1. ## Division of polynomials

Hi all

We're supposed to use the remainder theorem (f(p) = r) in this question.

When $\displaystyle x^3 + ax^2 + bx + 3$ is divided by (x-5), the remainder is 2. When it is divided by (x+3), the remainder is 7. What are the values of a and b?

I've set
f(5)=2 and f(-3)=7, and then isolated a and b, and used substitution to get 2839/270 for a, and 1909/30 for b.

Did I do this correctly?

2. You can always check the answer by plugging in the values of a and b, and the dividing by the given polynomials and see if the remainders match. To be honest, the numbers seem very ugly.

The easiest way is to plug in the values of x and then have a system of equations for a and b.

When $\displaystyle x^3 + ax^2 + bx + 3$ is divided by $\displaystyle (x-5)$, the remainder is 2.
When it is divided by $\displaystyle (x+3)$, the remainder is 7.
What are the values of $\displaystyle a$ and $\displaystyle b$ ?

I've set: .$\displaystyle f(5)=2\text{ and }f(-3)=7$, .and then isolated $\displaystyle a$ and $\displaystyle b$,
and used substitution to get: .$\displaystyle a = \frac{2839}{270},\;b = \frac{1909}{30}$

Did I do this correctly? . . . . Your game plan is correct, but . . .

$\displaystyle \begin{array}{ccccccccc}f(5) \:=\:2\!: & 125 + 25a + 5b + 3 &=& 5 & \Rightarrow & 25a + 5b &=& \text{-}123 & {\color{blue}[1]} \\ f(\text{-}3)\:=\: 7\!: & \text{-}27 + 9a - 3b + 3 &=& 7 & \Rightarrow & 9a - 3b &=& 31 & {\color{blue}[2]} \end{array}$

$\displaystyle \begin{array}{ccccc}\text{Multiply {\color{blue}[1]} by 3:} & 75a + 15b &=& \text{-}369 \\ \text{Multiply {\color{blue}[2]} by 5:} & 45a - 15b &=& 155 \end{array}$

. . Add: .$\displaystyle 120a \:=\:-214 \quad\Rightarrow\quad\boxed{ a \:=\:-\tfrac{107}{60}}$

Substitute into [1]: .$\displaystyle 25\left(\text{-}\tfrac{107}{60}\right) + 5b \:=\:\text{-}123 \quad\Rightarrow\quad \boxed{b \:=\:-\tfrac{941}{60}}$

The function is: .$\displaystyle f(x) \;=\;x^3 - \frac{107}{60}x^2 - \frac{941}{60}x + 3$