Let .

f(x) = x^{10}+5x^9-8x^8+7x^7-6x^6+12x^5-4x^4-8x^3+12x^2-5x-5

Without using long division, find the remainder when f(x) is divided by x^2 - 1.

Guest May 7, 2022

#1**+1 **

Let q(x) be the quotient when f(x) is divided by x^2 - 1.

Since the divisor is quadratic, the degree of the remainder must be at most 1. Let ax + b be the required remainder.

Then \(f(x) = (x^2- 1)q(x) + (ax + b)\).

Note that \(f(1) = (1^2 - 1)q(1) + a + b = a + b\), but on the other hand, we can find f(1) by substituting x = 1 into \(f(x) = x^{10}+5x^9-8x^8+7x^7-6x^6+12x^5-4x^4-8x^3+12x^2-5x-5\) directly, and that gives f(1) = 1.

So a + b = 1. Similarly, we can get -a + b = -21 by considering f(-1).

Now can you solve for the values of a and b with \(\begin{cases}a + b = 1\\-a + b = -21\end{cases}\)? After that, the remainder is just ax + b, with the values of a and b you got.

MaxWong May 8, 2022