Question #e2e13

Jun 8, 2017

${x}^{2} - 3 x - 28 = \left(x + 4\right) \left(x - 7\right) \mathmr{and} \left(x - 7\right) \left(x + 4\right)$ but they're both the same due to the Commutative Property of Multiplication.

Explanation:

To solve this problem, we need to identify an $a$ and a $b$ such that $a b = - 28$ and $a + b = - 3$. Let's first list out the factors of 28 and see which ones add up to -3 (Ones that don't work = red; one that works = blue):

$\textcolor{red}{1 \mathmr{and} 28}$

$\textcolor{red}{2 \mathmr{and} 14}$

$\textcolor{b l u e}{4 \mathmr{and} 7}$

We have found our pair! -4 and 7! So now, we need to write the original problem into a $\left(x + a\right) \left(x + b\right)$ format. We have two ways of writing this:

Either $\left(x + 4\right) \left(x - 7\right)$ or $\left(x - 7\right) \left(x + 4\right)$ but they're both the same due to the Commutative Property of Multiplication.