4
$\begingroup$

When $f(x)$ is divided by $x - 2$ and $x + 3$, the remainders are 5 and -1, respectively. Find the remainder when $f(x)$ is divided by $x^2 + x - 6$

My method: Since $x - 2$ and $x + 3$ are linears, dividing by quadratic will leave linear remainders.

Using the remainder theorem: $$ f(x) = g(x) q(x) + r(x) $$ Where $g(x)$ is a divisor, $q(x)$ is a remainder, and $r(x)$ is a remainder

I let $ax + b$ here be the remainder. So:

\begin{align} f(x) &= g(x) (x - 2)(x + 3) + ax + b\\ f(2) &= g(2) (0)(5) + 2a+ b\\ f(-3) &= g(-3) (-5)(0) - 3a + b\\ \\ &5 = 2a + b\\ &-1 = -3a + b\\ \\ &...\\ \\ &a = \frac{6}{5}, b = \frac{13}{5} \\ \end{align}

Then the remainder is $ax + b = \frac{6}{5}x + \frac{13}{5} $

Is it possible to find the $f(x)$ out of remainders?

$\endgroup$
3
$\begingroup$

No, there are infinite $f(x)$ which satisfy the given conditions: $$f(x)=q(x)(x^2 + x - 6)+\frac{6}{5}x + \frac{13}{5}$$ with any polynomial $q(x)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.