Question

Find s and r if (x+3) is a factor of sx^3 + rx^2 -28x +15 and has a remainder of -60 when divided by(x-3)

Answer 1

Let us find $s$ and $r$ if $(x+3)$ is a factor of $sx^3 + rx^2 -28x +15$ and has a remainder of $-60$ when divided by $(x-3)$.

By the polynomial remainder theorem (little Bézout's theorem), the remainder of the division of a polynomial $\displaystyle f(x)=sx^3 + rx^2 -28x +15$ by a linear polynomial $\displaystyle x-a$ is equal to $\displaystyle f(a)$. Therefore, we have the following system of linear equations

$\begin{cases} s(-3)^3 + r(-3)^2 -28(-3) +15=0\\ s\cdot 3^3 + r\cdot 3^2 -28\cdot 3 +15=-60 \end{cases}$

which is equivalent to the following systems:

$\begin{cases} -27s + 9r+99=0\\ 27s + 9r-9=0 \end{cases}$

$\begin{cases} -27s + 9r+99=0\\ 18r+90=0 \end{cases}$

$\begin{cases} -27s =-9r-99\\ r=-5 \end{cases}$

$\begin{cases} -27s =-54\\ r=-5 \end{cases}$

$\begin{cases} s =2\\ r=-5 \end{cases}$

Answer: $s=2,\ r =-5.$