# How do you express as a partial fraction (3x+18)/(x^2+5x+4)?

Apr 5, 2018

$- \frac{2}{x + 4} \mathmr{and} \frac{5}{x + 1}$

#### Explanation:

the expression in the denominator can be factorised to give the denominator of each partial fraction.

${x}^{2} + 5 x + 4 = \left(x + 4\right) \left(x + 1\right)$

the denominators of the partial fractions are $\left(x + 4\right)$ and $\left(x + 1\right)$.

the numerators are unknown; therefore, it is easiest to label them as two variables (e.g. $a$ and $b$).

if this is done, then the partial fractions are $\frac{a}{x + 4} \mathmr{and} \frac{b}{x + 1}$.

$\frac{a}{x + 4} + \frac{b}{x + 1} = \frac{3 x + 18}{{x}^{2} + 5 x + 4}$

to get rid of the denominator of all of the fractions, you can multiply them by $\left({x}^{2} + 5 x + 4\right)$, or $\left(x + 4\right) \left(x + 1\right)$.

$\frac{a}{x + 4} \cdot \left({x}^{2} + 5 x + 4\right) = a \left(x + 1\right)$

$\frac{b}{x + 1} \cdot \left({x}^{2} + 5 x + 4\right) = b \left(x + 4\right)$

$\frac{3 x + 18}{{x}^{2} + 5 x + 4} \cdot \left({x}^{2} + 5 x + 4\right) = 3 x + 18$

since $\frac{a}{x + 4} + \frac{b}{x + 1} = \frac{3 x + 18}{{x}^{2} + 5 x + 4}$,

$a \left(x + 1\right) + b \left(x + 4\right) = 3 x + 18$.

multiplying brackets out gives

$a x + a + b x + 4 b = 3 x + 18$

then you can group the $x$ terms and the constant terms:

$a x + b x + a + 4 b = 3 x + 18$

$\left(a + b\right) x + a + 4 b = 3 x + 18$

for both sides to be equal, the coefficients of $x$ and the coefficients of $1$ must be equal to each other.

$\left(a + b\right) x = 3 x$
$a + 4 b = 18$

$a + b = 3$
$a + 4 b = 18$

b can be solved for:

$a + 4 b = a + b + 15$

$4 b = b + 15$

$3 b = 15$

$b = 5$

and so can a:

$a + b = 3$

$a + 5 = 3$

$a = 3 - 5$

$a = - 2$

these values for $a$ and $b$ can be substituted into the numerators of the partial fractions $\frac{a}{x + 4}$ and $\frac{b}{x + 1}$,

giving $- \frac{2}{x + 4} \mathmr{and} \frac{5}{x + 1}$

to check:

$- \frac{2}{x + 4} + \frac{5}{x + 1} = \frac{- 2 \left(x + 1\right)}{{x}^{2} + 5 x + 4} + \frac{5 \left(x + 4\right)}{{x}^{2} + 5 x + 4}$

$= \frac{\left(- 2 x - 2\right) + \left(5 x + 20\right)}{{x}^{2} + 5 x + 4}$

$= \frac{3 x + 18}{{x}^{2} + 5 x + 4}$