If the remainders of the polynomial f(x) whendivided by x+1,x+2,x−2 are 6,15,3 respectively, then the remainder of f(x) when divided by (x+1)(x+2)(x−2) is
A
2x2−3x+1
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B
3x2−2x+1
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C
2x2−x−3
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D
3x2+2x+1
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Solution
The correct option is A2x2−3x+1 let f(x)=θ(x)(x+1)(x+2)(x−2)+γ(x) γ(x)=ax2+bx+c Since order of D(x)=3 So, f(x)=θ(x)(x+1)(x+2)(x−2)+ax2+bx+c f(−1)=a−b+c=b f(−2)=4a−2b+c=15 f(2)=4a+2b+c=3 Solving the 3 equations we get a=2;b=−3;c=1 So γ(x)=2x2−3x+1