Find $a$, $b$, $c$, $d$ such that $(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$ How would I find $a$, $b$, $c$, and $d$ in :
$$(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$$
I have already worked out that $a=2$, but I may be wrong. I am not sure how to find the values of the other letters. 
Thanks in advance!

Just do it:
$(ax+b)^2(x+c)$ expands to $(a^2x^2 + 2abx + b^2)(x+c) = a^2x^3  + 2abx^2 + b^2 x + a^2cx^2 + 2abcx + b^2c= a^2x^3 + (2ab + a^2c)x^2 + (b^2 +2abc)x + b^2 c$
So
$a^2x^3 + (2ab + a^2c)x^2 + (b^2 +2abc)x + b^2 c= 4x^3 + dx^2 + 55x - 100$
So you get three sets of equations:
$a^2 = 4$
$2ab + a^2c = d$
$b^2 +2abc=55$
and $b^2c = -100$.
So solve them:
So $a^2 = 4 \implies a = \pm 2$.  Now if $\pm2, b_1, c_1, d_1$ is a solutions set then $\mp2, -b_1,c_1, d_1$ is also a solution set so, wolog, we'll assume $a = 2$. And
$2ab + a^2c = d\implies d =4(b+c)$
$b^2 +2abc=55\implies b^2 + 4cb - 55 = 0 \implies b = \frac {- 4c \pm \sqrt{16c^2 + 220}}{2}= -2c \pm \sqrt{c^2 - 55}$
And finally $b^2c = -100$ means $c(-2c \pm \sqrt{c^2 - 55})^2 = -100$ 
$c(4c^2 \pm 4\sqrt{c^2 - 55} + c^2 - 55) = -100$
Which I don't envy you solving.
But $c < 0$ (because $cb^2 =-100 < 0$) and$c^2 > 55 $ so $|c|(4c^2 + c^2 - 55) > 100$ so that "$\pm$" must be "$-$"
