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$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$

Find the value of A, B and C.

I started it like this:

$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$

Solving the right hand side:

$ (ABx^2 - Ax - Bx + 1)(x - 1) + C$

$ ABx^3 - ABx^2 - Ax^2 + Ax - Bx^2 + Bx + x - 1 + C$

$ABx^3 - (AB + A + B)x^2 + (A + B + 1)x - 1 + C$

Comparing the coefficients:

$AB = 6$

$A = \frac6 B$

$AB + A + B = 11$

Then substitute the value of A in the above equation...is this right? Is there any error?

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2 Answers 2

up vote 1 down vote accepted

Equating x coefficients you will get A+B=5.You already have AB=6.Solving you will get A=2 or 3.This will be the easier approach.

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Hint $\ x=1\,\Rightarrow\,c = 6.\,$ Cancelling $\,x-1\,$ yields $6x^2-5x+1 = (Ax-1)(Bx-1)\ $ so $\,\ldots$

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