# How do you express (x^2 - 3x) / ((x-1)(x+2)) in partial fractions?

Apr 13, 2016

1-2/(3(x-1))-10/(3(x+2)

#### Explanation:

Firstly we must determine whether this type of fraction is proper fraction or improper fraction.

When we multiply the denominator of the fraction, the answer will be;

$\frac{{x}^{2} - 3 x}{{x}^{2} + x - 2}$

It is an improper fraction because the denominator (lower side of the fraction) is greater or the same as the numerator (upper side of the fraction, in terms of unknown power.

By then, to express it in partial fraction, we need to change it into proper fraction , which is by long division;

And we get the proper fraction of;

$\frac{{x}^{2} - 3 x}{\left(x - 1\right) \left(x + 2\right)} = 1 + \frac{- 4 x + 2}{\left(x - 1\right) \left(x + 2\right)}$

Until then, we can solve for the partial fraction by expressing form of the rational function into form of the partial fraction ;

$\frac{- 4 x + 2}{\left(x - 1\right) \left(x + 2\right)} = \frac{A}{x - 1} + \frac{B}{x + 2}$

We need to find the exact value of unknown A and B.

Equalize the denominator and we get;

(-4x+2)/((x-1)(x+2))=(A(x+2)+B(x-1))/((x-1)(x+2)

Cancel out the denominator on both side;

$- 4 x + 2 = A \left(x + 2\right) + B \left(x - 1\right)$

$- 4 x + 2 = A x + 2 A + B x - B$

Relate both side of equation in terms of unknown power by which;

In terms of unknown $x$;

$- 4 x = A x + B x$

$- 4 = A + B$ --- equation $\left(1\right)$

In terms of unknown ${x}^{0}$ or real number

$2 = 2 A - B$ --- equation $\left(2\right)$

From equation $\left(2\right)$;

$B = 2 A - 2$ --- equation $\left(3\right)$

Substitute equation $\left(3\right)$ into equation $\left(1\right)$

$- 4 = A + \left(2 A - 2\right)$

$A = - \frac{2}{3}$

Substitute $A = - \frac{2}{3}$ into equation $\left(3\right)$

$B = 2 \left(- \frac{2}{3}\right) - 2$

$B = - \frac{10}{3}$

Replace all the unknown in the partial fraction and we will get;

(x^2-3x)/((x-1)(x+2))=1-2/(3(x-1))-10/(3(x+2)