SOLUTION: If P(x)=(g(x^3))+(xh(x^3)) is divisible by (x^2)+x+1,then show that g(x) and h(x) are divisible by (x-1). Thanks

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: If P(x)=(g(x^3))+(xh(x^3)) is divisible by (x^2)+x+1,then show that g(x) and h(x) are divisible by (x-1). Thanks      Log On


   

Question 917729: If P(x)=(g(x^3))+(xh(x^3)) is divisible by (x^2)+x+1,then show that g(x) and h(x) are divisible by (x-1).

Thanks
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
P%28x%29=g%28x%5E3%29%2Bx%2Ah%28x%5E3%29
g%28x%29=%28x-1%29G%28x%29%2Br for some quotient polynomial G%28x%29 and some number remainder r .
(If r=0, then g%28x%29 is divisible %28x-1%29 ).
h%28x%29=%28x-1%29H%28x%29%2Bs for some quotient polynomial H%28x%29 and some number remainder s .
(If s=0, then h%28x%29 is divisible %28x-1%29 ).

Substituting, we get
P%28x%29=%28x%5E3-1%29G%28x%5E3%29%2Br%2Bx%2A%28%28x%5E3-1%29H%28x%5E3%29%2Bs%29
P%28x%29=%28x%5E3-1%29G%28x%5E3%29%2Br%2Bx%2A%28x%5E3-1%29H%28x%5E3%29%2Bsx
P%28x%29=%28x%5E3-1%29%28G%28x%5E3%29%2Bx%2AH%28x%5E3%29%29%2Bsx%2Br
Since x%5E3-1=%28x-1%29%28x%5E2%2Bx%2B1%29 ,
P%28x%29=%28x%5E2%2Bx%2B1%29%28x-1%29%28G%28x%5E3%29%2Bx%2AH%28x%5E3%29%29%2Bsx%2Br
That means that
when you divide P%28x%29 by the quadratic polynomial %28x%5E2%2Bx%2B1%29 ,
the quotient is %28x-1%29%28G%28x%5E3%29%2Bx%2AH%28x%5E3%29%29 ,
and the remainder is the linear polynomial rx%2Bs .
Since P%28x%29 is divisible by %28x%5E2%2Bx%2B1%29 ,
rx%2Bs must be zero for all values of x , meaning that r=s=0 .