Can someone please explain how this was factored?

Question

How was $x^2(x+1)-4(x+1)$ factored into $$(x^2-4)(x+1)?$$

I know this seems very basic but can someone please explain this?

Answer 1

Notice that in the original equation, $$ x^2(x+1)-4(x+1) $$ both the first and second term contain a $x+1$. So let's pull out the $x+1$. So we have $$ (x+1)(x^2-4) $$ Moreover, notice the term on the right is a difference of perfect squares so the fully factored form is $$ (x+1)(x+2)(x-2) $$

Answer 2

From the distributive law, $$(b + c)a = ba + ca$$ Choose $b = x^2$, $c = -4$ and $a = x + 1$.

Answer 3

Consider the general case: $$ u \cdot y + v \cdot y. $$ We go backwards. We have from the distributive law that $$ (u+v) \cdot y = u \cdot y + v \cdot y, $$ by definition. Now, if $ u = x^2 $ and $ v = -4 $, you get the desired result.