If frac{27x^2+32x+16}{(3x+2)^2(1-x)} = frac{A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the partial fraction decomposition: $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$.Multiplying b

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Given the partial fraction decomposition: $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$.Multiplying both sides by $(3x+2)^2(1-x)$,we get:$27x^2+32x+16 = A(3x+2)(1-x) + B(1-x) + C(3x+2)^2$.Setting $x = 1$: $27(1)^2 + 32(1) + 16 = C(3(1)+2)^2$ $\Rightarrow 75 = 25C$ $\Rightarrow C = 3$.Setting $x = -2/3$: $27(-2/3)^2 + 32(-2/3) + 16 = B(1 - (-2/3))$ $\Rightarrow 12 - 64/3 + 16 = B(5/3)$ $\Rightarrow 20/3 = 5B/3$ $\Rightarrow B = 4$.Setting $x = 0$: $16 = A(2)(1) + B(1) + C(2)^2$ $\Rightarrow 16 = 2A + 4 + 4(3)$ $\Rightarrow 16 = 2A + 16$ $\Rightarrow A = 0$.Finally,$AB + BC + CA = (0)(4) + (4)(3) + (3)(0) = 0 + 12 + 0 = 12$.

If $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$,then $AB+BC+CA =$

Similar Questions

Resolve the following expression into partial fractions: $\frac{x + 1}{(x - 1)(x - 2)(x - 3)}$

If $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ and $f(x) = Ax^3 + Bx^2 + 7x + C$, then $5C - f'(-2) = $

If $\frac{3x}{(x-a)(x-b)} = \frac{2}{x-a} + \frac{1}{x-b}$,then $a:b$ is equal to

If $\frac{x^4}{(x^2+1)(x-2)}=f(x)+\frac{Ax+B}{x^2+1}+\frac{C}{x-2}$, then $f(14)+2A-B=$ (in $C$)

If $\frac{x^2+5}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$, then $A+B+C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module