Question

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

• A. $8$
• B. $4$
• C. $-4$
• D. $-8$

Answer 1

The correct option is A $8$

Given,

$z^5=32$ can be factorized as

$(z-2)(z^2-pz+4)(z^2-qz+4)$ where $p>q.$

To find the value of $p^2-2q$

Solution,

$z^5=32z^5-32=0$

$z^5-2^5=0$
$z=2$

Will be one of the factor of given equation.

$z^4+{2z}^3+{4z}^2+8z+16\therefore z-2\sqrt{z^5-32}\frac{\pm z^5\mp {2z}^4}{{2z}^4-32}\frac{\pm {2z}^4\mp 4z^3}{4z^3-32}\frac{\pm 4z^3\mp 8z^2}{8z^2-32}\frac{\pm 8z^2\mp 16z}{16z-32}\frac{\pm 16z\mp 32}{0}(z-2)(z^4+2z^3+4z^2+8z+16)⟶\left(1\right)z^5-32=(z-2)(z^2-pz+4)(z^2-qz+4)z^5-32=(z-2)(z^4+(p+q)z^3+(8+pq)z^2+16-(4p+4q\left)z\right)⟶\left(2\right)$

On comparing $\left(1\right)\&\left(2\right)$ we get

$p=-2-q8+(-2-q)q=4-2q-q^2=-4q^2+2q-4=0d=4+16=20q=\frac{-2\pm 2\sqrt{5}}{5}q=-1\pm \sqrt{5}{\parallel }^{rly}p=-1\pm \sqrt{5}$

But$p>q\therefore p=-1+\sqrt{5}\&q=-1-\sqrt{5}$

value of $p^2-2q={(-1+\sqrt{5})}^2+2(-1-\sqrt{5})=1+5-2\sqrt{5}+2+2\sqrt{5}{p}^2-2q=8$