$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$
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?
$\endgroup$2 Answers
$\begingroup$Hint $\ x=1\,\Rightarrow\,c = 6.\,$ Cancelling $\,x-1\,$ yields $6x^2-5x+1 = (Ax-1)(Bx-1)\ $ so $\,\ldots$
$\endgroup$ $\begingroup$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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