1. ## A cubic question

The polynomial $\displaystyle P(x)$ has a remainder of 2 when divided by $\displaystyle x-1$ and a remainder of 3, when divided by $\displaystyle x-2$. The remainder when $\displaystyle P(x)$ is divided by$\displaystyle (x-1)(x-2)$ is $\displaystyle ax + b$, i.e. $\displaystyle P(x)$ can be written as $\displaystyle P(x) = (x-1)(x-2)Q(x)+ax+b$

Find a and b.

Well I got this already a= 1 and b=1

but the second question I don't know what to do...

Given that$\displaystyle P(x)$ is a cubic polynomial with coefficient of $\displaystyle x^3$ being 1, and −1 is a solution of the equation $\displaystyle P(x) = 0$, find $\displaystyle P(x)$.

I'm not too sure how to start... If -1 is a solution of the question when $\displaystyle P(x) = 0$ that means $\displaystyle (x+1)$ must be a factor right? I don't know what to do from here...

All replies are appreciated

2. Is it the same $\displaystyle \displaystyle P(x)$ as it is in Part 1? If so, $\displaystyle \displaystyle P(1) = 2$ and $\displaystyle \displaystyle P(2) = 3$ by the Remainder Theorem.

3. Oh wow, I really should think a bit before I start posting.

Using $\displaystyle P(x) = (x-1)(x-2)Q(x)+ax+b$

Let $\displaystyle Q(x) = x+1$ because -1 is a solution of x

Then multiply out and sub $\displaystyle a=b=1$

4. Originally Posted by jgv115
The polynomial $\displaystyle P(x)$ has a remainder of 2 when divided by $\displaystyle x-1$ and a remainder of 3, when divided by $\displaystyle x-2$. The remainder when $\displaystyle P(x)$ is divided by$\displaystyle (x-1)(x-2)$ is $\displaystyle ax + b$, i.e. $\displaystyle P(x)$ can be written as $\displaystyle P(x) = (x-1)(x-2)Q(x)+ax+b$

Find a and b.

Well I got this already a= 1 and b=1

but the second question I don't know what to do...

Given that$\displaystyle P(x)$ is a cubic polynomial with coefficient of $\displaystyle x^3$ being 1, and −1 is a solution of the equation $\displaystyle P(x) = 0$, find $\displaystyle P(x)$.

I'm not too sure how to start... If -1 is a solution of the question when $\displaystyle P(x) = 0$ that means $\displaystyle (x+1)$ must be a factor right? I don't know what to do from here...

All replies are appreciated
As $\displaystyle P(-1)=-6Q(-1)-1+1=-6Q(-1)=0$, this implies that $\displaystyle $$-1 is a root of \displaystyle$$Q(x)$..

CB