# Remainder/factor theorem question

1. Oct 4, 2012

### HerroFish

The question is
Find "a" and "b":

f(x) is
ax^3 - bx^2 + 2x - 12

Divisor is
x^2 - 5x +6

Remainder is
2x - 3

So what I tried to do was divide f(x) with the divisor and take the the remainder that I got and let it = 2x - 3 but I always get stuck in the end. My teacher suggested using the division statement but I'm not quite sure how that will help.

Thanks for helping!

2. Oct 4, 2012

### rcgldr

One method to solve this is use polynomial long division starting at the end, where you know the remainder, the divisor, and the last two terms of the dividend, which allows you to determine the last term of the quotient, then work backwards until you solve for b then a.

3. Oct 4, 2012

### HerroFish

ouu okay I see!
But is there a shorter method to this?

4. Oct 4, 2012

### rcgldr

What you have is

ax^3 - bx^2 + 2x - 12 = (cx + d) (x^2 - 5x + 6) + (2x - 3)

where a, b, c and d are unknown. You could subract (2x - 3) from both sides first, then use the long division method or inspection to determine the values, but subtracting (2x - 3) from both sides is the same as what you're doing in the first step of working the long division backwards. This backwards long division method is used in general for "division problems" when there are unknowns in the dividend, divisor, and/or remainder.

Last edited: Oct 4, 2012
5. Oct 4, 2012

### SammyS

Staff Emeritus
So, If f(x) divided by g(x) yields a quotient q(x) and remainder r(x), then

$f(x)=g(x)\cdot q(x)+r(x)\ .$

Therefore, $f(x)-r(x)=g(x)\cdot q(x)\ .$ So that g(x) divides (f(x)-r(x)) with a remainder of zero.

Notice that in your problem the divisor, x2 - 5x +6, is factorable. Apply the factor theorem to f(x)-r(x), with each of those factors.

6. Oct 4, 2012

### HerroFish

Okay I get it now!
Thanks everybody for all your help!!! :!!)