Factoring with the grouping method

Question

I'm trying to factor out using the grouping method the following polynomial: $$ a(a+6)-(a+6)+a(a-4)-(a-4). $$

The solution on the book is $2(x+1)(x-1)$.

Can someone explain the solution to me?

Answer 1

\begin{align*} a(a+6)-(a+6)+a(a-4)-(a-4)&=(a-1)(a+6)+(a-1)(a-4)\\ &=(a-1)[(a+6)+(a-4)]\\ &=(a-1)(2a+2)\\ &=2(a+1)(a-1) \end{align*}

Answer 2

You see a sum of four terms; two terms contain a factor $(a+6)$ ad two terms contain a factor $(a-4)$. We can group those two pairs as follows: $$a(a+6)-(a+6)+a(a-4)-(a-4)=(a-1)(a+6)+(a-1)(a-4).$$ Now we have a sum of two terms, and both contain a factor $(a-1)$. We can group them as follows: $$(a-1)(a+6)+(a-1)(a-4)=(a-1)(a+6+a-4)=(a-1)(2a+2).$$ This last factor can be written as $2a+2=2(a+1)$, and so rearranging the factors a bit we get $$(a-1)(2a+2)=2(a-1)(a+1).$$