How to Factor a Cubic Polynomial


This is an article about how to factorize a 3rd degree polynomial. We will explore how to factor using grouping as well as using the factors of the free term.

Part 1
Factoring By Grouping

Factoring By Grouping on How to Factor a Cubic Polynomial

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Grouping the polynomial into two sections will let you attack each section individually.[1] Say we're working with the polynomial x3 + 3x2 - 6x - 18 = 0. Let's group it into (x3 + 3x2) and (- 6x - 18)

Factoring By Grouping on How to Factor a Cubic Polynomial

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Looking at (x3 + 3x2), we can see that x2 is common. Looking at (- 6x - 18), we can see that -6 is common.

Factoring By Grouping on How to Factor a Cubic Polynomial

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Factoring out x2 from the first section, we get x2(x + 3). Factoring out -6 from the second section, you'll get -6(x + 3).

Factoring By Grouping on How to Factor a Cubic Polynomial

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[2] This gives you (x + 3)(x2 - 6).

Factoring By Grouping on How to Factor a Cubic Polynomial

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If you have an x2 in your roots, remember that both negative and positive numbers fulfill that equation.[3] The solutions are -3, √6 and -√6.

Part 2
Factoring Using the Free Term

Factoring Using the Free Term on How to Factor a Cubic Polynomial

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[4] Let's say you're working with the equation: x3 - 4x2 - 7x + 10 = 0.

Factoring Using the Free Term on How to Factor a Cubic Polynomial

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The constant "d" is going to be the number that doesn't have any variables, such as "x," next to it. Factors are the numbers you can multiply together to get another number. In your case, the factors of 10, or "d," are: 1, 2, 5, and 10.

Factoring Using the Free Term on How to Factor a Cubic Polynomial

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We want to determine which factor makes the polynomial equal zero when we substitute the factor for each "x" in the equation. Start by using your first factor, 1. Substitute "1" for each "x" in the equation: (1)3 - 4(1)2 - 7(1) + 10 = 0 This gives you: 1 - 4 - 7 + 10 = 0. Because 0 = 0 is a true statement, you know that x = 1 is a solution.

Factoring Using the Free Term on How to Factor a Cubic Polynomial

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If x = 1, you can rearrange the statement to look a bit different without changing what it means. "x = 1" is the same thing as "x - 1 = 0" or "(x - 1)". You've just subtracted a "1" from each side of the equation.

Factoring Using the Free Term on How to Factor a Cubic Polynomial

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"(x - 1)" is our root. See if you can factor it out of the rest of the equation. Take it one polynomial at a time. Can you factor (x - 1) out of the x3? No you can't. But you can borrow a -x2 from the second variable; then factor it: x2(x - 1) = x3 - x2. Can you factor (x - 1) out of what remains from your second variable? No, again you can't. You need to borrow another little bit from the third variable. You need to borrow a 3x from -7x. This gives you -3x(x - 1) = -3x2 + 3x. Since you took a 3x from -7x, our third variable is now -10x and our constant is 10. Can you factor this? You can! -10(x - 1) = -10x + 10. What you did was rearrange the variables so that you could factor out a (x - 1) out of the entire equation. Your rearranged equation looks like this: x3 - x2 - 3x2 + 3x - 10x + 10 = 0, but it's still the same thing as x3 - 4x2 - 7x + 10 = 0.

Factoring Using the Free Term on How to Factor a Cubic Polynomial

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Look at the numbers that you factored out using the (x - 1) in Step 5: x2(x - 1) - 3x(x - 1) - 10(x - 1) = 0. You can rearrange this to be a lot easier to factor one more time: (x - 1)(x2 - 3x - 10) = 0. You're only trying to factor (x2 - 3x - 10) here. This factors down into (x + 2)(x - 5).

Factoring Using the Free Term on How to Factor a Cubic Polynomial

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You can check whether your solutions actually work by plugging each one, individually, back into the original equation. (x - 1)(x + 2)(x - 5) = 0 This gives you solutions of 1, -2, and 5. Plug -2 back into the equation: (-2)3 - 4(-2)2 - 7(-2) + 10 = -8 - 16 + 14 + 10 = 0. Plug 5 back into the equation: (5)3 - 4(5)2 - 7(5) + 10 = 125 - 100 - 35 + 10 = 0.