(A) Let the remainder be $ax+b$ when $P(x)$ is divided by $(x-3)(x-5)$.$P(x) = (x-3)(x-5)Q(x) + (ax+b)$Given that $P(3) = 10$ and $P(5) = 6$.Substitut

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(A) Let the remainder be $ax+b$ when $P(x)$ is divided by $(x-3)(x-5)$.$P(x) = (x-3)(x-5)Q(x) + (ax+b)$Given that $P(3) = 10$ and $P(5) = 6$.Substituting $x=3$ in the equation: $3a+b = 10$ $(i)$Substituting $x=5$ in the equation: $5a+b = 6$ $(ii)$Subtracting equation $(i)$ from $(ii)$:$(5a+b) - (3a+b) = 6 - 10$$2a = -4 \Rightarrow a = -2$Substituting $a = -2$ in equation $(i)$:$3(-2) + b = 10$$-6 + b = 10 \Rightarrow b = 16$Thus, the remainder is $-2x+16$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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Let $P(x)$ be a polynomial, which when divided by $(x-3)$ and $(x-5)$ leaves remainders $10$ and $6$, respectively. If the polynomial is divided by $(x-3)(x-5)$, then the remainder is

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A
$-2x+16$
B
$16$
C
$2x-16$
D
$60$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Solution of quadratic equations and Nature of roots

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Let $P(x)$ be a polynomial, which when divided by $(x-3)$ and $(x-5)$ leaves remainders $10$ and $6$, respectively. If the polynomial is divided by $(x-3)(x-5)$, then the remainder is

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