Find $a$, $b$, $c$, $d$ such that $(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$

Question

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!

Answer 1

Let $f(x) = 4x^3 + dx^2 + 55x - 100 = (ax+b)^2(x+c)$.

It is trivial to see $a = \pm 2$. We will only consider the case $a = 2$. The key of this problem is $f(x)$ contains a squared factor $(ax+b)^2$. This means $f'(x) = 12x^2+2dx+55$ contain $(ax+b)$ as a factor.

Given any two polynomials $g(x), h(x) \in \mathbb{C}[x]$. If you compute their gcd over $\mathbb{C}[x]$ using Euclidean algorithm, the gcd you get will be an expression independent of $x$. It will be a polynomial in coefficients of $g(x)$ and $h(x)$. The most important property of this gcd is

$g(x)$ and $h(x)$ contains a common factor if and only if this gcd vanishes.

Up to some scaling factor, this gcd is known as the resultant of the two polynomials. If you throw the command Resultant[4*x^3 + d*x^2 + 55*x-100,12*x^2+2*d*x+55,x] to WA (wolfram alpha), you will obtain following condition for $d$:

$$D(d) \stackrel{def}{=} 16 d^3 + 121 d^2 - 15840 d - 279280 = 0\tag{*1}$$

Up to scaling factor again, the resultant of a polynomial $f(x)$ and its derivative $f'(x)$ is known as its discriminant. You can also use the command Discriminant[4*x^3+d*x^2+55*x-100,x] on WA to derive above condition.

To derive $b$, we will use the fact when you compute the gcd between $f(x)$ and $f'(x)$ using Euclidean algorithm. By the the time you obtain a linear factor, that factor will proportional to $ax + b$. Since

$$36f(x) - (12x+d)f'(x) = (1320-2d^2)x - (55d+3600)$$

we have

$$2 : b = 1320-2d^2 : 55d+3600 \quad\implies\quad b = B(d) \stackrel{def}{=} \frac{55d+3600}{d^2 - 660}$$

By comparing coefficients of $x^2$ in $4x^3 + dx^2 + 55x - 100 = (ax+b)^2(x+c)$, we obtain following expression for $c$:

$$c = \frac{d}{4} - b = C(d) \stackrel{def}{=} \frac{d}{4} - \frac{55d+3600}{d^2-660}$$

Using these expression of $b$ and $c$, one can use a CAS to verify

$$(2x+B(d))^2(x + C(b)) = f(x) + \frac{25 D(d) (36 d^2x-23760x+d^3+43200)}{4(d^2-660)^3}$$

As one can see, everything comes down to solving the cubic equation $D(d) = 0$.

Numerically, $D(d) = 0$ has a root at $d_0 \approx 34.98939913719146$. This leads to following approximate solution of the problem:

$$\begin{align}(a,b,c,d) &= (a,B(d_0),C(d_0),d_0)\\ &= ( 2,\; 9.790585944043832,-1.043236159745968,\; 34.98939913719146 )\end{align}$$

Answer 2

Hint: $$(ax+b)^2(x+c)={a}^{2}c{x}^{2}+{a}^{2}{x}^{3}+2\,abcx+2\,ab{x}^{2}+{b}^{2}c+x{b}^{2}=a^2x^3+x^2(a^2c+2ab)+x(2abc+b^2)+b^2c$$

Answer 3

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 $d$.

Answer 4

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.

Answer 5

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 "$-$"

Answer 6

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.