\"q88306\" class=\"hidden-answer\" style=\"display: none\"> We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. <\/p>\n<p>We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. <\/p>\n<p>In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\n<thead>\n<tr>\n<th>Factors of [latex]-15[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-15[\/latex]<\/td>\n<td>[latex]-14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,15[\/latex]<\/td>\n<td>[latex]14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-5[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,5[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].<\/p>\n<p>We can check our work by multiplying. <\/p>\n<p>Use FOIL to confirm that [latex]\\left(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex]. <\/p><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288370\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288370&theme=lumen&iframe_resize_id=ohm288370&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<h2>Factoring by Grouping<\/h2>\n<p>When we have trinomials with leading coefficients other than 1, they can often be <strong>factored by grouping<\/strong>. This method involves breaking the middle term into two terms that can be factored separately, and then extracting the greatest common factor (GCF).\u00a0<\/p>\n<section class=\"textbox example\">\n<p>The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be approached by writing it as [latex](2x^2+2x)+(3x+3)[\/latex] and then factoring each group separately. This gives us [latex]2x(x+1)+3(x+1)[\/latex]. We then factor out the common binomial [latex]\\left(x+1\\right)[\/latex] to get the final factored form [latex](2x+3)(x+1)[\/latex]<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Given a Trinomial in the Form [latex]a{x}^{2}+bx+c[\/latex], Factor by Grouping<\/strong><\/p>\n<ol>\n<li>Multiply [latex]a[\/latex] and [latex]c[\/latex] to find the key number.<\/li>\n<li>Find two numbers that multiply to the key number and add to [latex]b[\/latex].<\/li>\n<li>Split the middle term, [latex]bx[\/latex], using these two numbers and rewrite the trinomial.<\/li>\n<li>Group the terms into pairs and factor out the common factor from each group.<\/li>\n<li>Extract the common binomial factor from the groups.<\/li>\n<li>Write the original expression as the product of two binomials.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Factor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q806328\">Show Solution<\/button><\/p>\n<div id=\"q806328\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\n<thead>\n<tr>\n<th>Factors of [latex]-30[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-30[\/latex]<\/td>\n<td>[latex]-29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,30[\/latex]<\/td>\n<td>[latex]29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2,-15[\/latex]<\/td>\n<td>[latex]-13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,15[\/latex]<\/td>\n<td>[latex]13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-10[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,10[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So [latex]p=-3[\/latex] and [latex]q=10[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill & \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill & \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill & \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<p>We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].<\/p>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288372\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288372&theme=lumen&iframe_resize_id=ohm288372&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<h3>Factoring a Perfect Square Trinomial<\/h3>\n<p>A perfect square trinomial is one that can be expressed as a binomial squared. It occurs when you square a binomial, resulting in a trinomial where the first and last terms are perfect squares and the middle term is twice the product of the terms being squared.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}& =& {\\left(a+b\\right)}^{2} \\text{\u00a0 \u00a0 \u00a0 \u00a0 \u00a0When adding.} \\hfill \\\\ & \\text{and}& \\\\ \\hfill {a}^{2}-2ab+{b}^{2}& =& {\\left(a-b\\right)}^{2} \\text{\u00a0 \u00a0 \u00a0 \u00a0 When subtracting.}\\hfill \\end{array}[\/latex]<\/div>\n<div style=\"text-align: left;\">These formulas allow us to rewrite any perfect square trinomial in its factored form.<\/div>\n<section class=\"textbox example\">\n<p>The trinomial [latex]49{x}^{2}-14x+1[\/latex] factors into a binomial squared.<\/p>\n<p>The first term [latex]49x^2[\/latex] is the square of [latex]7x[\/latex], and the last term [latex]1[\/latex] is the square of [latex]1[\/latex]. The middle term, [latex]\u221214x[\/latex], is equal to twice the product of [latex]7x[\/latex] and [latex]\u22121[\/latex].<\/p>\n<p>Therefore, the trinomial is a perfect square and its factored form is [latex]{\\left(7x - 1\\right)}^{2}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><strong>How to: Factor a Perfect Square Trinomial:<\/strong><\/p>\n<ol>\n<li>Check that both the first and last terms are perfect squares.<\/li>\n<li>Verify that the middle term is double the product of the square roots of the first and last terms.<\/li>\n<li>Express the trinomial as a squared binomial, [latex]{\\left(a+b\\right)}^{2}[\/latex] or [latex]{\\left(a-b\\right)}^{2}[\/latex], based on the sign of the middle term.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Factor [latex]25{x}^{2}+20x+4[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q114092\">Show Solution<\/button><\/p>\n<div id=\"q114092\" class=\"hidden-answer\" style=\"display: none\">\n<p>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]. Then check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(5x+2\\right)}^{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288373\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288373&theme=lumen&iframe_resize_id=ohm288373&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<h3>Factoring a Difference of Squares<\/h3>\n<p>A <strong>difference of squares<\/strong> occurs when you subtract one perfect square from another. It&#8217;s a special pattern in algebra where two square terms are separated by a minus sign, and it can be factored into two binomials with 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<p>This equation represents the factored form of a difference of squares.<\/p>\n<section class=\"textbox example\">\n<p>Take the expression [latex]81y^2-100[\/latex], for example.<\/p>\n<p>Both [latex]81y^2[\/latex] and [latex]100[\/latex] are perfect squares, with [latex]81y^2[\/latex] being [latex](9y)^2[\/latex] and [latex]100[\/latex] being [latex]10^2[\/latex].<\/p>\n<p>This expression can be factored into binomials as follows:<\/p>\n<div style=\"text-align: center;\">The factored form of [latex]81{y}^{2}-100[\/latex] is [latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex].<\/div>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Factor a Difference of Squares<\/strong><\/p>\n<ol>\n<li><strong>Identify the Squares:<\/strong> Start by ensuring both terms are perfect squares. In other words, each term can be written as some expression squared, such as [latex]a^2[\/latex] and [latex]b^2[\/latex].<\/li>\n<li><strong>Determine the Roots:<\/strong> Find the square root of each term. The square root of [latex]a^2[\/latex] is [latex]a[\/latex], and the square root of [latex]b^2[\/latex] is [latex]b[\/latex].<\/li>\n<li><strong>Set Up Binomials:<\/strong> Create two binomial expressions. One binomial will have a plus sign, and the other will have a minus sign between the terms.<\/li>\n<li><strong>Write the Factored Form:<\/strong> Combine the binomials to form the factored expression: [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Factor [latex]9{x}^{2}-25[\/latex]. <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q830417\">Show Solution<\/button> <\/p>\n<div id=\"q830417\" class=\"hidden-answer\" style=\"display: none\"> 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]. The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex]. <\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288374\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288374&theme=lumen&iframe_resize_id=ohm288374&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<div class=\"glossary\"><span class=\"screen-reader-text\" id=\"definition\">definition<\/span><template id=\"term_4838_1526\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_4838_1526\"><div tabindex=\"-1\"><p>an algebraic expression with three terms<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><\/div>","protected":false},"author":15,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":503,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/4838"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/4838\/revisions"}],"predecessor-version":[{"id":4845,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/4838\/revisions\/4845"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/503"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/4838\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=4838"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=4838"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=4838"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=4838"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}