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Let $P(x)$ be a polynomial, which when $\begin{array}{llll}\text { divided by } & (x-3) & \text { and } & (x-5) & \text { leaves }\end{array}$ remainders 10 and 6, respectively. If the polynomial is divided by $(x-3)(x-5),$ thes the remainder is
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Verified Answer
The correct answer is:
$-2 x+16$
$\because P(x)=(x-3)(x-5)$
$=O(x)+(a x+b)$
Given, $\quad P(3)=10$ and $P(5)=6$
$\Rightarrow \quad 3 a+b=10$
and $\quad 5 a+b=6$
On solving Bqs.
(i) and (ii), we get $a=-2$ and $b=16$
$\therefore$ Remainder $=-2 x+16$
$=O(x)+(a x+b)$
Given, $\quad P(3)=10$ and $P(5)=6$
$\Rightarrow \quad 3 a+b=10$
and $\quad 5 a+b=6$
On solving Bqs.
(i) and (ii), we get $a=-2$ and $b=16$
$\therefore$ Remainder $=-2 x+16$
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