When x^4 + ax^3 + bx +c is divided by (x–1), (x+1), and (x+2), the remainders are 14, 0, and -28 respectively. Find the values of a, b, and c.

Guest Apr 23, 2022

#1**+1 **

By factor theorem, if a polynomial f(x) leaves a remainder of r when divided by x - a, then f(a) is the remainder.

Therefore, we have

\(\begin{cases}1^4 + a\cdot 1^3 + b\cdot 1 + c = 14\\ (-1)^4 + a\cdot (-1)^3 + b\cdot (-1) + c = 0\\ (-2)^4 + a\cdot (-2)^3 + b\cdot (-2) + c = -28\end{cases}\)

Simplifying,

\(\begin{cases}1+ a+b+c= 14\\ 1-a-b+c = 0\\ 16-8a-2b+ c = -28\end{cases}\)

If you solve the system, you will get \((a,b,c) = (6,1,6)\). Please try to fill in the solving process on your own.

P.S. You can solve the system by method of elimination, method of substitution, Gaussian elimination, or whatever method you know.

MaxWong Apr 23, 2022