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Question

If the polynomials 2x3+ax2+3x5 and x3+x24x+a leave the same remainder when divided by x2, 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 133
Given polynomial 2x3+ax2+3x5 and x3+x24x+a divided by x-2 the remainder is same
Then x+2= or x=2 replace x by 2 weget
p(x)=2x3+ax2+3x5
p(2)=2(2)3+a(2)2+3(2)5
p(2)=2×8+a×4+3×25
P(2)=16+4a+65
p(2)=4a+17
q(x)=x3+x24x+a
q(2)=(2)3+(2)24(2)+a
q(2)=8+48+a
A(2)=4+a
4a+17=4+a
Add -a and -17 both side we get
4a+17a17=4+aa17
3a=13
Divided both side by 3 we get
a=133

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