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Question

Find the value of a in the equation (x+1)(x+2)(x+3)=x3+6x2+(ax+b)

A
11
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B
5
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C
4
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D
1
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Solution

The correct option is A 11
The given equation is,
(x+1)(x+2)(x+3)=x3+6x2+ax+b (1)

Now, (x+1)(x+2)=x2+3x+2
Putting this value in equation (1), we get,
(x2+3x+2)(x+3)=x3+6x2+ax+b

x3+3x2+3x2+9x+2x+6=x3+6x2+ax+b

x3+6x2+11x+6=x3+6x2+ax+b

Comparing coefficients of x3, x2, x and constat terms on LHS and RHS, we get,
a=11 and b=6
Thus, Answer is option (A)


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