How do you solve #4x^2+28x-32=0# by factoring?

2 Answers

Answer:

Refer to explanation

Explanation:

We can write it as follows

#4x^2+28x-32=0=>4*(x^2+7x-8)=0=>x^2+7x-8=0=>x^2+8x-x-8=0= x^2-x+8(x-1)=0=>x(x-1)+8(x-1)=0=>(x-1)*(x+8)=0=> x=1 or x=-8#

Sep 30, 2015

Answer:

Solve #f(x) = 4x^2 + 28x - 32 = 0#

Ans: 1 and -8

Explanation:

#f(x) = 4y = 4(x^2 + 7x - 8) = 0#. Solve y = 0.
Because (a + b + c = 0), use Shortcut. The 2 real roost are: x = 1 and #x = c/a = -8#

REMINDER of the SHORTCUT.
1. When (a + b + c = 0), the 2 real roots are: x = 1 and #x = c/a#
2. When (a - b + c = 0), the 2 real roots are x = - 1 and #x = - c/a#