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# Polynomial

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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.

May 7, 2022

#1
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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.

May 8, 2022