Problem 2

Divide $36a^2+1-64a^4-12a$ by $6a-1-8a^{2.}$

Solution
a) Rewrite in standard form. The numerator becomes

$-64a^4+36a^2-12a+1$

and the denominator becomes

$-8a^2+6a-1.$

b) Factor the denominator by pulling out a negative sign:

$-8a^2+6a-1=-(8a^2-6a+1)$

Thus, the division is equivalent to

$\frac{-64a^4+36a^2-12a+1}{-(8a^2-6a+1)}=\frac{64a^4-36a^2+12a-1}{8a^2-6a+1}$

c)Set up the division. We expect a quadratic quotient $Q\left(a\right)=C{a}^2+Da+E$ such that

$(8a^2-6a+1)(C{a}^2+Da+E)=64a^4-36a^2+12a-1.$

d) Multiply out and equate coefficients. The product expands to

$8C{a}^4+(8D-6C)a^3+(8E-6D+C)a^2+(-6E+D)a+E$

Equate this with $64a^4+0a^3-36a^2+12a-1.$

For $a^4$: $8C=64$ so $C=8.$

For $a^3$: $8D-6C=0$ yields $8D-48=0$ so $D=6.$

For $a^2$: $8E-6D+C=-36$ gives $8E-36+8=-36$ so $8E=-8$ and $E=-1.$

For the $a$ term: $-6E+D=12$ with $E=-1$ and $D=6$ this checks.

The constant term gives $E=-1$ which matches.

Answer: $8a^2+6a-1$