Question 618110: Solve by factoring 4x^2+8x-5=0
4x²+8x-5 = 0
4x²+10x-2x-5 = 0
(4x²-2x)+(10x-5)
2x(2x-1)+5(2x-1)
(2x+5)(2x-1)
This is my answer but my instructor says "you need to solve for x. You factored correctly, but now you need to set each factor = 0 and solve for x". Confused.
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Solve by factoring 4x^2+8x-5=0
4x²+8x-5 = 0
4x²+10x-2x-5 = 0
(4x²-2x)+(10x-5)
2x(2x-1)+5(2x-1)
(2x+5)(2x-1)
This is my answer but my instructor says "you need to solve for x. You factored correctly, but now you need to set each factor = 0 and solve for x". Confused.
.
The last three lines should STILL be set to zero:
(4x²-2x)+(10x-5) = 0
2x(2x-1)+5(2x-1) = 0
(2x+5)(2x-1) = 0
.
Now, because you have:
(2x+5)(2x-1) = 0
IF the left term in parenthesis is zero the statement would be true.
Similarly, if you the right term in parenthesis is zero the statement is also true.
.
So, solve for x by setting each term to zero:
Left term:
2x+5 = 0
2x = -5
x = -5/2 (one solution)
.
Right term:
2x-1 = 0
2x = 1
x = 1/2 (second solution)
.
Answer: x = {-5/2, 1/2}
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