Substitution question Math

Question

Given that $f(x)=3x^5-11x^3+30x^2+36=(x-2)(x+3)Q(x)+ax+b$ for all values of $x$ and that $Q(x)$ is a polynomial,

i) Find the values of $a$ and of $b$.

ii) Hence, find the remainder when $f(x)+2$ is divided by $(x^2+x-6)$.

Hi, I'm stuck at part 2 please help, attached is the working and questions. :)\Why do we need to get rid of the polynomial? How do i find the Polynomial?

Answer 1

Now you have $f(x)=(x^2+x-6)Q(x)+58x+48$. So

$$f(x)+2=(x^2+x-6)Q(x)+58x+50$$

With this equality, you should be able to tell what the remainder will be when $f(x)+2$ is divided by $x^2+x-6$.

Answer 2

Using your information, we get:$$f(x)+2=(x^2+x-6)Q(x)+58x+50$$ Since $(x^2+x-6)Q(x)|(x^2+x-6)$, and $58x+50$ obviously cannot be divisible (because it has a smaller degree), leaving the remainder to be $\boxed{58x+50}$

Answer 3

$$ \left( x^{2} + x - 6 \right) \left( 3 x^{3} - 3 x^{2} + 10 x + 2 \right) + \left( 58 x + 48 \right) = \left( 3 x^{5} - 11 x^{3} + 30 x^{2} + 36 \right) $$

$$ \left( x^{2} + x - 6 \right) \left( 3 x^{3} - 3 x^{2} + 10 x + 2 \right) + \left( 58 x + 50 \right) = \left( 3 x^{5} - 11 x^{3} + 30 x^{2} + 38 \right) $$