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Question

# If the polynomials 2x3+ax2+3xâˆ’5 and x3+x2âˆ’4x+a leave the same remainder when divided by xâˆ’2, find the value of a

A
23
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B
133
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C
103
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D
None of the above
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Solution

## The correct option is A −133Given polynomial 2x3+ax2+3x−5 and x3+x2−4x+a divided by x-2 the remainder is sameThen x+2= or x=2 replace x by 2 wegetp(x)=2x3+ax2+3x−5 p(2)=2(2)3+a(2)2+3(2)−5⇒p(2)=2×8+a×4+3×2−5⇒P(2)=16+4a+6−5⇒p(2)=4a+17q(x)=x3+x2−4x+a⇒q(2)=(2)3+(2)2−4(2)+a⇒q(2)=8+4−8+a⇒A(2)=4+a∴4a+17=4+aAdd -a and -17 both side we get4a+17−a−17=4+a−a−17⇒3a=−13Divided both side by 3 we geta=−133

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