If $x^2 - 3x + 2$ is a factor of $x^4 - px^2 +q$, then find the value of $p$ and $q$ If $x^2 - 3x +2$ is a factor of $x^4-px^2+q$ then find the value of $p$ and $q$.
My attempt:
$$x^2-3x+2$$
$$x^2-2x-x+2$$
$$x(x-2)-1(x-2)$$
$$(x-1)(x-2)$$
How do I proceed further?
P.S: Edit after Deepak's comment!
 A: Since $1$ and $2$ are zeros of $x^2 - 3x + 2$, they must also be zeros of $x^4-px^2+q.$

Substituting $x=1\,$ into $x^4-px^2+q\,$ yields $1-p+q = 0$.

Substituting $x=2\,$ into $x^4-px^2+q\,$ yields $16-4p+q = 0$.

So now you have two equations in two unknowns.
A: Using long division:
$$ \frac{x^4 - px^2 + q}{x^2-3x+2} = x^2 + 3x + \frac{(7-p)x^2 - 6x +q}{x^2-3x+2}.$$
Look at the last fraction. By inspection, the ratio will be equal to $2$ if
$$ 7-p = 2 \implies p = 5$$
$$q = 4$$
Note: Previously, I mistook $q$ for $9$.
A: Hint: use polynomial division to and divide the two polynomials, then, since $x^2-3x+2$ is a factor, the remainder has to be $0$
A: EDIT1 
Factors of $ (x^2 - 3 x + 2) $ are $(x-1),(x-2)$  and each factor produces a zero remainder. Each factor zero remainder requires that
$$ 1-p+q = 0, \, 16-4p+q = 0  $$ 
Solving
$$ p=5, q=4 $$
and original expression was
$$ x^4 - 5 x^2 + 4 =(x-1)(x+1)(x-2)(x+2)$$
BTW the answer is same even if problem is set in any of six ways..
$$(x^2-3x+2),(x^2-x-2),..,..,..$$
is a factor of $ x^4-px^2+ q $  etc.
