SOLUTION: 57x^2-36x^3+4x^4-16x+63/x-7

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Question 1052158: 57x^2-36x^3+4x^4-16x+63/x-7
Found 2 solutions by ikleyn, Boreal:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Using synthetic division with 7 (the sign changes)
7/4===-36====57===-16===63; order the coefficients beginning with the highest power
4*7=28 (bring down the 4, use the next smallest power, and first term is 4x^3 in the quotient.
28====28
=======-8 add
=============-56 The polynomial is 4x^3-8x^2
==============1 Plus x
===================-7
=====================-9 and minus 9
========================-63 which leaves 0, meaning it divides evenly.
The polynomial that is the quotient is comprised of the multiples beginning with one less power, here, x^3.
(x-7)(4x^3-8x^2+x-9).
The second is not factorable. x=2 is almost a root, but f(2)=32-32+2-9 and not equal to 0.
graph%28300%2C300%2C-10%2C10%2C-100%2C100%2C4x%5E4-36x%5E3%2B57x%5E2-16x%2B63%29