# Tripos question 1903

• Aug 29th 2010, 09:47 AM
berachia
Tripos question 1903
x^6+x^5-2x^4-x^3+x^2+1=0
Y=x^4-x^2+x-1

show y satisfies a quadratic equation with rational coefficients.

I've tried and have not be able to do so, so far.

Thanks.
• Aug 29th 2010, 10:53 AM
Opalg
Quote:

Originally Posted by berachia
$x^6+x^5-2x^4-x^3+x^2+1=0$
$y=x^4-x^2+x-1$

show y satisfies a quadratic equation with rational coefficients.

I think this is just an exercise in division of polynomials. Square y, then divide by the expression that is given to be zero:

. . . . . \begin{aligned}y^2 &= (x^4-x^2+x-1)^2 \\
&= x^8 - 2x^6 + 2x^5 - x^4 - 2x^3 + 3x^2 - 2x + 1 \\
&= (x^2 - x + 1)(x^6 + x^5-2x^4-x^3+x^2+1) - x^4 + x^2 - x \\
&= -y - 1\end{aligned}

Therefore y satisfies the quadratic equation $y^2+y+1=0$.

(I'll leave you to convince yourself that this is an occasion where division by zero is justified!)