How do I factorise the following expression?

Question

How do I go from the left expression to the right one?

$$ (2-x)^2 \cdot (-2-x) - (-2-x) = - (x+2)(x-3)(x-1) $$

I'm guessing that I have to solve the third degree equation. What are the steps for that?

Answer 1

You don't need to solve the third degree equation.$$\begin{align}(2-x)^2\color{red}{(-2-x)}-\color{red}{(-2-x)}&=\color{red}{(-2-x)}((2-x)^2-1)\\&=-(2+x)((x-2)^2-1^2)\\&=-(x+2)(x-2-1)(x-2+1)\\&=-(x+2)(x-3)(x-1)\end{align}$$

Answer 2

There is a trick. Note that $(-2-x) = -(x+2)$ is a factor of both summands so $$\begin{align*} (2-x)^2 \cdot (-2-x) - (-2-x) & = -(x+2)\cdot((2-x)^2 - 1) \\ & = -(x+2)(x^2 - 4x + 4 - 1) \\ & = -(x+2)(x^2-4x+3) \\ & = -(x+2)(x-3)(x-1) \end{align*}$$ Where we only had to factor the quadratic $x^2-4x+3$

Answer 3

Hint: $$a^2-b^2=(a-b)(a+b)$$ Apply this to $$ (2-x)^2 \cdot (-2-x) - (-2-x) = - (x+2)((2-x)^2-1^2) $$