Need help finding factor of a polynomial.

Question

I have two polynomials:

$p \left( z \right) ={z}^{4}+2\,{z}^{3}-7\,z-5$

and

$d \left( z \right) ={z}^{2}-2\,z-13$

I need to (by hand) find two other polynomials (q(z) and r(z)) that satisfies:

$p \left( z \right) =d \left( z \right) q \left( z \right) +r \left( z \right) $ (Where r(z) has a max degree of 1)

What I've tried:

Is start with rewriting d(z)q(z) into (my Maple latex output is messing up, so I just take screenshots here)

Where $b_{1}$ and $b_{0}$ are the unknown coefficients of q(z). I know that $b_{2}=1$ since the highest degree of p(z) is $z^{4}$

Now I can say (I've colored different expressions that are related to each other):

Then:

So we can see that $b_{1}=4$

Then:

And $b_{0} = 21$

Then:

So in the end I have

$r(z)=35z+320$

and

$q(z)=b_{2}\,{z}^{2}+b_{1}\,z+b_{0}=z^{2}+4z+21$

However:

$d(z)q(z)+r(z)= \left( {z}^{2}-2\,z-13 \right) \left( {z}^{2}+4\,z+21 \right) +35\,z +320={z}^{4}+2\,{z}^{3}-59\,z+47 $

Which does not satisfy my equation.If this isn't too much of a mess, can anyone see where I went wrong?

Answer 1

I got $$z^4+2z^3-7z-5=z^4-2z^3-13z^2+4z^3-8z^2-52z+21z^2-42z-273+87z+268=$$ $$=(z^2-2z-13)(z^2+4z+21)+87z+268.$$

Answer 2

You are asked to perform a long division of polynomials. This is done in a way similar to the written division of integers.

The leading terms of the dividend and divisor are $z^4,z^2$, so that the tentative quotient is $\color{green}{z^2}$. Then

$$z^4+2z^3-7z-5-z^2(z^2-2z-13)=4z^3+13z^2-7z-5.$$

The next term in the quotient is $\color{green}{4z}$, and

$$4z^3+13z^2-7z-5-4z(z^2-2z-13)=21z^2+45z-5.$$

And finally $\color{green}{21}$,

$$21z^2+45z-5-21(z^2-2z-13)=87z+268.$$

In conclusion,

$$z^4+2z^3-7z-5=(z^2+4z+21)(z^2-2z-13)+(87z+268).$$