Let $ f(x) \in C [x] .$
Suppose $ f(-1+i) = 2+5i $ and $ f(-2-i)=-3. $
Determine the remainder of f(x) divided by $(x+1-i)(x+2+i). $
How would i begin with this question, like how would i determine what f(x) is to begin with?
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$\begingroup$ Do you know how to approach this question if all if the coeffient of complex numbers were 0? $\endgroup$– Calvin LinMar 29, 2014 at 5:21
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$\begingroup$ Yeah then its regular long division right? $\endgroup$– user136088Mar 29, 2014 at 5:42
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1 Answer
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HINT:
Let $\displaystyle f(x)=A(x+1-i)(x+2+i)+B(x+2+i)+C(x+1-i)$
Put $x=-1+i$ and $-2-i$ one by one to find $B,C$
Alternatively,
let $\displaystyle f(x)=A(x+1-i)(x+2+i)+Bx+C$
Put $x=-1+i$ and $-2-i$ one by one to find $B,C$
In either cases, $B,C$ are arbitrary constants and $A$ is a polynomial
answered Mar 29, 2014 at 5:13
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$\begingroup$ What are $A,B,C$? Numbers? Polynomials? $\endgroup$ Mar 29, 2014 at 6:02
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$\begingroup$ @GerryMyerson, arbitrary constants independent of $x$ $\endgroup$ Mar 29, 2014 at 6:03
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$\begingroup$ So, you are assuming $f$ is a quadratic? $\endgroup$ Mar 29, 2014 at 6:03
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$\begingroup$ @GerryMyerson, sorry $B,C$ are arbitrary constants, $A$ is a polynomial $\endgroup$ Mar 29, 2014 at 6:05
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$\begingroup$ Maybe that information should be edited into your answer. $\endgroup$ Mar 29, 2014 at 6:06