This question arose while I was tutoring a student on the topic of the Remainder Theorem. Now, the Remainder Theorem tells us that when a polynomial $p(x)$ is divided by a linear factor $(x-a)$, the remainder is simply $p(a)$. However, in this case we have a product of linear factors.
Using the Remainder Theorem we can see that neither $(x-1)$ nor $(x+2)$ is a factor of $f(x)$. Also, if we try to find the remainder using long division, we get a relatively ugly remainder of
$$
3(14x – 13)
$$
I assume this is not the correct approach as all other questions in this topic used the Remainder Theorem. So perhaps there is a more elegant approach?
Best Answer
Hint: the remainder will be a polynomial of degree (at most) $1$ so:
$$f(x) = (x-1)(x+2)q(x) + ax + b$$
Substitute $x=1,-2$ in the above and you get two equations in $a,b$.
[ EDIT ] For a less conventional approach (justified in the answer here) note that $(x-1)(x+2)=0 \iff x^2=-x+2$. Repeatedly using the latter substitution:
$$ \begin{align} 3x^5 - 5x^2 + 4x + 1 &= 3 (x^2)^2 \cdot x - 5(x^2) + 4x + 1 \\ &= 3(x^2-4x+4)x - 5(-x+2) + 4x +1 \\ &= 3(-x+2-4x+4)x + 9x -9 \\ &= -15(x^2)+ 18x + 9x - 9 \\ &= -15(-x+2) + 27 x - 9 \\ &= 42 x -39 \end{align} $$