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Question

# On dividing 2x2+3x+1 by a linear polynomial g(x), the quotient is 2x – 1 and remainder is ‘r’, where rϵR, then g(x) is

A
x – 1
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B
x + 1
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C
x + 2
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D
x + 4
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Solution

## The correct option is C x + 2f(x)=2x2+3x+1=(2x−1)g(x)+r, where g (x) is the divisor. Let g(x)=(x−a) When f(x) is divided by x−a, the remainder is f(a) Thus, r=2a2+3a+1 Hence, 2x2+3x+1=(2x−1)(x−a)+2a2+3a+1 ⇒2x2+3x+1=2x2−2ax−x+a+2a2+3a+1 ⇒2x2+3x+1=2x2−(2a+1)x+2a2+4a+1 Comparing the coefficient of x on both sides, we get: 3=−(2a+1) ⇒4=−2a ⇒a=−2 Thus, g(x)=(x+2)

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