1. ## find values

hi, i've got the question:

the polynomials f(x) and g(x) are defined by $\displaystyle f(x) = x^3 + px^2 - x + 5$, $\displaystyle g(x) = x^3 - x^2 + px + 1$ where p is a constant. when f(x) and g(x) are divided by x - 2 the remainder is R in each case. find the values of p and R

i got it down to
R = f(2) = 8 + 4p - 2 + 5 = 11 + 4p and
R = g(2) = 8 - 4 + 2p + 1 = 5 + 2p but i'm not sure if either of those are right.

can someone show me the way to do it please? thanks

2. Originally Posted by mark
hi, i've got the question:

the polynomials f(x) and g(x) are defined by $\displaystyle f(x) = x^3 + px^2 - x + 5$, $\displaystyle g(x) = x^3 - x^2 + px + 1$ where p is a constant. when f(x) and g(x) are divided by x - 2 the remainder is R in each case. find the values of p and R

i got it down to
R = f(2) = 8 + 4p - 2 + 5 = 11 + 4p and
R = g(2) = 8 - 4 + 2p + 1 = 5 + 2p but i'm not sure if either of those are right.

can someone show me the way to do it please? thanks
I would start by dividing each by x-2, then set the remainders equal.

3. $\displaystyle \frac{x^{3}+px^{2}-x+5}{x-2}=\underbrace{\frac{4p+11}{x-2}}_{\text{remainder}}$

$\displaystyle \frac{x^{3}-x^{2}+px+1}{x-2}=\underbrace{\frac{2p+5}{x-2}}_{\text{remainder}}$

$\displaystyle 4p+11=2p+5\Rightarrow p=-3$

$\displaystyle R=-1$

Therefore, the remainder in each case would be $\displaystyle \frac{-1}{x-2}$

$\displaystyle \frac{x^{3}-x^{2}-3x+1}{x-2}=\boxed{\frac{-1}{x-2}}+x^{2}+x-1$

$\displaystyle \frac{x^{3}-3x^{2}-x+5}{x-2}=\boxed{\frac{-1}{x-2}}+x^{2}-x-3$

4. Galactus, your latexing method is as good as that of soroban, excellent!

5. ok i understand it now up to $\displaystyle \frac {-1}{x - 2}$, when you get to that, wouldn't x have to be 3 for the final answer to be -1? how do you know the value of x?

6. Yes, p=-3. Sub that in and we get R=-1