Algebra #11

by sjaelee, Aug 15, 2011, 1:08 PM

Quote:
Let $x_i$ be the roots of $x^5+x^2+1=0$.Let $q(x)=x^2-2$. Find $q(x_1)q(x_2)q(x_3)q(x_4)q(x_5)$.

Source: USAMTS
This is a scary problem. Oh boy..... Lets see what we want:

$(x_1^2-2)(x_2^2-2)...(x_5^2-2)$. Hmmm... Aha! We can find a

polynomial with $x_i^2$ as roots! Let $y=x^2$. Then

$xy^2+y+1=0$, so $xy^2=-(y+1)$. To get rid of the x, we square!

$x^2y^4=y^2+2y+1$, and $y=x^2$, so $y^5-y^2-2y-1=0$. By

the Fundamental Theorem of Algebra, $y^5-y^2-2y-1=0$ in factored

form (for any y and roots $x_i^2$) is

$(y-x_1^2)(y-x_2^2)...(y-x_5^2)$. Aha! If $y=2$, we have:

$(2-x_1^2)(2-x_2^2)...(2-x_5^2)$, which is $(-1)^5$ times what

we want [since $-1(2-x_i^2)=x_i^2-2$]. That means all we have to

do is plug in $2$ into our new polynomial and multiply $-1$ to cancel

the factor $(-1)^5$ :

$2^5-2^2-2(2)-1=23$, then $-1(23)=-23$.

Thus, we have $-23$.
This post has been edited 3 times. Last edited by djmathman, Sep 5, 2013, 1:23 AM

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