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Question

The polynomial ax3+bx2+x – 6 has (x+2) as a factor and leaves a remainder 4 when divided by (x–2). Find a and b.


A

0, 2

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B

0, 4

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C

2, 4

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D

2, 2

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Solution

The correct option is A

0, 2


Let p(x)=ax3+bx2+x6

By factor theorem, (x+2) can be a factor of p(x) only when p(2)=0
p(2)=a(2)3+b(2)2+(2)6=0
8a+4b8=0
2a+b=2...(i)

Also, when p(x) is divided by (x2) the remainder is 4.

p(2)=4

a(2)3+b(2)2+26=4
8a+4b+26=4
8a+4b=8
2a+b=2...(ii)

Adding equations (i) and (ii), we get (2a+b)+(2a+b)=2+2

2b=4b=2

Putting b=2 in (i), we get,

2a+2=2

2a=0a=0

Hence, a=0 and b=2


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