The sum of the partial fraction is [tex]\frac{2x + 4}{x(x + 2)(x -4)} = -\frac 12 + \frac{1}{2(x -4)}[/tex]
How to express as partial fractions?
The expression is given as:
[tex]\frac{2x + 4}{x(x + 2)(x -4)}[/tex]
As a partial fraction, we have:
[tex]\frac{2x + 4}{x(x + 2)(x -4)} = \frac Ax + \frac{B}{x + 2} + \frac{C}{x -4}[/tex]
Take the LCM
[tex]\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)}[/tex]
This gives
2x + 4 = A(x + 2)(x -4) +Bx(x -4) + Cx(x + 2)
Expand
[tex]2x + 4 = A(x^2 - 2x - 8) + B(x^2 - 4x) + C(x^2 + 2x)[/tex]
Further expand
[tex]2x + 4 = Ax^2 - 2Ax - 8A + Bx^2 - 4Bx + Cx^2 + 2Cx[/tex]
Collect like terms
[tex]2x + 4 = Ax^2 + Bx^2 + Cx^2 - 2Ax - 4Bx + 2Cx - 8A[/tex]
By comparing the coefficients, we have:
A + B + C = 0
-2A - 4B + 2C = 2
-8A = 4
Divide both sides of -8A = 4 by -8
[tex]A = -\frac 12[/tex]
Substitute [tex]A = -\frac 12[/tex] in the other equations
[tex]-\frac 12 + B + C = 0[/tex]
[tex]B + C = \frac 12[/tex]
[tex]C = \frac 12 - B[/tex]
[tex]-2*- \frac 12 - 4B + 2C = 2[/tex]
1 - 4B + 2C = 2
- 4B + 2C = 1
Substitute [tex]C = \frac 12 - B[/tex] in - 4B + 2C = 1
[tex]- 4B + 2*\frac12 = 1[/tex]
- 4B + 1 = 1
Subtract 1 from both sides
-4B = 0
This gives
B = 0
Substitute B = 0 in [tex]C = \frac 12 - B[/tex]
[tex]C = \frac 12 - 0[/tex]
[tex]C = \frac 12[/tex]
So, we have:
[tex]A = -\frac 12[/tex], B = 0 and [tex]C = \frac 12[/tex]
The equation [tex]\frac{2x + 4}{x(x + 2)(x -4)} = \frac Ax + \frac{B}{x + 2} + \frac{C}{x -4}[/tex] becomes
[tex]\frac{2x + 4}{x(x + 2)(x -4)} = -\frac 12 + \frac{0}{x + 2} + \frac{1}{2(x -4)}[/tex]
Evaluate
[tex]\frac{2x + 4}{x(x + 2)(x -4)} = -\frac 12 + \frac{1}{2(x -4)}[/tex]
Hence, the sum of the partial fraction is [tex]\frac{2x + 4}{x(x + 2)(x -4)} = -\frac 12 + \frac{1}{2(x -4)}[/tex]
Read more about partial fraction at:
https://brainly.com/question/18958301
#SPJ1
