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(52810)   (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)
~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

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) =  =


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