Question

$f\left(y\right)=2y^4+3y^3-2y^2-9y-12$
$g\left(y\right)=y^2-3$

Check whether g(y) is a factor of f(y) by applying the division algorithm. if YES type 1 as answer and if NO type 0 as answer

Answer 1

$f\left(y\right)=2y^4+3y^3-2y^2-9y-12$

$g\left(y\right)=y^2-3$
Applying Division algorithm, we get
$f\left(y\right)=g\left(y\right)q+r$ where $q$ and $r$ are unique integer and $0<=r<g\left(y\right)$.
If $r=0$, then $g\left(y\right)$ is a factor $f\left(y\right)$
Thus,
$q=\frac{f\left(y\right)}{g\left(y\right)}$

	$2y^2+3y+4$
$y^2-3$	$2y^4+3y^3-2y^2-9y-12$ $2y^4$ $-6y^2$
	$3y^3+4y^2-9y$ $3y^3-9y$
	$4y^2-12$ $4y^2-12$
	$0$

Here remainder $r=0$, thus, $g\left(y\right)$ is a factor $f\left(y\right)$