### 6 Improper fractions

When calculating the partial fractions of improper fractions an extra polynomial is added to any partial fractions that would normally arise. The added polynomial has degree $n-d$ where $n$ is the degree of the numerator and $d$ is the degree of the denominator. Recall that

a polynomial of degree 0 is a constant, $A$ say,

a polynomial of degree 1 has the form $Ax+B$ ,

a polynomial of degree 2 has the form $A{x}^{2}+Bx+C$ ,

and so on.

If, for example, the improper fraction is such that the numerator has degree 5 and the denominator has degree 3, then $n-d=2$ , and we need to add a polynomial of the form $A{x}^{2}+Bx+C$ .

##### Key Point 17

If a fraction is improper an additional term is included taking the form of a polynomial of degree $n-d$ , where $n$ is the degree of the numerator and $d$ is the degree of the denominator.

##### Example 42

Express as partial fractions

$\phantom{\rule{2em}{0ex}}\frac{2{x}^{2}-x-2}{x+1}$

##### Solution

The fraction is improper because $n=2$ , $d=1$ and so $d\le n$ . Here $n-d=1$ , so we need to include as an extra term a polynomial of the form $Bx+C$ , in addition to the usual partial fractions. The linear term in the denominator gives rise to a partial fraction $\frac{A}{x+1}$ . So altogther we have

$\phantom{\rule{2em}{0ex}}\frac{2{x}^{2}-x-2}{x+1}=\frac{A}{x+1}+\left(Bx+C\right)$

Multiplying both sides by $x+1$ we find

$\phantom{\rule{2em}{0ex}}2{x}^{2}-x-2=A+\left(Bx+C\right)\left(x+1\right)=B{x}^{2}+\left(C+B\right)x+\left(C+A\right)$

Equating coefficients of ${x}^{2}$ gives $B=2$ .

Equating coefficients of $x$ gives $-1=C+B$ and so $C=-1-B=-3$ .

Equating the constant terms gives $-2=C+A$ and so $A=-2-C=-2-\left(-3\right)=1$ .

Finally, we have

$\phantom{\rule{2em}{0ex}}\frac{2{x}^{2}-x-2}{x+1}=\frac{1}{x+1}+2x-3$

##### Exercise

Express each of the following improper fractions in terms of partial fractions.

- $\frac{x+3}{x+2}$ ,
- $\frac{3x-7}{x-3}$ ,
- $\frac{{x}^{2}+2x+2}{x+1}$ ,
- $\frac{2{x}^{2}+7x+7}{x+2}$
- $\frac{3{x}^{5}+4{x}^{4}-21{x}^{3}-40{x}^{2}-24x-29}{{\left(x+2\right)}^{2}\left(x-3\right)}$ ,
- $\frac{4{x}^{5}+8{x}^{4}+23{x}^{3}+27{x}^{2}+25x+9}{\left({x}^{2}+x+1\right)\left(2x+1\right)}$

- $1+\frac{1}{x+2}$ ’
- $3+\frac{2}{x-3}$ ,
- $1+x+\frac{1}{x+1}$
- $2x+3+\frac{1}{x+2}$ ,
- $\frac{1}{{\left(x+2\right)}^{2}}+\frac{1}{x+2}+\frac{1}{x-3}+3{x}^{2}+x+2$ ,
- $2{x}^{2}+x+7+\frac{1}{2x+1}+\frac{1}{{x}^{2}+x+1}$