# 11.4 Partial fractions  (Page 5/7)

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## Decomposing a rational function with a repeated irreducible quadratic factor in the denominator

Decompose the given expression that has a repeated irreducible factor in the denominator.

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

The factors of the denominator are $\text{\hspace{0.17em}}x,\left({x}^{2}+1\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\left({x}^{2}+1\right)}^{2}.\text{\hspace{0.17em}}$ Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form $\text{\hspace{0.17em}}Ax+B.\text{\hspace{0.17em}}$ So, let’s begin the decomposition.

$\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\frac{A}{x}+\frac{Bx+C}{\left({x}^{2}+1\right)}+\frac{Dx+E}{{\left({x}^{2}+1\right)}^{2}}$

We eliminate the denominators by multiplying each term by $\text{\hspace{0.17em}}x{\left({x}^{2}+1\right)}^{2}.\text{\hspace{0.17em}}$ Thus,

${x}^{4}+{x}^{3}+{x}^{2}-x+1=A{\left({x}^{2}+1\right)}^{2}+\left(Bx+C\right)\left(x\right)\left({x}^{2}+1\right)+\left(Dx+E\right)\left(x\right)$

Expand the right side.

Now we will collect like terms.

${x}^{4}+{x}^{3}+{x}^{2}-x+1=\left(A+B\right){x}^{4}+\left(C\right){x}^{3}+\left(2A+B+D\right){x}^{2}+\left(C+E\right)x+A$

Set up the system of equations matching corresponding coefficients on each side of the equal sign.

We can use substitution from this point. Substitute $\text{\hspace{0.17em}}A=1\text{\hspace{0.17em}}$ into the first equation.

Substitute $\text{\hspace{0.17em}}A=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B=0\text{\hspace{0.17em}}$ into the third equation.

Substitute $\text{\hspace{0.17em}}C=1\text{\hspace{0.17em}}$ into the fourth equation.

Now we have solved for all of the unknowns on the right side of the equal sign. We have $\text{\hspace{0.17em}}A=1,\text{\hspace{0.17em}}$ $B=0,\text{\hspace{0.17em}}$ $C=1,\text{\hspace{0.17em}}$ $D=-1,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}E=-2.\text{\hspace{0.17em}}$ We can write the decomposition as follows:

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

Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.

$\frac{{x}^{3}-4{x}^{2}+9x-5}{{\left({x}^{2}-2x+3\right)}^{2}}$

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

Access these online resources for additional instruction and practice with partial fractions.

## Key concepts

• Decompose $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}\text{\hspace{0.17em}}$ by writing the partial fractions as $\text{\hspace{0.17em}}\frac{A}{{a}_{1}x+{b}_{1}}+\frac{B}{{a}_{2}x+{b}_{2}}.\text{\hspace{0.17em}}$ Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. See [link] .
• The decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}\text{\hspace{0.17em}}$ with repeated linear factors must account for the factors of the denominator in increasing powers. See [link] .
• The decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}\text{\hspace{0.17em}}$ with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in $\text{\hspace{0.17em}}\frac{A}{x}+\frac{Bx+C}{\left(a{x}^{2}+bx+c\right)}.\text{\hspace{0.17em}}$ See [link] .
• In the decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)},\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}Q\left(x\right)\text{\hspace{0.17em}}$ has a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers as
$\frac{Ax+B}{\left(a{x}^{2}+bx+c\right)}+\frac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\cdots \text{+}\frac{{A}_{n}x+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}.$

## Verbal

Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction

No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example, $\text{\hspace{0.17em}}\frac{1}{{x}^{2}+1}\text{\hspace{0.17em}}$ cannot be decomposed because the denominator cannot be factored.

Can you explain why a partial fraction decomposition is unique? (Hint: Think about it as a system of equations.)

Can you explain how to verify a partial fraction decomposition graphically?

Graph both sides and ensure they are equal.

You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could double-check your answer.

Once you have a system of equations generated by the partial fraction decomposition, can you explain another method to solve it? For example if you had $\text{\hspace{0.17em}}\frac{7x+13}{3{x}^{2}+8x+15}=\frac{A}{x+1}+\frac{B}{3x+5},\text{\hspace{0.17em}}$ we eventually simplify to $\text{\hspace{0.17em}}7x+13=A\left(3x+5\right)+B\left(x+1\right).\text{\hspace{0.17em}}$ Explain how you could intelligently choose an $\text{\hspace{0.17em}}x$ -value that will eliminate either $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B.$

If we choose $\text{\hspace{0.17em}}x=-1,\text{\hspace{0.17em}}$ then the B -term disappears, letting us immediately know that $\text{\hspace{0.17em}}A=3.\text{\hspace{0.17em}}$ We could alternatively plug in $\text{\hspace{0.17em}}x=-\frac{5}{3},\text{\hspace{0.17em}}$ giving us a B -value of $\text{\hspace{0.17em}}-2.$

## Algebraic

For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors.

$\frac{5x+16}{{x}^{2}+10x+24}$

$\frac{3x-79}{{x}^{2}-5x-24}$

$\frac{8}{x+3}-\frac{5}{x-8}$

$\frac{-x-24}{{x}^{2}-2x-24}$

$\frac{10x+47}{{x}^{2}+7x+10}$

$\frac{1}{x+5}+\frac{9}{x+2}$

$\frac{x}{6{x}^{2}+25x+25}$

$\frac{32x-11}{20{x}^{2}-13x+2}$

$\frac{3}{5x-2}+\frac{4}{4x-1}$

$\frac{x+1}{{x}^{2}+7x+10}$

$\frac{5x}{{x}^{2}-9}$

$\frac{5}{2\left(x+3\right)}+\frac{5}{2\left(x-3\right)}$

$\frac{10x}{{x}^{2}-25}$

$\frac{6x}{{x}^{2}-4}$

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

$\frac{2x-3}{{x}^{2}-6x+5}$

$\frac{4x-1}{{x}^{2}-x-6}$

$\frac{9}{5\left(x+2\right)}+\frac{11}{5\left(x-3\right)}$

$\frac{4x+3}{{x}^{2}+8x+15}$

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

$\frac{8}{x-3}-\frac{5}{x-2}$

For the following exercises, find the decomposition of the partial fraction for the repeating linear factors.

$\frac{-5x-19}{{\left(x+4\right)}^{2}}$

$\frac{x}{{\left(x-2\right)}^{2}}$

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

$\frac{7x+14}{{\left(x+3\right)}^{2}}$

$\frac{-24x-27}{{\left(4x+5\right)}^{2}}$

$-\frac{6}{4x+5}+\frac{3}{{\left(4x+5\right)}^{2}}$

$\frac{-24x-27}{{\left(6x-7\right)}^{2}}$

$\frac{5-x}{{\left(x-7\right)}^{2}}$

$-\frac{1}{x-7}-\frac{2}{{\left(x-7\right)}^{2}}$

$\frac{5x+14}{2{x}^{2}+12x+18}$

$\frac{5{x}^{2}+20x+8}{2x{\left(x+1\right)}^{2}}$

$\frac{4}{x}-\frac{3}{2\left(x+1\right)}+\frac{7}{2{\left(x+1\right)}^{2}}$

$\frac{4{x}^{2}+55x+25}{5x{\left(3x+5\right)}^{2}}$

$\frac{54{x}^{3}+127{x}^{2}+80x+16}{2{x}^{2}{\left(3x+2\right)}^{2}}$

$\frac{4}{x}+\frac{2}{{x}^{2}}-\frac{3}{3x+2}+\frac{7}{2{\left(3x+2\right)}^{2}}$

$\frac{{x}^{3}-5{x}^{2}+12x+144}{{x}^{2}\left({x}^{2}+12x+36\right)}$

For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor.

$\frac{4{x}^{2}+6x+11}{\left(x+2\right)\left({x}^{2}+x+3\right)}$

$\frac{x+1}{{x}^{2}+x+3}+\frac{3}{x+2}$

$\frac{4{x}^{2}+9x+23}{\left(x-1\right)\left({x}^{2}+6x+11\right)}$

$\frac{-2{x}^{2}+10x+4}{\left(x-1\right)\left({x}^{2}+3x+8\right)}$

$\frac{4-3x}{{x}^{2}+3x+8}+\frac{1}{x-1}$

$\frac{{x}^{2}+3x+1}{\left(x+1\right)\left({x}^{2}+5x-2\right)}$

$\frac{4{x}^{2}+17x-1}{\left(x+3\right)\left({x}^{2}+6x+1\right)}$

$\frac{2x-1}{{x}^{2}+6x+1}+\frac{2}{x+3}$

$\frac{4{x}^{2}}{\left(x+5\right)\left({x}^{2}+7x-5\right)}$

$\frac{4{x}^{2}+5x+3}{{x}^{3}-1}$

$\frac{1}{{x}^{2}+x+1}+\frac{4}{x-1}$

$\frac{-5{x}^{2}+18x-4}{{x}^{3}+8}$

$\frac{3{x}^{2}-7x+33}{{x}^{3}+27}$

$\frac{2}{{x}^{2}-3x+9}+\frac{3}{x+3}$

$\frac{{x}^{2}+2x+40}{{x}^{3}-125}$

$\frac{4{x}^{2}+4x+12}{8{x}^{3}-27}$

$-\frac{1}{4{x}^{2}+6x+9}+\frac{1}{2x-3}$

$\frac{-50{x}^{2}+5x-3}{125{x}^{3}-1}$

$\frac{-2{x}^{3}-30{x}^{2}+36x+216}{{x}^{4}+216x}$

$\frac{1}{x}+\frac{1}{x+6}-\frac{4x}{{x}^{2}-6x+36}$

For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor.

$\frac{3{x}^{3}+2{x}^{2}+14x+15}{{\left({x}^{2}+4\right)}^{2}}$

$\frac{{x}^{3}+6{x}^{2}+5x+9}{{\left({x}^{2}+1\right)}^{2}}$

$\frac{x+6}{{x}^{2}+1}+\frac{4x+3}{{\left({x}^{2}+1\right)}^{2}}$

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

$\frac{{x}^{2}+5x+5}{{\left(x+2\right)}^{2}}$

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

$\frac{{x}^{3}+2{x}^{2}+4x}{{\left({x}^{2}+2x+9\right)}^{2}}$

$\frac{{x}^{2}+25}{{\left({x}^{2}+3x+25\right)}^{2}}$

$\frac{1}{{x}^{2}+3x+25}-\frac{3x}{{\left({x}^{2}+3x+25\right)}^{2}}$

$\frac{2{x}^{3}+11x+7x+70}{{\left(2{x}^{2}+x+14\right)}^{2}}$

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

$\frac{1}{8x}-\frac{x}{8\left({x}^{2}+4\right)}+\frac{10-x}{2{\left({x}^{2}+4\right)}^{2}}$

$\frac{{x}^{4}+{x}^{3}+8{x}^{2}+6x+36}{x{\left({x}^{2}+6\right)}^{2}}$

$\frac{2x-9}{{\left({x}^{2}-x\right)}^{2}}$

$-\frac{16}{x}-\frac{9}{{x}^{2}}+\frac{16}{x-1}-\frac{7}{{\left(x-1\right)}^{2}}$

$\frac{5{x}^{3}-2x+1}{{\left({x}^{2}+2x\right)}^{2}}$

## Extensions

For the following exercises, find the partial fraction expansion.

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

$\frac{1}{x+1}-\frac{2}{{\left(x+1\right)}^{2}}+\frac{5}{{\left(x+1\right)}^{3}}$

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

For the following exercises, perform the operation and then find the partial fraction decomposition.

$\frac{7}{x+8}+\frac{5}{x-2}-\frac{x-1}{{x}^{2}-6x-16}$

$\frac{5}{x-2}-\frac{3}{10\left(x+2\right)}+\frac{7}{x+8}-\frac{7}{10\left(x-8\right)}$

$\frac{1}{x-4}-\frac{3}{x+6}-\frac{2x+7}{{x}^{2}+2x-24}$

$\frac{2x}{{x}^{2}-16}-\frac{1-2x}{{x}^{2}+6x+8}-\frac{x-5}{{x}^{2}-4x}$

$-\frac{5}{4x}-\frac{5}{2\left(x+2\right)}+\frac{11}{2\left(x+4\right)}+\frac{5}{4\left(x+4\right)}$

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