Factorization of cubic polynamials


#1

How can we factorize a cubic polynomial?


#2

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
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.
3

Find one factor that causes the polynomial to equal to zero. 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.
4
Do a little rearranging. 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.
5

Factor your root out of the rest 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.
6
Continue to substitute by the factors of the free term. 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).
7
ur solutions will be the factored roots. 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.