SOLUTION: How do you factor 5x^2+4x-96=0?

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Question 859522: How do you factor 5x^2+4x-96=0?
Found 2 solutions by Fombitz, rothauserc:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Complete the square.
5x%5E2%2B4x-96=0
5%28x%5E2%2B%284%2F5%29x%29=96
5%28x%5E2%2B%284%2F5%29x%2B4%2F25%29=96%2B5%284%2F25%29
5%28x%2B2%2F5%29%5E2=480%2F5%2B4%2F5
5%28x%2B2%2F5%29%5E2=484%2F5
%28x%2B2%2F5%29%5E2=484%2F25
x%2B2%2F5=0+%2B-+22%2F5
x=-2%2F5+%2B-+22%2F5
x=-24%2F5 and x=4
So then
5x=-24
5x%2B24=0
and
x=4
x-4=0
So,
5x%5E2%2B4x-96=%285x%2B24%29%28x-4%29

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
First write down the prime factors of 96
2*48
2*2*24
2*2*2*12
2*2*2*2*6
2*2*2*2*2*6
from the list we see that 2*2*24 or 4*24 satisfies our factorization
5x^2+4x-96 = (5x+24)*(x-4) = 0