If $z^5-32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5-32)=(z-2)(z^2-pz+4)(z^2-qz+4)$,then find $p^2+2p.$

Question

If the expression $z^5-32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5-32)=(z-2)(z^2-pz+4)(z^2-qz+4)$,then find the value of $p^2+2p.$

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I used $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$ to get $z^5-2^5=(z-2)(z^4+2z^3+4z^2+8z+16)$
Then i compared $z^4+2z^3+4z^2+8z+16$ with $(z^2-pz+4)(z^2-qz+4)$
to get $p+q=-2$ and $p^2q^2=2$

$pq=\pm\sqrt2$
But when i solve the two equations $pq=\pm\sqrt2$ and $p+q=-2$,i get the value of $p$ and $q$ whch are non-manageable difficult to simplify.And i am not able to calculate the final desired result $p^2+2p$.
Is my approach not correct?What is the simple and elegant method to do this?Please help me.

Answer 1

Equating the coefficients of $z^2,$

$$4+pq+4=4\iff pq=-4$$

Equating the coefficients of $z^3,$

$$-2=p+q$$ So, $p,q$ are the roots of $$t^2+2t-4=0$$

Answer 2

HINT:

$z^5=32=2^5,$

$z_r=2e^{2r\pi/5}$ where $r=0,\pm1,\pm2$

Now $(z-z_s)(z-z_{-s})=z^2-2z\cos\dfrac{2s\pi}5+4$

So, the values of $p,q$ will be reached by setting $s=1,2$