Say we’re working with the polynomial x3 + 3x2 - 6x - 18 = 0. Let’s group it into (x3 + 3x2) and (- 6x - 18)

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).

The solutions are -3, √6 and -√6.

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

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.

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.

“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.

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.

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).

(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.