How to Factor a Cubic Polynomial

Step-by-Step Guide

1. 1

   Step 1: Group the polynomial into two sections.

   Grouping the polynomial into two sections will let you attack each section individually.
   
   Say, "We're working with the polynomial." x3 + 3x2
   - 6x
   - 18 =
   0.
   
   Let's group it into (x3 + 3x2) and (- 6x
   - 18)
2. 2

   Step 2: Find what's the common in each section.

   Looking at (x3 + 3x2), we can see that x2 is common.
   
   Looking at (- 6x
   - 18), we can see that
   -6 is common. , 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). , This gives you (x + 3)(x2
   - 6). , If you have an x2 in your roots, remember that both negative and positive numbers fulfill that equation.
   
   The solutions are
   -3, √6 and
   -√6. , Let's say you're working with the equation: x3
   - 4x2
   - 7x + 10 =
   0. , 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. , 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. , 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. , "(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. , 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). , 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.
3. 3

   Step 3: Factor the commonalities out of the two terms.

4. 4

   Step 4: If each of the two terms contains the same factor

5. 5

   Step 5: you can combine the factors together.

6. 6

   Step 6: Find the solution by looking at the roots.

7. 7

   Step 7: Rearrange the expression so it's in the form of aX3+bX2+cX+d.

8. 8

   Step 8: Find the all of the factors of "d".

9. 9

   Step 9: Find one factor that causes the polynomial to equal to zero.

10. 10

    Step 10: Do a little rearranging.

11. 11

    Step 11: Factor your root out of the rest of the equation.

12. 12

    Step 12: Continue to substitute by the factors of the free term.

13. 13

    Step 13: Your solutions will be the factored roots.

Detailed Guide

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