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Let $$P(x)=x^{6}-x^{5}-x^{3}-x^{2}-x$$ $$Q(x)=x^{4}-x^{3}-x^{2}-1$$

Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the roots of Q(x) Prove that $$P\left (z_{1} \right )+ P\left (z_{2} \right )+ P\left (z_{3} \right )+ P\left (z_{4} \right )=6$$

please delete me
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    You meant to not put a 4th power term in the expression for $P(x)$, right? – Eric Auld Aug 09 '13 at 11:08
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    Since you know what $Q(z_i)$ is, this should lead you to try to express as much as possible of $P(x)$ in terms of $Q(x)$. You can start out with $x^2 Q(x)$, and work with what's left. Also, if I recall correctly you eventually need to know what $z_1 + z_2 + z_3 + z_4$ is, maybe even $z_1^2 + z_2^2 + z_3^2 + z_4^2$. – Arthur Aug 09 '13 at 11:26

2 Answers2

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$P(x)=Q(x)(x^2+1)+(x^2-x+1)$, hence $\sum P(z_i)= \sum z_i^2 -\sum z_i +4 $. By Vieta's formulas $\sum z_i=1$ and $\sum z_i^2=(\sum z_i)^2-2\sum z_iz_j=1+2=3$. So $\sum P(z_i)=6$.

Boris Novikov
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HINT:

So, $$z_1^4-z_1^3-z_1^2-1=0\ \ \ \ \ (1)$$

$$P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1$$ $$=z_1^2(z_1^4-z_1^3-z_1^2-1)+z_1^4-z_1^3-z_1=z_1^2-z_1+1$$

Let $z_1^2-z_1+1=y$

$$z_1^4-z_1^3-z_1^2-1=z_1^2(z_1^2-z_1+1)-2z_1^2-1=z_1^2y-2(y-1+z_1)-1$$ $$=(y+z_1-1)y-2(y-1+z_1)-1$$

$$\implies (y+z_1-1)y-2(y-1+z_1)-1=0$$

Express $z_1$ in terms of $y$ and put the value in $(1)$

Then use Vieta's formulas 1,2,3

lab bhattacharjee
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