Question 1209683: Let f(x) = x^{10}+5x^9-8x^8+4x^7-x^6-33x^5+4x^4-22x^3+8x^2-3x-5.
Without using long division (which would be horribly nasty!), find the remainder when f(x) is divided by x^2 - 4.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let $f(x) = x^{10}+5x^9-8x^8+4x^7-x^6-33x^5+4x^4-22x^3+8x^2-3x-5$.
We want to find the remainder when $f(x)$ is divided by $x^2 - 4$.
Since the divisor is of degree 2, the remainder will be of the form $ax+b$.
Let $f(x) = (x^2-4)q(x) + ax+b$, where $q(x)$ is the quotient.
The roots of $x^2-4=0$ are $x=2$ and $x=-2$.
Substituting $x=2$ into the equation, we get
$f(2) = 2^{10} + 5(2^9) - 8(2^8) + 4(2^7) - 2^6 - 33(2^5) + 4(2^4) - 22(2^3) + 8(2^2) - 3(2) - 5 = 2a+b$
$f(2) = 1024 + 5(512) - 8(256) + 4(128) - 64 - 33(32) + 4(16) - 22(8) + 8(4) - 6 - 5$
$f(2) = 1024 + 2560 - 2048 + 512 - 64 - 1056 + 64 - 176 + 32 - 6 - 5 = 1024 + 2560 - 2048 + 512 - 64 - 1056 + 64 - 176 + 32 - 11 = 1321$
So, $2a+b = 1321$.
Substituting $x=-2$ into the equation, we get
$f(-2) = (-2)^{10} + 5(-2)^9 - 8(-2)^8 + 4(-2)^7 - (-2)^6 - 33(-2)^5 + 4(-2)^4 - 22(-2)^3 + 8(-2)^2 - 3(-2) - 5 = -2a+b$
$f(-2) = 1024 - 5(512) - 8(256) - 4(128) - 64 + 33(32) + 4(16) + 22(8) + 8(4) + 6 - 5$
$f(-2) = 1024 - 2560 - 2048 - 512 - 64 + 1056 + 64 + 176 + 32 + 1 = -2287$
So, $-2a+b = -2287$.
Adding the two equations, we get
$2b = 1321 - 2287 = -966$
$b = -483$
Substituting $b=-483$ into $2a+b=1321$, we get
$2a - 483 = 1321$
$2a = 1804$
$a = 902$
Therefore, the remainder is $902x - 483$.
Final Answer: The final answer is $\boxed{902x - 483}$
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