Author Topic: POLYNOMIAL HELP (Read 4183 times) Tweet Share

sarahtorrington · « **on:** July 26, 2016, 09:33:48 am »

Hey guys, I'm stuck on a question from the CSSA 2015 trial paper! Any help would be greatly appreciated!!

RuiAce · « **Reply #1 on:** July 26, 2016, 09:39:25 am »

Quote from: sarahtorrington on July 26, 2016, 09:33:48 am

Hey guys, I'm stuck on a question from the CSSA 2015 trial paper! Any help would be greatly appreciated!!

(Image removed from quote.)

$\text{By the division transformation, division by a quadratic factor yields a remainder of at most degree 1.}\\ \text{Hence, let }P(x)=(x-1)(x-2)Q(x) + (ax+b)$
$\text{By the remainder theorem, we know that }\\ \begin{align*} P(1)=3 &\implies 3=a+b \\ P(2)=5&\implies 5=2a+b \end{align*}\\ \text{Note: Upon substitution of }1\text{ or }2, (x-1)(x-2)Q(x)\text{ goes to 0}$
$\text{Solving these equations simultaneously we get }a=2, b=1$
$\text{So }P(x)=(x-1)(x-2)Q(x) + (2x+1)\\ \text{Hence the remainder is }2x+1$

sarahtorrington · « **Reply #2 on:** July 26, 2016, 09:45:17 am »

Thanks for the quick reply, but does that mean the answer is simply 2x +b or otherwise?

RuiAce · « **Reply #3 on:** July 26, 2016, 09:49:25 am »

b is equal to 1

But yeah.

Quote from: RuiAce on July 26, 2016, 09:39:25 am

$\text{So }P(x)=(x-1)(x-2)Q(x) + (2x+1)\\ \text{Hence the remainder is }2x+1$

sarahtorrington · « **Reply #4 on:** July 26, 2016, 09:57:54 am »

Quote from: RuiAce on July 26, 2016, 09:49:25 am

b is equal to 1

But yeah.

Thank you!!

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Author Topic: POLYNOMIAL HELP (Read 4183 times) Tweet Share

sarahtorrington

POLYNOMIAL HELP

RuiAce

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sarahtorrington

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RuiAce

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