# How do you solve (x + 3) ( x - 1) = 32?

May 13, 2018

$x = - 7 , 5$

#### Explanation:

Solve:

$\left(x + 3\right) \left(x - 1\right) = 32$

FOIL the left-hand side.
https://www.mathsisfun.com/definitions/foil-method.html

${x}^{2} + 2 x - 3 = 32$

Move all terms to the left-hand side.

${x}^{2} + 2 x - 3 - 32 = 0$

${x}^{2} + 2 x - 35 = 0$

Find two numbers that when added equal $2$ and when multiplied equal $- 35$. The numbers $7$ and $- 5$ meet the criteria.

$\left(x + 7\right) \left(x - 5\right) + 0$

$x + 7 = 0$

$x = - 7$

$x - 5 = 0$

$x = 5$

$x = - 7 , 5$

First of all, you do the multiplications, so you get:
${x}^{2} + 2 x - 3 = 32$
Then you put the 32 on the first class just like:
${x}^{2} + 2 x - 35 = 0$
Then you do $D = {2}^{2} - 4 \cdot \left(- 35\right) = 144$ (Using the method $D = {b}^{2} - 4 a c$)
so your one root is ${x}_{1} = \frac{b - \sqrt{D}}{2 a}$
and the other one is ${x}_{2} = \frac{b + \sqrt{D}}{2 a}$
Do the maths and you get ${x}_{1} = - 7$ , ${x}_{2} = 5$