Given 2 roots and a remainder of a cubic polynomial, find the remainder of $f(3)$

Question

$f(x)$ is a cubic polynomial and the coefficient of $x^3$ is $k$. If the polynomial has exactly $2$ roots and these roots are $2$ and $-5$ and $f(1) = -48$, find the possible values of $f(3)$.

I've gotten part of the cubic equation to be $(kx+a)(x^2+3x-10)$, then I let $f(x) = (kx+a)(x^2+3x-10)$, substitute in $f(1)$ to end up with $-6k-a=-48$, and I am stuck from here onward. Any hint will be appreciated.

Thanks!

Answer 1

We have also $2k+a=0$ or $-5k+a=0$.

In the first case $f(x)=k(x-2)^2(x+5)$, which gives $-48=6k$ or $k=-8$ and

$f(3)=-8\cdot1^2\cdot8=-64$.

In the second case $f(x)=k(x+5)^2(x-2)$, which gives $-48=-36k$ or $k=\frac{4}{3}$ and $f(3)=\frac{4}{3}\cdot8^2\cdot1=\frac{256}{3}$.

Answer 2

You know that either

$$f(x)=k(x-2)(x+5)^2\tag{1}$$ or $$f(x)=k(x-2)^2(x+5)\tag{2}$$

So either $f(1)=-36k=-48$ or $f(1)=6k=-48$ so $k=\frac{4}{3}$ in $(1)$ or $k=-8$ in $(2)$.

So there are two possibilities for $f(x)$

1. $f(x)=\frac{4}{3}(x-2)(x+5)^2$
2. $f(x)=-8(x-2)^2(x+5)$

Use these to find the two possible values of $f(3)$.