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!
 A: Hint: $$(ax+b)^2(x+c)={a}^{2}c{x}^{2}+{a}^{2}{x}^{3}+2\,abcx+2\,ab{x}^{2}+{b}^{2}c+x{b}^{2}=a^2x^3+x^2(a^2c+2ab)+x(2abc+b^2)+b^2c$$
A: 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 "$-$"
A: You can expand out the left side.  The coefficient of each power of $x$ on the left must match the corresponding coefficient on the right.
But you'll end up solving a cubic equation for $b$, $c$ or $d$.
A: There are four unknown in $$(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$$, so by assigning four values to  $x$ and solving the resulting equations you find your unknowns. 
For example for $x=0$ we get  
$$2bc = - 100$$
For x=1, you get  $$(a+b)^2(1+c) = 4 + d + 55 - 100$$
The rest is simple algebra and you can manage it.
A: It cannot be done, assuming these coefficients are supposed to be integers.
You have established that $a$ may be assumed to be $2$, assuming $a$ is positive.
Note that $b^2c=-100$, so $|b|$ is either $1$, $2$, $5$, or $10$. But $b$ cannot be even, or all coeffficients of the right side would be even. And there is that $55$ countering that. So $|b|$ is either $1$ or $5$.
$4x^3+55x-100$ is monotonically increasing with one root. Subtracting a positive multiple of $x^2$ would not change the root count. So $d$ must be positive. Furthermore, the doubled root will be negative and $b>0$. So $b$ itself is $1$ or $5$ (with $c$ being $-100$ or $-4$).
Now it is narrowed down enough to check:
$$(2x+1)^2(x-100)\stackrel{?}{=}4x^3+dx+55x-100$$
$$(2x+5)^2(x-4)\stackrel{?}{=}4x^3+dx+55x-100$$
Expanded, you have:
$$4x^3-396x^2-99x-100\stackrel{?}{=}4x^3+dx+55x-100$$
$$4x^3+4x^2-35x-100\stackrel{?}{=}4x^3+dx+55x-100$$
And in both cases the linear coefficient is not working out to match.
