Remainder & Factor Thm

**acc100jt** · September 9th 2008, 10:51 PM

The expression $g(x)$ has the same remainder constant $k$ when divided by $(x-a)$ or $(x-b)$ . Determine, with reasons, whether it is always true that $g(x)$ also has remainder constant $k$ when divided by $(x-a)(x-b), k \neq 0$ .

My proof:

$g(x)=(x-a)Q_{1}(x)+k$
$g(x)=(x-b)Q_{2}(x)+k$

where $Q_{1}(x)$ and $Q_{2}(x)$ are quotients.

Let $f(x)=g(x)-k$ ,
then $f(x)=(x-a)Q_{1}(x)$ and $f(x)=(x-b)Q_{2}(x)$ .

So $(x-a)$ and $(x-b)$ are factors of $f(x)$

Thus,
$f(x)=(x-a)(x-b)Q_{3}(x)$
$\Rightarrow g(x)-k=(x-a)(x-b)Q_{3}(x)$
$\Rightarrow g(x)=(x-a)(x-b)Q_{3}(x)+k$

Hence, it is TRUE.

Is my proof ok? Is there a better one?
Many thanks for those who help

**Moo** · September 10th 2008, 03:18 AM

Hello,

Your proof is ok

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