SOLUTION: Solve for X X^2(2x-5)^2+x(5-2x)^3 Possible answers: a. -13 b. -13/2 c. 13/2 d. 13

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solve for X X^2(2x-5)^2+x(5-2x)^3 Possible answers: a. -13 b. -13/2 c. 13/2 d. 13      Log On


   

Question 40435: Solve for X
X^2(2x-5)^2+x(5-2x)^3
Possible answers:
a. -13
b. -13/2
c. 13/2
d. 13
Found 2 solutions by Fermat, Nate:
Answer by Fermat(136) About Me  (Show Source):
You can put this solution on YOUR website!
I assume that your expression is equated to zero, yes?
x%5E2%282x-5%29%5E2%2Bx%285-2x%29%5E3+=+0
The term %285-2x%29%5E3 can be rewritten as: -%282x-5%29%5E3, giving
x%5E2%282x-5%29%5E2-x%282x-5%29%5E3+=+0
Take out the common factor of %282x-5%29%5E2,
%282x-5%29%5E2%28x%5E2+-+x%282x-5%29%29+=+0
%282x-5%29%5E2%28x%5E2+-+2x%5E2+%2B+5x%29+=+0
%282x-5%29%5E2%285x+-+x%5E2%29+=+0
x%282x-5%29%5E2%285+-+x%29+=+0
Since the product of the three terms, x,%282x-5%29 and %285-x%29 is zero, then we equate each of the individual terms to zero as a solution to the problem. This gives us,
x+=+0, 2x-5+=+0, 5-x+=+0
Ans: x+=+0, x+=+5%2F2, x+=+5
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Unfortunately, none of the solutions match your possible answers. I can't see anything wrong with my working. Is the expression supposed to be equated to zero ?

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%282x-5%29%5E2+%2B+x%285-2x%29%5E3=0
x%5E2%284x%5E2+-+20x+%2B+25%29+%2B+x%28-8x%5E3+%2B+60x%5E2+-+150x+%2B+125%29+=+0
4x%5E4+-+20x%5E3+%2B+25x%5E2+%2B+-8x%5E4+%2B+60x%5E3+-+150x%5E2+%2B+125x+=+0
-4x%5E4+%2B+40x%5E3+-+125x%5E2+%2B+125x+=+0
Plug in your available answers, and none work.