How to factor this degree 3 polynomial.

Can someone explain how to factor this: x^3 - 3x^2 - 9x +27 ?
I can easily factor any ax^2 + bx + c polynomial, but when it's a degree 3, I get confused.

First find multiples of the last number that will make this equation = 0.
Factors of 27 are +-1,+-3,+-9,+-27

f(x) = x^3 - 3x^2 - 9x +27
f(3) = 3^3 - 3(3)^2 - 9(3) + 27
f(3) = 27 - 27 - 27 + 27
f(3) = 0

Since f(3) = 0, (x - 3) is a factor. Now use synthetic division since its faster to find the next factors.

3 l 1 -3 -9 27
3 l 0 3 0 -27
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3 l 1 0 -9 0

Therefore, (x^2 - 9) is a factor.

f(x) = (x^2 - 9)(x-3)
f(x) = (x+3)(x-3)(x-3)
f(x) = (x-3)^2(x+3) <---Fully factorized.

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sorry didn't realize you can factor by grouping also, it's way easier.

f(x) = x^3 - 3x^2 - 9x +27
f(x) = [x^3 - 3x^2] - [9x + 27] <--Take a common factor out of each brackets.
f(x) = x^2(x - 3) - 9 (x - 3)
f(x) = (x^2 - 9)(x-3) <-- Difference of squares.
f(x) = (x+3)(x-3)(x-3)
f(x) = (x-3)^2(x+3)

OK. When confronted with a 3rd degree polynomial the first option is to look at "factoring by grouping". If that doesn't work then you have to resort to rational roots theorem and Descartes rule of signs...tedious.

So, look at what you've got. group the x^3 -9x together and -3x^2+27 together and let's see what happens. In the first group you can factor x out to give x(x^2-9). In the second you can factor out a -3 to give -3(x^2-9). Looking at the terms in ( ) you can see that (x^2-9) is a common factor so you can reduce to (x^2-9)*(x-3). Now, x^2-9 has two factors...(x-3)*(x+3)...so in the end you have
(x+3)*(x-3)^2 as your answer.

x^3 - 3x^2 - 9x +27
=x^2(x-3)-9(x-3)
=(x-3(x^2-9)
=(x-3)(x-3)(x+3)
=(x-3)^2(x+3)

x^3 - 3x^2 - 9x +27 =x^2(x-3)-9(x-3)=(x-3)(x^-9)=(x-3)^2(x+3…