Question

reslove into partial fraction 2x+5 /(x-3) (x+2)​

Answer 1

Given :-

$\sf\dfrac{2x + 5}{(x - 3)(x + 2)}$

To find :-

We have to resolve to partial Fraction

Solution:-

We have to resolve into partial Fraction So, Lets equation be

$\sf\dfrac{2x+5}{(x-3)(x+2)} = \sf\dfrac{A}{(x-3)}+\sf\dfrac{B}{(x+2)}$

Take L.C.M To the RHS

$\sf\dfrac{2x+5}{(x-3)(x+2)}= \sf\dfrac{A(x+2)+B(x-3)}{(x-3)(x+2)}$

2x + 5 = A(x+2)+B(x-3)

2x + 5 = Ax + 2A + Bx - 3B

2x + 5 = Ax + Bx + 2A - 3B

2x + 5 = x(A + B ) + 2A - 3B

Comparing both LHS And RHS

Comparing "x" terms

A + B = 2 ----1

Comparing constant term

2A - 3B = 5 ----- 2

Multiplying equation 1 with 2

Multiplying equation 2 with 1

(A + B = 2 )× 2

(2A - 3B= 5) × 1

2A + 2B = 4

2A - 3B = 5

Subtracting both equations

2A + 2B -( 2A -3B ) = 4-5

2A + 2B - 2A + 3B = -1

5B = -1

B = -1/5

Substitute value of B in equation 1

A + B = 2

A - 1/5 = 2

A = 2+1/5

A = 11/5

So, the value of A = 11/5 B = -1/5

Now, substitute value what we consider the equation

$\sf\dfrac{2x+5}{(x-3)(x+2)}= \sf\dfrac{A}{(x-3)}+\sf\dfrac{B}{(x+2)}$

$\sf\dfrac{2x+5}{(x-3)(x+2)} = \sf\dfrac{11/5}{x-3}+\sf\dfrac{-1/5}{x+2}$

$\sf\dfrac{2x+5}{(x-3)(x+2)} = \sf\dfrac{11}{5(x-3)}- \sf\dfrac{1}{5(x+2)}$

Hence Resolved !