How to decompose in partial fractions 1/(2x^2+4x)?

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The fraction `1/(2x^2+4x)` has to be decomposed into partial fractions. First factorize the denominator.

`2x^2+4x`

= `2x(x +2)`

Now take two constants a and b

`1/(2x^2+4x) = a/(2x) + b/(x + 2)`

This has to be solved for a and b.

`1/(2x^2+4x) = (a*(x+2) + b*2x)/((2x)(x + 2))`

`1/(2x^2+4x) = (ax + 2a + 2bx)/((2x)(x + 2))`

Equate the terms containing x.

a + 2b = 0

Another equation that can be derived is 2a = 1

a = 1/2

Substituting in a + 2b = 0

b = -1/4

The partial fraction decomposition is `1/(4x) - 1/(4*(x+2))`

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll factorize by 2x the denominator:

1/(2x^2+4x) = 1/2x(x+2)

We'll suppose that the fraction 1/2x(x+2) is the result of addition or subtraction of 2 irreducible partial fractions:

1/2x(x+2) = A/2x + B/(x+2) (1)

We'll multiply the ratio A/2x by (x+2) and we'll multiply the ratio B/(x+2) by 2x.

1/2x(x+2)= [A(x+2) + 2Bx]/2x(x+2)

Since the denominators of both sides are matching, we'll write the numerators, only.

1 = A(x+2) + 2Bx

We'll remove the brackets:

1 = Ax + 2A + 2Bx

We'll factorize by x to the right side:

1 = x(A+2B) + 2A

Comparing, we'll get:

A+2B = 0

2A = 1 => A = 1/2

1/2 + 2B = 0

B = -1/4

We'll substitute A and B into the expression (1):

The result of decomposition into partial fractions is: 1/2x(x+2) = 1/4x - 1/4(x+2)

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