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Question

If x3âˆ’6x2+ax+b is exactly divisible by x2âˆ’3x+2, then 12a+22b=?

A
132
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B
0.0
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C
264
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D
- 66
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Solution

The correct option is B 0.0Given, f(x)=x3−6x2+ax+b is exactly divisible by g(x)=x2−3x+2. Let the remainder be r(x). Then r(x)=0 x−3x2−3x+2 )¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x3−6x2+ax+b −x3∓3x2±2x––––––––––––––––– −3x2+(a−2)x+b∓3x2±9x∓6–––––––––––––––––––––– (a−11)x+b+6 ∴r(x)=(a−11)x+(b+6) (a−11)x+(b+6)=0 The polynomial will be zero if a−11=0 & b+6=0 ∴a=11 & b=−6 ∴12a+22b=12×11+22×(−6) =0

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