# Quick Math Partial Fractions

## Quick Math Partial Fractions

1
##### Enter expression in this form

Example: (7x+1)/((x-1)(x+2))

### Partial Fractions in Algebra

Partial fractions is a technique used to decompose a complex rational expression into a sum of simpler fractions. This is especially useful in calculus for integrating rational functions. The basic idea is to express a rational function as a sum of fractions whose denominators are the factors of the original denominator.

#### Steps for Decomposing into Partial Fractions:

1. Factor the Denominator: Factor the denominator of the rational expression into irreducible factors.
2. Set Up the Partial Fractions: Write the expression as a sum of fractions with unknown coefficients. The form of these fractions depends on the factors of the denominator:
– For each linear factor $$(ax + b)$$, use $$\frac{A}{ax + b}$$.
– For each repeated linear factor $$(ax + b)^n$$, use $$\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$$.
– For each irreducible quadratic factor $$(ax^2 + bx + c)$$, use $$\frac{Ax + B}{ax^2 + bx + c}$$.
3. Combine and Solve: Combine the partial fractions over a common denominator and solve for the unknown coefficients by equating the numerator to the original numerator.

#### Detailed Example:

Decompose $$\frac{7x + 1}{(x – 1)(x + 2)}$$ into partial fractions.

1. Factor the Denominator:
– The denominator is already factored as $$(x – 1)(x + 2)$$

2. Set Up the Partial Fractions:
$\frac{7x + 1}{(x – 1)(x + 2)} = \frac{A}{x – 1} + \frac{B}{x + 2}$

3. Combine and Solve:
$7x + 1 = A(x + 2) + B(x – 1)$
Expand and combine like terms:
$7x + 1 = Ax + 2A + Bx – B$
$7x + 1 = (A + B)x + (2A – B)$
Equate coefficients:
– For $$x$$: $$A + B = 7$$
– For the constant term: $$2A – B = 1$$

Solve the system of equations:
1. $$A + B = 7$$
2. $$2A – B = 1$$

$(A + B) + (2A – B) = 7 + 1$
$3A = 8 \implies A = \frac{8}{3}$
Substitute $$A = \frac{8}{3}$$ into $$A + B = 7$$:
$\frac{8}{3} + B = 7 \implies B = 7 – \frac{8}{3} = \frac{21}{3} – \frac{8}{3} = \frac{13}{3}$
$\frac{7x + 1}{(x – 1)(x + 2)} = \frac{\frac{8}{3}}{x – 1} + \frac{\frac{13}{3}}{x + 2}$
$\frac{7x + 1}{(x – 1)(x + 2)} = \frac{8}{3(x – 1)} + \frac{13}{3(x + 2)}$
So, the partial fraction decomposition of $$\frac{7x + 1}{(x – 1)(x + 2)}$$ is:
$\frac{7x + 1}{(x – 1)(x + 2)} = \frac{8}{3(x – 1)} + \frac{13}{3(x + 2)}$