# Partial Fraction $\frac{x^3-x^2+5x-1}{2x^2+x-6}$

How to solve this ?

Express in partial fractions of improper algebraic fractions $\frac{x^3-x^2+5x-1}{2x^2+x-6}$?

I tried long division which I've learned last time, but it couldn't be solved? Can anyone explain to me? Thanks

• Doing long division gives ${x^3-x^2+5x-1 \over 2x^2+x-6}={1 \over 2}x - {3 \over 4} + {{35 \over 4}x - {11 \over 2} \over 2x^2+x-6}$. Then, note that $2x^2+x-6 = (2x-3)(x+2)$. Hopefully, you should be able to do the rest. – eloiprime Jun 1 '15 at 10:11

## 2 Answers

Note that the degree of the numerator is not less than that of the denominator. Using long division we have $$x^3-x^2+5x-1 = (\frac{1}{2}x - \frac{3}{4})(2x^2+x-6)+(\frac{35}{4}x-\frac{11}{2})$$ Now it suffices to perform partial fraction on the remainder $\frac{35}{4}x-\frac{11}{2}$. First we factorize the denominator as $$2x^2+x-6 = (2x-3)(x+2)$$ Then we write $$\frac{\frac{35}{4}x-\frac{11}{2}}{(2x-3)(x+2)} = \frac{A}{2x-3} + \frac{B}{x+2}$$ where the equal sign should be interpreted as an identity, i.e. holds true for all $x$. Multiplying the denominator $(2x-3)(x+2)$ to both sides yields $$\frac{35}{4}x-\frac{11}{2} = A(x+2)+B(2x-3) = (A+2B)x+(2A-3B)$$ Since it is true for all $x$, the coefficients must match. Comparing like terms we have $$A+2B = \frac{35}{4}$$ $$2A-3B = -\frac{11}{2}$$ and solving this system of linear equations we have $$A = \frac{61}{28}$$ $$B = \frac{23}{7}$$ and finally $$\frac{x^3-x^2+5x-1}{2x^2+x-6} = \frac{1}{2}x - \frac{3}{4} + \frac{61}{28(2x-3)} + \frac{23}{7(x+2)}$$

To do partial fraction decomposition, you need to know the root of denominator. Here is the result (I got it by running Apart in Mathematica) $$-\frac{3}{4}+\frac{x}{2}+\frac{23}{7 (x+2)}+\frac{61}{28 (2 x-3)}$$