Question

$4yz(z^2+6z-16)÷2y(z+8)=$

• A. $z-2$
• B. $z(z-2)$
• C. $2z(z-2)$

Answer 1

The correct option is C $2z(z-2)$

Given $4yz(z^2+6z-16)$
First, we will factorise $(z^2+6z-16)$
Here, we need the product to be
$-16$ and sum OR difference to be $+6,$
Hence, $6z$ can be written as $-2z+8z$
$\Rightarrow 4yz(z^2+6z-16)=4yz[z^2-2z+8z-16]$
$=4yz\left[z\right(z-2)+8(z-2\left)\right]$
$=4yz(z-2)(z+8)$
Therefore, $4yz(z^2+6z-16)÷2y(z+8)$
$=\frac{4yz(z-2)(z+8)}{2y(z+8)}$
$=2z(z-2)$