SOLUTION: If ax³ + bx² + x - a = 0 is exactly divisible by (x+3) and by (2x-1), find a and b.

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Question 1132102: If ax³ + bx² + x - a = 0 is exactly divisible by (x+3) and by (2x-1), find a and b.
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
that means that 2 roots are -3 and +1/2
-27a+9b-3-a=0
-28a+9b=3
1/8 a+1/4b+1/2-a=0
-7a+2b=-4
28a-8b=16
-28a+9b=3
b=19
a=6
6x^3+19x^2+x-6=0
graph%28300%2C300%2C-5%2C5%2C-50%2C100%2C6x%5E3%2B19x%5E2%2Bx-6%29

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


From the given information, we know that two of the roots are -3 and 1/2.

Vieta's Theorem tells us the product of the three roots of a cubic polynomial is (opposite of constant term) divided by (leading coefficient): a/a = 1.

Therefore, the third root is -2/3.

So the polynomial is

%28x%2B3%29%282x-1%29%283x%2B2%29+=+6x%5E3%2B19x%5E2%2Bx-6