How to find a substitution that transforms an equation into a particular equation. (Roots of polynomials).

Question

Find the value of $c$ so that the substitution $x=u+c$ transforms the equation $x^3-12x^2+45x-54=0$ into the equation $u^3-3u^2=0$

My first idea is to find the roots of $u^3-3u^2=0$ And getting $u=0$ and $u=3$, however I am not sure how to continue.

If someone could explain how to tackle this sort of problem, I would greatly appreciate it. Thanks.

Answer 1

$$f(u+c)=(u+c)^3-12(u+c)^2+45(u+c)-54$$ In the final equation the coefficient of $u$ is $0$.
That means want to eliminate coefficient of $u$ in $f(u+c)$,
so collect the coefficients of $u$ and set it to $0$: $$\begin{align}&3c^2-12(2c) + 45=0\\&c^2-8c+15=0\\&c=3,5\end{align}$$

Plug each of these in $f(u+c)$ and see if the constant term also gets eliminated.

Answer 2

You want a factorization of the form $u^2(u-3)$. This means, the given polynomial should also factor into $(x-c)^2(x-c-3)$.

So, the first thing to do, whether you can factor the given polynomial.

Based on the rational root test you find quickly that your polynomial factors into

$$(x-3)^2(x-6) \Rightarrow u = x-3 \Leftrightarrow x= u+3$$

Answer 3

Another more systematic way without factorization uses Taylor polynomials.

Set $f(x) = x^3-12x^2+45x-54$. Then according Taylor you have around any $a \in \mathbb{R}$ $$f(x) = \sum_{k=0}^{3}\frac{f^{(k)}(a)}{k!}(x-a)^k$$ $$\stackrel{a=-c}{\Longrightarrow}f(x) = \frac{f^{(2)}(-c)}{2!}(x+c)^2 + \frac{f^{(3)}(-c)}{3!}(x+c)^3$$

Since the coefficient in front of $u^2 = (x+c)^2$ is $-3$, the following equation must hold true $$\frac{f^{(2)}(-c)}{2!} \stackrel{!}{=} -3 \Leftrightarrow \frac{6(-c)-24}{2!}=-3 \Leftrightarrow \boxed{c=-3}$$

Answer 4

Here's a relatively fast approach that avoids factoring a cubic.

Notice that $u^3 - 3 u^2 = u^2 (u - 3)$ has a double root at $u = 0$, so if $x = u + c$, the given polynomial, $$p(x) := x^3 − 12 x^2 + 45 x − 54$$ has a double root at $x = (0) + c = c$.

Since $c$ is a double root, it is also a root of $$p'(x) = 3 (x^2 - 8 x + 15) = 3 (x - 3) (x - 5) ,$$ so $c = 3$ or $c = 5$. But $5 \not\mid 54$, so the Rational Root Theorem implies that $p(5) \neq 0$, leaving $$\color{#df0000}{\boxed{c = 3}} .$$