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

2 Answers
May 13, 2018

Answer:

#x=-7,5#

Explanation:

Solve:

#(x+3)(x-1)=32#

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

#x^2+2x-3=32#

Move all terms to the left-hand side.

#x^2+2x-3-32=0#

#x^2+2x-35=0#

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

#(x+7)(x-5)+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+2x-3=32#
Then you put the 32 on the first class just like:
#x^2+2x-35=0#
Then you do #D=2^2-4*(-35) = 144# (Using the method #D = b^2-4ac#)
so your one root is #x_1=(b-sqrt(D))/(2a)#
and the other one is #x_2=(b+sqrt(D))/(2a)#
Do the maths and you get #x_1=-7# , #x_2=5#