Question

Someone pls help me ASAP! Express the following as the sum of its partial fraction: 2x+4/x(x+2)(x-4)​

Answer 1

The sum of the partial fraction is $\frac{2x + 4}{x(x + 2)(x -4)} = -\frac 12 + \frac{1}{2(x -4)}$

How to express as partial fractions?

The expression is given as:

$\frac{2x + 4}{x(x + 2)(x -4)}$

As a partial fraction, we have:

$\frac{2x + 4}{x(x + 2)(x -4)} = \frac Ax + \frac{B}{x + 2} + \frac{C}{x -4}$

Take the LCM

$\frac{2x + 4}{x(x + 2)(x -4)} = \frac {A(x + 2)(x -4) +Bx(x -4) + Cx(x + 2)}{x(x + 2)(x -4)}$

This gives

2x + 4 = A(x + 2)(x -4) +Bx(x -4) + Cx(x + 2)

Expand

$2x + 4 = A(x^2 - 2x - 8) + B(x^2 - 4x) + C(x^2 + 2x)$

Further expand

$2x + 4 = Ax^2 - 2Ax - 8A + Bx^2 - 4Bx + Cx^2 + 2Cx$

Collect like terms

$2x + 4 = Ax^2 + Bx^2 + Cx^2 - 2Ax - 4Bx + 2Cx - 8A$

By comparing the coefficients, we have:

A + B + C = 0

-2A - 4B + 2C = 2

-8A = 4

Divide both sides of -8A = 4 by -8

$A = -\frac 12$

Substitute $A = -\frac 12$ in the other equations

$-\frac 12 + B + C = 0$

$B + C = \frac 12$

$C = \frac 12 - B$

$-2*- \frac 12 - 4B + 2C = 2$

1 - 4B + 2C = 2

- 4B + 2C = 1

Substitute $C = \frac 12 - B$ in - 4B + 2C = 1

$- 4B + 2*\frac12 = 1$

- 4B + 1 = 1

Subtract 1 from both sides

-4B = 0

This gives

B = 0

Substitute B = 0 in $C = \frac 12 - B$

$C = \frac 12 - 0$

$C = \frac 12$

So, we have:

$A = -\frac 12$, B = 0 and $C = \frac 12$

The equation $\frac{2x + 4}{x(x + 2)(x -4)} = \frac Ax + \frac{B}{x + 2} + \frac{C}{x -4}$ becomes

$\frac{2x + 4}{x(x + 2)(x -4)} = -\frac 12 + \frac{0}{x + 2} + \frac{1}{2(x -4)}$

Evaluate

$\frac{2x + 4}{x(x + 2)(x -4)} = -\frac 12 + \frac{1}{2(x -4)}$

Hence, the sum of the partial fraction is $\frac{2x + 4}{x(x + 2)(x -4)} = -\frac 12 + \frac{1}{2(x -4)}$

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