# Thread: finding P(x) and the remainder

1. ## finding P(x) and the remainder

If P(x) has the remainder -5 when divided by x-3 and the remainder 3 when divided by x+2, find the reminder when P(x) is divided by (x-3)(x+2)
how do you find remainder from the details provided above?

2. Imagine you write the euclidean division :

There exists a polynomial Q such that :
$\displaystyle P(x)=(x-3)Q(x)-5$
Now let $\displaystyle x=3$
This gives $\displaystyle P(3)=0-5=-5$

Same thing for the division by (x+2) : we get $\displaystyle P(-2)=3$

Now what if we consider the division by $\displaystyle (x-3)(x+2)$ ?
There exists polynomials R and S (S is the remainder) such that :
$\displaystyle P(X)=(x-3)(x+2)R(x)+S(x)$
Now once again, if you let $\displaystyle x=3$, you get :

$\displaystyle \underbrace{P(3)}_{-5}=0+S(3)$ so we have $\displaystyle S(3)=-5$
Similarly, if you let $\displaystyle x=-2$, you get : $\displaystyle S(-2)=3$

So the remainder of P when divided by (x-2)(x+3) will be a polynomial S such that $\displaystyle S(3)=-5$ and $\displaystyle S(-2)=3$

But the degree of S cannot exceed 2, since (x-2)(x+3) has a degree 2. (if S has a degree equal or superior to 2, then it still can be divided by (x-2)(x+3))
Hence we're looking for a and b in :
$\displaystyle S(x)=ax+b$

Now the previous working gives you the following system :
$\displaystyle \left\{\begin{array}{ll} 3a+b=-5 \\ -2a+b=3 \end{array} \right.$
Which is very basic algebra

3. how if P(x) was a cubic equation? is it still possible to find P(x)

4. Originally Posted by hungrybarts
how if P(x) was a cubic equation? is it still possible to find P(x)
You're too fast >< I've edited my post !

5. Originally Posted by hungrybarts
how do you find remainder from the details provided above?