How do you express $\frac{- 2 x - 3}{x^{2} - x}$ $(-2x-3)/(x^2-x)$ in partial fractions?

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Answer 1 · 2016-02-20T09:53:58

$\frac{3}{x} - \frac{5}{x - 1}$ $3/x-5/(x-1)$

To write algebraic expression $\frac{- 2 x - 3}{x^{2} - x}$ $(−2x−3)/(x^2−x)$ ,

we first factorize $(x^{2} - x)$ $(x^2−x)$ i.e. $x (x - 1)$ $x(x−1)$ .

Hence partial fractions will be of he type

$\frac{A}{x} + \frac{B}{x - 1}$ $A/x+B/(x-1)$ or $\frac{A x - A + B x}{x (x - 1)}$ $(Ax-A+Bx)/(x(x-1))$ or

$\frac{(A + B) x - A}{x (x - 1)}$ $((A+B)x-A)/(x(x-1))$

Hence $A + B = - 2$ $A+B=-2$

and $- A = - 3$ $-A=-3$ i.e. $A = 3$ $A=3$ and $B = - 5$ $B=-5$

Hence partial fractions are

$\frac{3}{x} - \frac{5}{x - 1}$ $3/x-5/(x-1)$

How do you express −2x−3x2−x(-2x-3)/(x^2-x) in partial fractions?