Question

Find the remainder in the following division problem:
x^5-4x^4+x^3-7x+1/x+2

Answer 1

$P(x)=x^5-4x^4+x^3-7x+1,\ Q(x)=x+2\\\\P(x)=H(x)\cdot Q(x)+R$

Since the polynomial degree Q(x) is 1, the remainder of the division must be a number. Therefore we only need to calculate the value of polynomial for x = -2:

$P(-2)=(-2)^5-4(-2)^4+(-2)^3-7(-2)+1\\\\=-32-4(16)+(-8)+14+1\\\\=-32-64-8+15\\\\=-89$

Answer: The ramainder is equal -89.

$x^5-4x^4+x^3-7x+1=H(x)(x+2)-89$

Calculated in the program:

$x^5-4x^4+x^3-7x+1=(x^4-6x^3+13x^2-26x+45)(x+2)-89\\\Downarrow\\\dfrac{x^5-4x^4+x^3-7x+1}{x+2}=x^4-6x^3+13x^2-26x+45\ \ \ r(-89)$