Question 762066: Given a polynomial and one of its factors, find the remaining factors of the polynomial. Some of the factors may not be binomials.
64x^3-768x^2-9x+8 ; x-12
A. (8x-3)
B. (64x^2-9)
C. (8x-3)(8x+3)
D. (8x-3)(8x-3)
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Given a polynomial and one of its factors, find the remaining factors of the polynomial. Some of the factors may not be binomials.
64x^3-768x^2-9x+8 ; x-12
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Using synthetic division:
12).....64....-768....-9....8
........64.....0.....-9...|-100
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Quotient: 64x^2 -9
Remainder: -100
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A. (8x-3)
B. (64x^2-9)
C. (8x-3)(8x+3)
D. (8x-3)(8x-3)
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Ans: B
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Cheers,
Stan H.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
I think this problem was botched. The other tutor's answer in incorrect because
you don't have a factor if the remainder isn't 0. And certainly a remainder of
100 is not a remainder of 0.
I had to change your last term +8 to +108, because otherwise none of those,
not even x-12, would have been factors. So I assumed the last term should have
been +108, not +8.
64x³-768x²-9x+108; x-12
12|64 -768 -9 108
| 768 0 -108
64 0 -9 0
So 64x³-768x²-9x+108 factors as
(x-12)(64x²-9)
Then 64x²-9 factors as (8x-3)(8x+3)
So the complete factorization is
(x-12)(8x-3)(8x+3)
So all 7 factors are:
1. 1
2. x-12
3. 8x-3
4. 8x+3
5. (x-12)(8x-3) = 8x²-99x+36
6. (x-12)(8x+3) = 8x²-93x-36
7. (8x-3)(8x+3) = 64x²-9
On your list, A, B, and C are factors, but D is not a factor.
(Incidentally, B and C are the same.)
Edwin
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