{"id":2944,"date":"2024-02-09T20:31:05","date_gmt":"2024-02-09T20:31:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&#038;p=2944"},"modified":"2026-03-24T19:21:06","modified_gmt":"2026-03-24T19:21:06","slug":"9-7-solving-polynomial-equations-using-the-zero-product-principle","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/9-7-solving-polynomial-equations-using-the-zero-product-principle\/","title":{"raw":"9.7: Solving Polynomial Equations Using the Zero Product Principle","rendered":"9.7: Solving Polynomial Equations Using the Zero Product Principle"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h1>Learning Outcomes<\/h1>\r\n<ul>\r\n \t<li>Use the zero product\u00a0<span style=\"font-size: 14.4px;\">principle <\/span>to solve polynomial equations<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h1>Key words<\/h1>\r\n<ul>\r\n \t<li><strong>Zero product principle<\/strong>: if two or more factors are multiplied to a product of zero, at least one of the factors is zero<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Zero Product Principle<\/h2>\r\nIf we multiply two numbers together and get an answer of zero, what can we say about the two numbers? The only way to get a product of zero is if we multiply by [latex]0[\/latex]. This means that one of the factors\u00a0<em>has\u00a0<\/em>to be zero. This idea is called the <em><strong>zero product principle<\/strong><\/em>, and it is useful for solving certain kinds of equations.\r\n<div class=\"textbox shaded\">\r\n<h3>Zero Product PRINCIPLE<span style=\"font-size: 1.2em;\">\u00a0<\/span><\/h3>\r\nThe Zero Product Principle <span style=\"font-size: 1rem; text-align: initial;\">states that if the product of two or more factors is [latex]0[\/latex], then at least one of the factors must be [latex]0[\/latex].\u00a0<\/span>\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">If [latex]ab=0[\/latex], then either [latex]a=0[\/latex] or [latex]b=0[\/latex], or [latex]\\text{both } a\\text{ and }b=0[\/latex].<\/span>\r\n\r\n<\/div>\r\nWhen we say\u00a0<span style=\"font-size: 1rem; text-align: initial;\">[latex]a=0[\/latex] or [latex]b=0[\/latex],\u00a0[latex]\\text{both } a\\text{ and }b=0[\/latex] is implied. i.e. at least one of the factors must equal zero.<\/span>\r\n\r\nThe zero product principle can be used to solve factored equations that are equal to zero.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse the zero product principle<span style=\"font-size: 1rem; text-align: initial;\">\u00a0to solve [latex]5y=0[\/latex]<\/span>\r\n<h4>Solution<\/h4>\r\nBy the zero product\u00a0<span style=\"font-size: 1rem; text-align: initial;\">principle, when two factors are multiplied and the result is zero at least one of them is equal to zero. Therefore, either [latex]5=0[\/latex], or [latex]y=0[\/latex].<\/span>\r\n\r\nIn this case, we know that [latex]5[\/latex] is not equal to zero, so [latex]y[\/latex] must be equal to zero.\r\n\r\nWe can verify this with algebra.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}5y=0\\\\\\text{}\\\\\\frac{5y}{5}=\\frac{0}{5}\\\\\\text{}\\\\y=0\\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n[latex]y=0[\/latex]\r\n\r\n<\/div>\r\nWe can extend this idea to products of more than just two factors.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]5x(x-4)(3x+2)=0[\/latex]\r\n<h4>Solution<\/h4>\r\nThere are three factors with a product of zero, so at least one of the factors must equal zero:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}5x(x-4)(3x+2)&amp;=0 \\\\ 5x=0\\text{ or }x-4=0\\text{ or }3x+2&amp;=0 \\\\ x=0\\text{ or }x=4\\text{ or }3x&amp;=-2 \\\\x=0\\text{ or }x=4\\text{ or }x&amp;=-\\frac{2}{3} \\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n<p style=\"text-align: left;\">[latex]x=0\\text{ or }x=4\\text{ or }x=-\\frac{2}{3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n1. Solve the equation: [latex]-7y(3y-2)(4y+1)=0[\/latex]\r\n\r\n[reveal-answer q=\"hjm385\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm385\"][latex]y=0[\/latex] or [latex]y=\\frac{2}{3}[\/latex] or [latex]y=-\\frac{1}{4}[\/latex][\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n2. Solve the equation:\u00a0 [latex]x^2(2x+1)(5x-2)(x+2)=0[\/latex]\r\n\r\n[reveal-answer q=\"hjm717\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm717\"][latex]x=0[\/latex] or [latex]x=-\\frac{1}{2}[\/latex] or [latex]x=\\frac{2}{5}[\/latex] or [latex]x=-2[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nLet's consider the equation [latex]t(5-t)=0[\/latex]. If we set each factor to zero and solve, we get two solutions [latex]t=0[\/latex] or [latex]t=5[\/latex].\r\n\r\nWhy don't we just use the distributive property and the properties of equations to solve this kind of equation? Let's try using the distributive property on this example to explain why this can be problematic.\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}t\\left(5-t\\right)&amp;=0 \\\\ 5t-t^2&amp;=0 \\\\5t&amp;=t^2 \\\\ \\frac{5t}{t}&amp;=\\frac{t^2}{t} \\\\ 5&amp;=t \\end{aligned}\\end{equation}[\/latex]<\/p>\r\nWait, our original solution was\u00a0[latex]t=0\\text{ or }t=5[\/latex]. \u00a0How did we lose one of the solutions?\u00a0 The problem occurred when we divided both sides of the equation by [latex]t[\/latex].\u00a0 Remember, that we can divide both sides of an equation by the same\u00a0<strong>non-zero<\/strong> term. So if we want to divide by\u00a0[latex]t[\/latex], we must ensure that\u00a0 [latex]t\\ne 0[\/latex]. SInce we don't know whether[latex]t=0[\/latex] or not, we cannot divide by [latex]t[\/latex].\u00a0We lost the solution\u00a0<span style=\"font-size: 1em;\">[latex]t=0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">when we divided by [latex]t[\/latex].<\/span>\r\n<p style=\"text-align: left;\">When we are solving polynomial equations, we need to use some different\u00a0methods than we used to solve linear equations to make sure we get <em>all<\/em> of the<em> correct<\/em> answers. \u00a0The zero product principle is one tool that allows us to do this.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n\r\n<img class=\" wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"65\" height=\"57\" \/>Caution! \u00a0It is important to remember that the principle of zero products <strong>only<\/strong> works when we have an equation with zero on one side, and only a product on the other side. The table below gives examples of equations for which we can and cannot apply the principle of zero products.\r\n<table style=\"height: 36px; width: 934px;\">\r\n<thead>\r\n<tr style=\"height: 12px;\">\r\n<th style=\"height: 12px; width: 274.594px;\">YES Zero Product Principle Works to Solve<\/th>\r\n<th style=\"height: 12px; width: 320.688px;\">NO Zero Product Principle Does Not Work to Solve<\/th>\r\n<th style=\"height: 12px; width: 303.688px;\">WHY NOT?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px; width: 274.594px;\">[latex]\\frac{1}{2}\\left(x-2\\right)=0[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 320.688px;\">[latex]\\frac{1}{2}\\left(x-2\\right)=28[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 303.688px;\">There is\u00a0a product on the left, but it is not equal to zero.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px; width: 274.594px;\">[latex]s\\left(9+s\\right)=0[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 320.688px;\">[latex]s^2+9=0[\/latex]<\/td>\r\n<td style=\"height: 12px; width: 303.688px;\">There is a sum equal to zero but no product equal to zero.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nThe following video presents more examples of how to use the zero product principle to solve polynomial equations that are in factored form.\r\n\r\nhttps:\/\/youtu.be\/yCcMCPHFrVc\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.7-1.docx\">Transcript-9.7-1<\/a>\r\n<h2>Solving Polynomial Equations<\/h2>\r\nIn this section we will solve polynomial equations that can be factored using the zero product principle.\r\n\r\nWe will begin with an example where the polynomial is already equal to zero.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve:\u00a0 [latex]-t^2+t=0[\/latex]\r\n<h4>Solution<\/h4>\r\nTo solve this equation, we need to factor the left side. Each term has a common factor of [latex]t[\/latex] and the leading term is negative,\u00a0so we can factor out [latex]-t[\/latex] and\u00a0use the zero product principle:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-t^2+t&amp;=0 \\\\-t(t-1)&amp;=0\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-t(t-1)&amp;=0 \\\\ -t&amp;=0\\text{ OR }t-1=0 \\\\ t&amp;=0\\text{ OR }t=1\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n[latex]t=0\\text{ OR }t=1[\/latex]\r\n\r\n<\/div>\r\nThe following video presents more examples of solving equations by factoring. A polynomial of degree two is often referred to as a\u00a0<em><strong>quadratic<\/strong><\/em>. i.e. a quadratic is any polynomial of the form [latex]ax^2+bx+c[\/latex], where [latex]a, b, c[\/latex] are real numbers.\r\n\r\nhttps:\/\/youtu.be\/Hpb8DVYBDzA\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.7-2.docx\">Transcript-9.7-2<\/a>\r\n\r\nWhen we don't have a zero on one side of the equation, we can subtract those terms from both sides of the equation to force a zero on one side.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]6t=3t^2-12t[\/latex]\r\n<h4>Solution<\/h4>\r\nOur first goal is to try and see if we can use the zero product principle, since that is currently the only tool we know for solving polynomial equations. So, let's move all the terms to one side, leaving zero on the other side.\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}6t&amp;=3t^2-12t \\\\0&amp;=3t^2-12t-6t \\\\ 0&amp;=3t^2-18t\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We now have all the terms on the right side, and zero on the other side. Each term on the right side has a common factor of [latex]3t[\/latex], so we can factor and use the zero product principle:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned} 0&amp;=3t^2-18t \\\\ 0&amp;=3t(t-6)\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Set each factor to zero and solve the equations:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}0&amp;=3t(t-6) \\\\3t&amp;=0\\text{ OR }t-6=0 \\\\ t&amp;=0\\text{ OR }t=6\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n[latex]t=6\\text{ OR }t=0[\/latex]\r\n\r\n<\/div>\r\nThe video that follows provides another example of solving a polynomial equation using the zero product principle and factoring.\r\n\r\nhttps:\/\/youtu.be\/oYytjgbd6Q0\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.7-3.docx\">Transcript-9.7-3<\/a>\r\n\r\nWe will work through one more example that is similar to the ones above, except this example has fractions.\u00a0 If we were asked to factor an <em><strong>expression<\/strong><\/em> containing fractions, we would have no choice but to work with the fractions and pull out a GCF that is a fraction. But, since an <em><strong>equation<\/strong><\/em> has two sides, we can eliminate the fractions by multiplying the whole equation by the least common multiple. Let's do the same example both ways to see the difference.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^2[\/latex]\r\n<h4>Solution<\/h4>\r\nTo work with the fractions, we first find a common denominator, then factor:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}\\frac{1}{2}y&amp;=-4y-\\frac{1}{2}y^2\\\\0&amp;=-\\frac{1}{2}y-4y-\\frac{1}{2}y^2\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nTo combine like terms, we use the common denominator [latex]2[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}0&amp;=-\\frac{1}{2}y-4y-\\frac{1}{2}y^2\\\\\\text{ }\\\\0&amp;=-\\frac{1}{2}y-\\frac{8y}{2}-\\frac{1}{2}y^2\\\\\\text{}\\\\0&amp;=-\\frac{9}{2}y-\\frac{1}{2}y^2\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Find the greatest common factor of the terms of the polynomial:<\/p>\r\n<p style=\"text-align: left;\">Factors of [latex]-\\frac{9}{2}y[\/latex] are [latex]-\\frac{1}{2}\\cdot{3}\\cdot{3}\\cdot{y}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Factors of [latex]-\\frac{1}{2}y^2[\/latex] are [latex]-\\frac{1}{2}\\cdot{y}\\cdot{y}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Both terms have [latex]-\\frac{1}{2}\\text{ and }y[\/latex] in common.<\/p>\r\n<p style=\"text-align: left;\">Rewrite each term as the product of the GCF and the remaining terms:<\/p>\r\n<p style=\"text-align: center;\">[latex]-\\frac{9}{2}y=-\\frac{1}{2}y\\left(3\\cdot{3}\\right)=-\\frac{1}{2}y\\left(9\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-\\frac{1}{2}y^2=-\\frac{1}{2}y\\left(y\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Rewrite the polynomial equation\u00a0using the factored terms in place of the original terms. Remember to pay attention to signs when we factor. Notice that we end up with a sum as a factor because the common factor is a negative number. [latex]\\left(9+y\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}0&amp;=-\\frac{9}{2}y-\\frac{1}{2}y^2\\\\\\text{}\\\\0&amp;=-\\frac{1}{2}y\\left(9\\right)-\\frac{1}{2}y\\left(y\\right)\\\\\\text{}\\\\0&amp;=-\\frac{1}{2}y\\left(9+y\\right)\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Solve the two equations.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-\\frac{1}{2}y&amp;=0\\text{ or }y+9=0\\\\ y&amp;=0\\text{ or }y=-9\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n<p style=\"text-align: left;\">[latex]y=0\\text{ or }y=-9[\/latex]<\/p>\r\n\r\n<\/div>\r\nIn this last example, we used <em>many<\/em> skills to solve one equation. \u00a0Let's summarize them:\r\n<ul>\r\n \t<li>We needed a common denominator to combine the like terms [latex]-4y\\text{ and }-\\frac{1}{2}y[\/latex], after we moved all the terms to one side of the equation<\/li>\r\n \t<li>We found the GCF\u00a0of the terms\u00a0[latex]-\\frac{9}{2}y\\text{ and }-\\frac{1}{2}y^2[\/latex]<\/li>\r\n \t<li>We used the GCF to factor the polynomial [latex]-\\frac{9}{2}y-\\frac{1}{2}y^2[\/latex]<\/li>\r\n \t<li>We used the zero product principle to solve the polynomial equation [latex]0=-\\frac{1}{2}y\\left(9+y\\right)[\/latex]<\/li>\r\n<\/ul>\r\nSometimes solving an equation requires the combination of many algebraic principles and techniques. \u00a0The last facet of solving the polynomial equation [latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^2[\/latex] that we should talk about is negative signs.\r\n\r\nWhat follows is the same example completed by first eliminating the fractions by multiplying the whole equation by the least common multiple.\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nSolve the equation:\u00a0\u00a0[latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^2[\/latex]\r\n<h4>Solution<\/h4>\r\nFirst multiply both sides of the equation by the LCD [latex]2[\/latex]:\u00a0 [latex]2\\cdot\\frac{1}{2}y=2\\cdot \\left (-4y-\\frac{1}{2}y^2\\right )\\\\y=-8y-y^2[\/latex]\r\n\r\nAdd [latex]y^2+8y[\/latex] to both sides of the equation:\u00a0 [latex]y^2+8y+y=0[\/latex]\r\n\r\nCombine like terms:\u00a0 [latex]y^2+9y=0[\/latex]\r\n\r\nFactor:\u00a0 [latex]y(y+9)=0[\/latex]\r\n\r\nSet each factor to zero and solve the equations:\u00a0 [latex]y=0\\text{ or }y+9=0\\\\y=0\\text{ or }y=-9[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex]y=0\\text{ or }y=-9[\/latex]\r\n\r\n<\/div>\r\nWe get the same solution whether we work with the fractions or eliminate the fractions. We get to choose which way we prefer to complete such problems with fractions, but generally speaking, it is more efficient to eliminate the fractions as a first step.\r\n\r\nThe following video presents another example of solving an equation with fractional coefficients using factoring and the zero product principle.\r\n\r\nhttps:\/\/youtu.be\/wm6DJ1bnaJs\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.7-4.docx\">Transcript-9.7-4<\/a>\r\n<div class=\"textbox shaded\"><img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"44\" height=\"39\" \/>If the GCF of a polynomial is\u00a0negative, pay attention to the signs that are left when we factor it from the terms of a polynomial.<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve the equation: 5x^2=4+2x^2\r\n\r\n[reveal-answer q=\"hjm680\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm680\"][latex]x=0\\text{ or }x=-\\frac{4}{3}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nOften, there is no GCF to pull out of an equation.\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nSolve the equation:\u00a0 [latex]3x^2-4x-4=0[\/latex]\r\n<h4>Solution<\/h4>\r\nFactor the left side of the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x^2-4x-4 &amp;=0\\\\ac=-12\\text{ and }b=-4\\text{ so split }-4x\\text{ into }-6x+2x\\\\3x^2-6x+2x-4 &amp;=0\\\\3x(x-2)+2(x-2)&amp;=0\\\\(3x+2)(x-2)&amp;=0\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nSet each factor to zero and solve:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x+2&amp;=0\\text{ or }x-2=0\\\\3x&amp;=-2\\text{ or }x=2\\\\ x &amp;=-\\frac{2}{3}\\text{ or }x=2\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=-\\frac{2}{3}\\text{ or }x-2=0[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nSolve the equation:\u00a0 [latex]9x^4-24x^2=-30x^3[\/latex]\r\n<h4><strong>Solution<\/strong><\/h4>\r\nGet all non-zero terms on the same side of the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}9x^4-24x^2&amp;=15x^3\\\\9x^4+30x^3-24x^2 &amp;=0\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nLook for a GCF in all three terms: [latex]3x^2[\/latex]\r\n\r\nFactor:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x^2\\left (3x^2+10x-8 \\right ) &amp;=0\\;\\;\\;\\;\\;ac=-24\\text{ and }b=10\\text{ so split }10x\\text{ into }12x-2x\\\\3x^2\\left ( 3x^2+12x-2x-8\\right ) &amp;=0\\\\3x^2\\left[ 3x(x+4)-2(x+4)\\right ] &amp;=0\\\\ 3x^2(x+4)(3x-2)&amp;=0\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nSet each factor to zero and solve:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x^2(x+4)(3x-2)&amp;=0\\\\3x^2=0\\text{ or }x+4 &amp;=0\\text{ or }3x-2=0\\\\x=0\\text{ or }x &amp;=-4\\text{ or }3x=2\\\\x=0\\text{ or }x &amp;=-4\\text{ or }x=-\\frac{2}{3}\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=0\\text{ or }x=-4\\text{ or }x=-\\frac{2}{3}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve the equation:\u00a0 [latex]42x^4-49x^3=35x^2[\/latex]\r\n\r\n[reveal-answer q=\"hjm908\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm908\"][latex]7x^2(2x+1)(3x-5)=0[\/latex]\r\n\r\n[latex]x=0\\text{ or }x=-\\frac{1}{2}\\text{ or }x=\\frac{5}{3}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h1>Learning Outcomes<\/h1>\n<ul>\n<li>Use the zero product\u00a0<span style=\"font-size: 14.4px;\">principle <\/span>to solve polynomial equations<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h1>Key words<\/h1>\n<ul>\n<li><strong>Zero product principle<\/strong>: if two or more factors are multiplied to a product of zero, at least one of the factors is zero<\/li>\n<\/ul>\n<\/div>\n<h2>The Zero Product Principle<\/h2>\n<p>If we multiply two numbers together and get an answer of zero, what can we say about the two numbers? The only way to get a product of zero is if we multiply by [latex]0[\/latex]. This means that one of the factors\u00a0<em>has\u00a0<\/em>to be zero. This idea is called the <em><strong>zero product principle<\/strong><\/em>, and it is useful for solving certain kinds of equations.<\/p>\n<div class=\"textbox shaded\">\n<h3>Zero Product PRINCIPLE<span style=\"font-size: 1.2em;\">\u00a0<\/span><\/h3>\n<p>The Zero Product Principle <span style=\"font-size: 1rem; text-align: initial;\">states that if the product of two or more factors is [latex]0[\/latex], then at least one of the factors must be [latex]0[\/latex].\u00a0<\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">If [latex]ab=0[\/latex], then either [latex]a=0[\/latex] or [latex]b=0[\/latex], or [latex]\\text{both } a\\text{ and }b=0[\/latex].<\/span><\/p>\n<\/div>\n<p>When we say\u00a0<span style=\"font-size: 1rem; text-align: initial;\">[latex]a=0[\/latex] or [latex]b=0[\/latex],\u00a0[latex]\\text{both } a\\text{ and }b=0[\/latex] is implied. i.e. at least one of the factors must equal zero.<\/span><\/p>\n<p>The zero product principle can be used to solve factored equations that are equal to zero.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use the zero product principle<span style=\"font-size: 1rem; text-align: initial;\">\u00a0to solve [latex]5y=0[\/latex]<\/span><\/p>\n<h4>Solution<\/h4>\n<p>By the zero product\u00a0<span style=\"font-size: 1rem; text-align: initial;\">principle, when two factors are multiplied and the result is zero at least one of them is equal to zero. Therefore, either [latex]5=0[\/latex], or [latex]y=0[\/latex].<\/span><\/p>\n<p>In this case, we know that [latex]5[\/latex] is not equal to zero, so [latex]y[\/latex] must be equal to zero.<\/p>\n<p>We can verify this with algebra.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}5y=0\\\\\\text{}\\\\\\frac{5y}{5}=\\frac{0}{5}\\\\\\text{}\\\\y=0\\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p>[latex]y=0[\/latex]<\/p>\n<\/div>\n<p>We can extend this idea to products of more than just two factors.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]5x(x-4)(3x+2)=0[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>There are three factors with a product of zero, so at least one of the factors must equal zero:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}5x(x-4)(3x+2)&=0 \\\\ 5x=0\\text{ or }x-4=0\\text{ or }3x+2&=0 \\\\ x=0\\text{ or }x=4\\text{ or }3x&=-2 \\\\x=0\\text{ or }x=4\\text{ or }x&=-\\frac{2}{3} \\end{aligned}\\end{equation}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p style=\"text-align: left;\">[latex]x=0\\text{ or }x=4\\text{ or }x=-\\frac{2}{3}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>1. Solve the equation: [latex]-7y(3y-2)(4y+1)=0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm385\">Show Answer<\/span><\/p>\n<div id=\"qhjm385\" class=\"hidden-answer\" style=\"display: none\">[latex]y=0[\/latex] or [latex]y=\\frac{2}{3}[\/latex] or [latex]y=-\\frac{1}{4}[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>2. Solve the equation:\u00a0 [latex]x^2(2x+1)(5x-2)(x+2)=0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm717\">Show Answer<\/span><\/p>\n<div id=\"qhjm717\" class=\"hidden-answer\" style=\"display: none\">[latex]x=0[\/latex] or [latex]x=-\\frac{1}{2}[\/latex] or [latex]x=\\frac{2}{5}[\/latex] or [latex]x=-2[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Let&#8217;s consider the equation [latex]t(5-t)=0[\/latex]. If we set each factor to zero and solve, we get two solutions [latex]t=0[\/latex] or [latex]t=5[\/latex].<\/p>\n<p>Why don&#8217;t we just use the distributive property and the properties of equations to solve this kind of equation? Let&#8217;s try using the distributive property on this example to explain why this can be problematic.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}t\\left(5-t\\right)&=0 \\\\ 5t-t^2&=0 \\\\5t&=t^2 \\\\ \\frac{5t}{t}&=\\frac{t^2}{t} \\\\ 5&=t \\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Wait, our original solution was\u00a0[latex]t=0\\text{ or }t=5[\/latex]. \u00a0How did we lose one of the solutions?\u00a0 The problem occurred when we divided both sides of the equation by [latex]t[\/latex].\u00a0 Remember, that we can divide both sides of an equation by the same\u00a0<strong>non-zero<\/strong> term. So if we want to divide by\u00a0[latex]t[\/latex], we must ensure that\u00a0 [latex]t\\ne 0[\/latex]. SInce we don&#8217;t know whether[latex]t=0[\/latex] or not, we cannot divide by [latex]t[\/latex].\u00a0We lost the solution\u00a0<span style=\"font-size: 1em;\">[latex]t=0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">when we divided by [latex]t[\/latex].<\/span><\/p>\n<p style=\"text-align: left;\">When we are solving polynomial equations, we need to use some different\u00a0methods than we used to solve linear equations to make sure we get <em>all<\/em> of the<em> correct<\/em> answers. \u00a0The zero product principle is one tool that allows us to do this.<\/p>\n<div class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"65\" height=\"57\" \/>Caution! \u00a0It is important to remember that the principle of zero products <strong>only<\/strong> works when we have an equation with zero on one side, and only a product on the other side. The table below gives examples of equations for which we can and cannot apply the principle of zero products.<\/p>\n<table style=\"height: 36px; width: 934px;\">\n<thead>\n<tr style=\"height: 12px;\">\n<th style=\"height: 12px; width: 274.594px;\">YES Zero Product Principle Works to Solve<\/th>\n<th style=\"height: 12px; width: 320.688px;\">NO Zero Product Principle Does Not Work to Solve<\/th>\n<th style=\"height: 12px; width: 303.688px;\">WHY NOT?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px; width: 274.594px;\">[latex]\\frac{1}{2}\\left(x-2\\right)=0[\/latex]<\/td>\n<td style=\"height: 12px; width: 320.688px;\">[latex]\\frac{1}{2}\\left(x-2\\right)=28[\/latex]<\/td>\n<td style=\"height: 12px; width: 303.688px;\">There is\u00a0a product on the left, but it is not equal to zero.<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px; width: 274.594px;\">[latex]s\\left(9+s\\right)=0[\/latex]<\/td>\n<td style=\"height: 12px; width: 320.688px;\">[latex]s^2+9=0[\/latex]<\/td>\n<td style=\"height: 12px; width: 303.688px;\">There is a sum equal to zero but no product equal to zero.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>The following video presents more examples of how to use the zero product principle to solve polynomial equations that are in factored form.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Solve an Equation in Factored Form\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/yCcMCPHFrVc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.7-1.docx\">Transcript-9.7-1<\/a><\/p>\n<h2>Solving Polynomial Equations<\/h2>\n<p>In this section we will solve polynomial equations that can be factored using the zero product principle.<\/p>\n<p>We will begin with an example where the polynomial is already equal to zero.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve:\u00a0 [latex]-t^2+t=0[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>To solve this equation, we need to factor the left side. Each term has a common factor of [latex]t[\/latex] and the leading term is negative,\u00a0so we can factor out [latex]-t[\/latex] and\u00a0use the zero product principle:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-t^2+t&=0 \\\\-t(t-1)&=0\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-t(t-1)&=0 \\\\ -t&=0\\text{ OR }t-1=0 \\\\ t&=0\\text{ OR }t=1\\end{aligned}\\end{equation}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p>[latex]t=0\\text{ OR }t=1[\/latex]<\/p>\n<\/div>\n<p>The following video presents more examples of solving equations by factoring. A polynomial of degree two is often referred to as a\u00a0<em><strong>quadratic<\/strong><\/em>. i.e. a quadratic is any polynomial of the form [latex]ax^2+bx+c[\/latex], where [latex]a, b, c[\/latex] are real numbers.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 1:  Factor and Solve a Quadratic Equation - GCF\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Hpb8DVYBDzA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.7-2.docx\">Transcript-9.7-2<\/a><\/p>\n<p>When we don&#8217;t have a zero on one side of the equation, we can subtract those terms from both sides of the equation to force a zero on one side.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]6t=3t^2-12t[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Our first goal is to try and see if we can use the zero product principle, since that is currently the only tool we know for solving polynomial equations. So, let&#8217;s move all the terms to one side, leaving zero on the other side.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}6t&=3t^2-12t \\\\0&=3t^2-12t-6t \\\\ 0&=3t^2-18t\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p style=\"text-align: left;\">We now have all the terms on the right side, and zero on the other side. Each term on the right side has a common factor of [latex]3t[\/latex], so we can factor and use the zero product principle:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned} 0&=3t^2-18t \\\\ 0&=3t(t-6)\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p style=\"text-align: left;\">Set each factor to zero and solve the equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}0&=3t(t-6) \\\\3t&=0\\text{ OR }t-6=0 \\\\ t&=0\\text{ OR }t=6\\end{aligned}\\end{equation}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p>[latex]t=6\\text{ OR }t=0[\/latex]<\/p>\n<\/div>\n<p>The video that follows provides another example of solving a polynomial equation using the zero product principle and factoring.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Factor and Solve a Quadratic Equation - GCF Only\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/oYytjgbd6Q0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.7-3.docx\">Transcript-9.7-3<\/a><\/p>\n<p>We will work through one more example that is similar to the ones above, except this example has fractions.\u00a0 If we were asked to factor an <em><strong>expression<\/strong><\/em> containing fractions, we would have no choice but to work with the fractions and pull out a GCF that is a fraction. But, since an <em><strong>equation<\/strong><\/em> has two sides, we can eliminate the fractions by multiplying the whole equation by the least common multiple. Let&#8217;s do the same example both ways to see the difference.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^2[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>To work with the fractions, we first find a common denominator, then factor:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}\\frac{1}{2}y&=-4y-\\frac{1}{2}y^2\\\\0&=-\\frac{1}{2}y-4y-\\frac{1}{2}y^2\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>To combine like terms, we use the common denominator [latex]2[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}0&=-\\frac{1}{2}y-4y-\\frac{1}{2}y^2\\\\\\text{ }\\\\0&=-\\frac{1}{2}y-\\frac{8y}{2}-\\frac{1}{2}y^2\\\\\\text{}\\\\0&=-\\frac{9}{2}y-\\frac{1}{2}y^2\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p style=\"text-align: left;\">Find the greatest common factor of the terms of the polynomial:<\/p>\n<p style=\"text-align: left;\">Factors of [latex]-\\frac{9}{2}y[\/latex] are [latex]-\\frac{1}{2}\\cdot{3}\\cdot{3}\\cdot{y}[\/latex]<\/p>\n<p style=\"text-align: left;\">Factors of [latex]-\\frac{1}{2}y^2[\/latex] are [latex]-\\frac{1}{2}\\cdot{y}\\cdot{y}[\/latex]<\/p>\n<p style=\"text-align: left;\">Both terms have [latex]-\\frac{1}{2}\\text{ and }y[\/latex] in common.<\/p>\n<p style=\"text-align: left;\">Rewrite each term as the product of the GCF and the remaining terms:<\/p>\n<p style=\"text-align: center;\">[latex]-\\frac{9}{2}y=-\\frac{1}{2}y\\left(3\\cdot{3}\\right)=-\\frac{1}{2}y\\left(9\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-\\frac{1}{2}y^2=-\\frac{1}{2}y\\left(y\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">Rewrite the polynomial equation\u00a0using the factored terms in place of the original terms. Remember to pay attention to signs when we factor. Notice that we end up with a sum as a factor because the common factor is a negative number. [latex]\\left(9+y\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}0&=-\\frac{9}{2}y-\\frac{1}{2}y^2\\\\\\text{}\\\\0&=-\\frac{1}{2}y\\left(9\\right)-\\frac{1}{2}y\\left(y\\right)\\\\\\text{}\\\\0&=-\\frac{1}{2}y\\left(9+y\\right)\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p style=\"text-align: left;\">Solve the two equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-\\frac{1}{2}y&=0\\text{ or }y+9=0\\\\ y&=0\\text{ or }y=-9\\end{aligned}\\end{equation}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p style=\"text-align: left;\">[latex]y=0\\text{ or }y=-9[\/latex]<\/p>\n<\/div>\n<p>In this last example, we used <em>many<\/em> skills to solve one equation. \u00a0Let&#8217;s summarize them:<\/p>\n<ul>\n<li>We needed a common denominator to combine the like terms [latex]-4y\\text{ and }-\\frac{1}{2}y[\/latex], after we moved all the terms to one side of the equation<\/li>\n<li>We found the GCF\u00a0of the terms\u00a0[latex]-\\frac{9}{2}y\\text{ and }-\\frac{1}{2}y^2[\/latex]<\/li>\n<li>We used the GCF to factor the polynomial [latex]-\\frac{9}{2}y-\\frac{1}{2}y^2[\/latex]<\/li>\n<li>We used the zero product principle to solve the polynomial equation [latex]0=-\\frac{1}{2}y\\left(9+y\\right)[\/latex]<\/li>\n<\/ul>\n<p>Sometimes solving an equation requires the combination of many algebraic principles and techniques. \u00a0The last facet of solving the polynomial equation [latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^2[\/latex] that we should talk about is negative signs.<\/p>\n<p>What follows is the same example completed by first eliminating the fractions by multiplying the whole equation by the least common multiple.<\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Solve the equation:\u00a0\u00a0[latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^2[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>First multiply both sides of the equation by the LCD [latex]2[\/latex]:\u00a0 [latex]2\\cdot\\frac{1}{2}y=2\\cdot \\left (-4y-\\frac{1}{2}y^2\\right )\\\\y=-8y-y^2[\/latex]<\/p>\n<p>Add [latex]y^2+8y[\/latex] to both sides of the equation:\u00a0 [latex]y^2+8y+y=0[\/latex]<\/p>\n<p>Combine like terms:\u00a0 [latex]y^2+9y=0[\/latex]<\/p>\n<p>Factor:\u00a0 [latex]y(y+9)=0[\/latex]<\/p>\n<p>Set each factor to zero and solve the equations:\u00a0 [latex]y=0\\text{ or }y+9=0\\\\y=0\\text{ or }y=-9[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=0\\text{ or }y=-9[\/latex]<\/p>\n<\/div>\n<p>We get the same solution whether we work with the fractions or eliminate the fractions. We get to choose which way we prefer to complete such problems with fractions, but generally speaking, it is more efficient to eliminate the fractions as a first step.<\/p>\n<p>The following video presents another example of solving an equation with fractional coefficients using factoring and the zero product principle.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Factor and Solve a Quadratic Equation with Fractions - GCF Only\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/wm6DJ1bnaJs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.7-4.docx\">Transcript-9.7-4<\/a><\/p>\n<div class=\"textbox shaded\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"44\" height=\"39\" \/>If the GCF of a polynomial is\u00a0negative, pay attention to the signs that are left when we factor it from the terms of a polynomial.<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve the equation: 5x^2=4+2x^2<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm680\">Show Answer<\/span><\/p>\n<div id=\"qhjm680\" class=\"hidden-answer\" style=\"display: none\">[latex]x=0\\text{ or }x=-\\frac{4}{3}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Often, there is no GCF to pull out of an equation.<\/p>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Solve the equation:\u00a0 [latex]3x^2-4x-4=0[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Factor the left side of the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x^2-4x-4 &=0\\\\ac=-12\\text{ and }b=-4\\text{ so split }-4x\\text{ into }-6x+2x\\\\3x^2-6x+2x-4 &=0\\\\3x(x-2)+2(x-2)&=0\\\\(3x+2)(x-2)&=0\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Set each factor to zero and solve:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x+2&=0\\text{ or }x-2=0\\\\3x&=-2\\text{ or }x=2\\\\ x &=-\\frac{2}{3}\\text{ or }x=2\\end{aligned}\\end{equation}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=-\\frac{2}{3}\\text{ or }x-2=0[\/latex]<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Solve the equation:\u00a0 [latex]9x^4-24x^2=-30x^3[\/latex]<\/p>\n<h4><strong>Solution<\/strong><\/h4>\n<p>Get all non-zero terms on the same side of the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}9x^4-24x^2&=15x^3\\\\9x^4+30x^3-24x^2 &=0\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Look for a GCF in all three terms: [latex]3x^2[\/latex]<\/p>\n<p>Factor:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x^2\\left (3x^2+10x-8 \\right ) &=0\\;\\;\\;\\;\\;ac=-24\\text{ and }b=10\\text{ so split }10x\\text{ into }12x-2x\\\\3x^2\\left ( 3x^2+12x-2x-8\\right ) &=0\\\\3x^2\\left[ 3x(x+4)-2(x+4)\\right ] &=0\\\\ 3x^2(x+4)(3x-2)&=0\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Set each factor to zero and solve:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x^2(x+4)(3x-2)&=0\\\\3x^2=0\\text{ or }x+4 &=0\\text{ or }3x-2=0\\\\x=0\\text{ or }x &=-4\\text{ or }3x=2\\\\x=0\\text{ or }x &=-4\\text{ or }x=-\\frac{2}{3}\\end{aligned}\\end{equation}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=0\\text{ or }x=-4\\text{ or }x=-\\frac{2}{3}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve the equation:\u00a0 [latex]42x^4-49x^3=35x^2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm908\">Show Answer<\/span><\/p>\n<div id=\"qhjm908\" class=\"hidden-answer\" style=\"display: none\">[latex]7x^2(2x+1)(3x-5)=0[\/latex]<\/p>\n<p>[latex]x=0\\text{ or }x=-\\frac{1}{2}\\text{ or 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