1. ## Remainder Theorem

Given that $x^5 + ax^3 + bx^2 - 3 = (x^2 - 1)Q(x) - x - 2$, where $Q(x)$ is a polynomial.
State the degree of $Q(x)$ and find the value of $a$ and of $b$.

(I am able to deduce the degree i.e. 3
I am also able to find the value of a ,i.e -2 and the value of b, i.e. 1.)

However, the following question linked to the above is what I am not able to solve:

Find also the remainder when $Q(x)$ is divided by $x + 2$.

2. Originally Posted by Ilsa
Given that $x^5 + ax^3 + bx^2 - 3 = (x^2 - 1)Q(x) - x - 2$, where $Q(x)$ is a polynomial.
State the degree of $Q(x)$ and find the value of $a$ and of $b$.

(I am able to deduce the degree i.e. 3
I am also able to find the value of a ,i.e -2 and the value of b, i.e. 1.)

However, the following question linked to the above is what I am not able to solve:

Find also the remainder when $Q(x)$ is divided by $x + 2$.

Hint: Given a polynomial P(x), when you divide it by x - r, the remainder is equal to P(r).

-Dan

3. How do we get the answer -5 then?
If the remainder is Q(-2), how do I further calculate the remainder?

4. Let's proceed step by step.

(i) Since $Q(x)$ is of degree 3, assume $Q(x)=Ax^3+Bx^2+Cx+D$.

(ii) Put this in RHS. Then by comparing the coefficients of RHS to the coefficients of LHS find the values of $A,B,C$ and $D$.

(iii) After getting the values of $A,B,C,D$ you know the exact expression of $Q(x)$. So you can easily put $-2$ in place of $x$ and calculate $Q(-2)$.

5. That, actually finding Q, seems to me a very difficult way to do it. Topsquark's suggestion is best. You are told that
$x^5 + ax^3 + bx^2 - 3 = (x^2 - 1)Q(x) - x - 2$ and you say you have determined that a= -2 and b= 1 so you have $x^5 - 2x^3 + x^2 - 3 = (x^2 - 1)Q(x) - x - 2$.

Setting x= -2, $(-2)^5- 2(-2)^3+ (-2)^2- 3= ((-2)^2- 1)Q(-2)- (-2)- 2$

$-32+ 16+ 4- 3= -15= 3Q(-2)$
so it is easy to find Q(-2) which is, as Topsquark said, the remainder when Q(x) is divided by x+ 2.