$f(x) =x^ 3 +ax^ 2 +b$ and $g(x) = ax^ 3 +bx^ 2 +x-a$. find a common linear factor.

Question

I dont know how to find the common linear factor. They also have said to prove that the above common factor is a factor of the polynomial $h(x)=(b-a^ 2 )x^ 2 +x-a(1+b)$. Can someone explain how to do this?

Answer 1

The general statement seems false to me, when $a,b,c\in\mathbb R$.

We consider $a≠0\wedge b=0$. Let $x-m$ be a common factor of $f(x)$ and $g(x)$. Then $m$ is a common root. This implies:

$$\begin {cases} m^3+am^2=0\\am^3+m-a=0\end{cases} $$

If $m$ is a common root, then $m\in\{0,-a\}$.

Obviously, $m=0$ gives a contradiction. Therefore, $m=-a$. We have:

$$ -a^4-a-a=0,\;a≠0\\ \implies a^3=-2. $$

This tells us, if $a≠-\sqrt [3]{2}\wedge b=0$, then the common factor doesn't exist.

But, the second part of the question seems correct me. If $f(x)$ and $g(x)$ have a common factor, then:

$$ \begin{align} &\begin {cases} m^3+am^2+b=0\\am^3+bm^2+m-a=0\end{cases} \\ \implies &a(-am^2-b)+bm^2+m-a=0\\ \implies &(b-a^2)m^2+m-a(1+b)=0 \end{align} $$

This means that, $x-m$ is indeed a factor of

$$(b-a^2)x^2+x-a(1+b).$$

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Comments:

1. The statement fails, when $a,b,c\in\mathbb Z$. Because, we showed that if $a\in\mathbb Z, b=0$ then $f(x)$ and $g(x)$ doesn't have a common factor.

2. The statement also fails, when $a,b,c\in\mathbb Z^{+}$. Because, if $a=b=1$, then we have $2$ different irreducible cubics with one real root. This means, also in this case, $f(x)$ and $g(x)$ don't have a common linear factor.

Answer 2

COMMENT.- I suspect that the real problem might be perhaps determining $a$ and $b$ such that $f$ and $g$ have a linear factor in common. In general there is no possibility unless you work on complicated algebraic extensions for the coefficients. For example if $$f(x)=(x+r)(x^2+sx+t)$$ we have for the coefficient $s$ the equation $s^3-2as^2+a^2s+b=0$ whose real root have a tedious expression; one has

so the other coefficients $r,t$ must be similar. For this I subscribe to the affirmation of @lone student's answer.