SOLUTION: Find the other factors of P(x)=x^5-x^4-20x^3+20x^2+64x-64 if you are given that x-4 and x+4 are factors of P(x)

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Question 1043391: Find the other factors of P(x)=x^5-x^4-20x^3+20x^2+64x-64 if you are given that x-4 and x+4 are factors of P(x)
Found 3 solutions by josgarithmetic, solver91311, ikleyn:
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website!
Try synthetic division to break P one factor at a time. You could try instead divisor x^2-16, but using synthetic division with first x-4 and then x+4 divisors could be easier. You still then need to handle the cubic result, whichever way you choose.

4 | 1 -1 -20 20 64 -64 | | 4 12 32 104 672 ---------------------------------------- 1 3 8 52 168 608

NO. x-4 AND x+4 ARE NOT BOTH FACTORS OF P(x).

Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!

4 | 1 -1 -20 20 64 -64 | 4 12 -32 -48 64 ------------------------------- 1 3 -8 -12 16 0 -4 | 1 3 -8 -12 16 | -4 4 16 -16 -------------------------- 1 -1 -4 4 0

Hence,

Factor

So the complete factorization is:

John

My calculator said it, I believe it, that settles it

Answer by ikleyn(52809)   (Show Source):
You can put this solution on YOUR website!
.
Find the other factors of P(x)=x^5-x^4-20x^3+20x^2+64x-64 if you are given that x-4 and x+4 are factors of P(x)
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Both (x+4) and (x-4) are the factors. I checked it. If and when you divide the given polynomial P(x) = by (x+4)*(x-4) = , you will get a quotient q(x) = . In turn, it has the root x = 1. You can easily check it. Hence (according to the Remainder theorem), it has the factor (x-1). After dividing q(x) by (x-1) you get a quotient r(x) = = . Hence, the original polynomial has factoring f(x) = =