# How do you simplify 5(x+2)(x-3)-2(x^2+3x-4)?

Sep 30, 2017

See a solution process below:

#### Explanation:

First, expand the terms in the two sets of parenthesis on the left by multiplying each term in the parenthesis on the left by each term in the parenthesis on the right:

$5 \left(\textcolor{red}{x} + \textcolor{red}{2}\right) \left(\textcolor{b l u e}{x} - \textcolor{b l u e}{3}\right) - 2 \left({x}^{2} + 3 x - 4\right) \implies$

$5 \left[\left(\textcolor{red}{x} \times \textcolor{b l u e}{x}\right) - \left(\textcolor{red}{x} \times \textcolor{b l u e}{3}\right) + \left(\textcolor{red}{2} \times \textcolor{b l u e}{x}\right) - \left(\textcolor{red}{2} \times \textcolor{b l u e}{3}\right)\right] - 2 \left({x}^{2} + 3 x - 4\right) \implies$

$5 \left({x}^{2} - 3 x + 2 x - 6\right) - 2 \left({x}^{2} + 3 x - 4\right) \implies$

$5 \left({x}^{2} + \left[- 3 + 2\right] x - 6\right) - 2 \left({x}^{2} + 3 x - 4\right) \implies$

$5 \left({x}^{2} + \left[- 1\right] x - 6\right) - 2 \left({x}^{2} + 3 x - 4\right) \implies$

$5 \left({x}^{2} - 1 x - 6\right) - 2 \left({x}^{2} + 3 x - 4\right) \implies$

$5 \left({x}^{2} - x - 6\right) - 2 \left({x}^{2} + 3 x - 4\right)$

Next, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{5} \left({x}^{2} - x - 6\right) - \textcolor{b l u e}{2} \left({x}^{2} + 3 x - 4\right) \implies$

$\left(\textcolor{red}{5} \times {x}^{2}\right) - \left(\textcolor{red}{5} \times x\right) - \left(\textcolor{red}{5} \times 6\right) - \left(\textcolor{b l u e}{2} \times {x}^{2}\right) - \left(\textcolor{b l u e}{2} \times 3 x\right) + \left(\textcolor{b l u e}{2} \times 4\right) \implies$

$5 {x}^{2} - 5 x - 30 - 2 {x}^{2} - 6 x + 8$

Now, group and combine like terms:

$5 {x}^{2} - 2 {x}^{2} - 5 x - 6 x - 30 + 8 \implies$

$\left(5 - 2\right) {x}^{2} + \left(- 5 - 6\right) x + \left(- 30 + 8\right) \implies$

$3 {x}^{2} + \left(- 11\right) x + \left(- 22\right) \implies$

$3 {x}^{2} - 11 x - 22$