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Question

A polynomial in x of degree greater than 3, leaves remainders 2,1 and −1 when divided by (x−1),(x+2) and (x+1) respectively. What will be the remainder when it is divided by (x−1)(x+2)(x+1).

A
32x+23
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B
32x23
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C
76x232x23
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D
76x2+32x23
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Solution

The correct option is D 76x2+32x23
Given: 2,1 and 1 are remainders when polynomial is divided by (x1),(x+2) and (x+1)

Now, let the required polynomial be f(x)=p(x)(x1)(x+2)(x+1)+a0x2+a1x+a2

By remainder theorem, we know f(1)=2,f(2)=1,f(1)=1

f(1)=p(1)(11)(1+2)(1+1)+a0(1)2+a1(1)+a2=2
f(1)=a0+a1+a2=2(i)

f(2)=p(2)(21)(2+2)(2+1)+a0(2)2+a1(2)+a2=1
f(2)=4a02a1+a2=1(ii)

f(1)=p(1)(11)(1+2)(1+1)+a0(1)2+a1(1)+a2=1
f(1)=a0a1+a2=1(iii)

On solving (i),(ii),(iii) we get, a0=76,a1=32,a2=23

f(x)=p(x)(x1)(x+2)(x1)+76x2+32x23

Now if we divide f(x) by (x1)(x+2)(x+1) remainder will be 76x2+32x23

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