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Decompose the given expression that has a repeated irreducible factor in the denominator.
The factors of the denominator are $\text{\hspace{0.17em}}x,({x}^{2}+1),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{({x}^{2}+1)}^{2}.\text{\hspace{0.17em}}$ 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.
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,
Expand the right side.
Now we will collect like terms.
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 $\text{\hspace{0.17em}}A=1,\text{\hspace{0.17em}}$ $B=0,\text{\hspace{0.17em}}$ $C=1,\text{\hspace{0.17em}}$ $D=\mathrm{-1},\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}E=\mathrm{-2.}\text{\hspace{0.17em}}$ We can write the decomposition as follows:
Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.
$\frac{x\mathrm{-2}}{{x}^{2}\mathrm{-2}x+3}+\frac{2x+1}{{\left({x}^{2}\mathrm{-2}x+3\right)}^{2}}$
Access these online resources for additional instruction and practice with partial fractions.
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 $\text{\hspace{0.17em}}7x+13=A(3x+5)+B(x+1).\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}x=\mathrm{-1},\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}\mathrm{-2.}$
For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors.
$\frac{5x+16}{{x}^{2}+10x+24}$
$\frac{3x\mathrm{-79}}{{x}^{2}\mathrm{-5}x\mathrm{-24}}$
$\frac{8}{x+3}-\frac{5}{x\mathrm{-8}}$
$\frac{-x\mathrm{-24}}{{x}^{2}\mathrm{-2}x\mathrm{-24}}$
$\frac{10x+47}{{x}^{2}+7x+10}$
$\frac{1}{x+5}+\frac{9}{x+2}$
$\frac{x}{6{x}^{2}+25x+25}$
$\frac{32x\mathrm{-11}}{20{x}^{2}\mathrm{-13}x+2}$
$\frac{3}{5x\mathrm{-2}}+\frac{4}{4x\mathrm{-1}}$
$\frac{x+1}{{x}^{2}+7x+10}$
$\frac{5x}{{x}^{2}\mathrm{-9}}$
$\frac{5}{2\left(x+3\right)}+\frac{5}{2\left(x\mathrm{-3}\right)}$
$\frac{10x}{{x}^{2}\mathrm{-25}}$
$\frac{6x}{{x}^{2}\mathrm{-4}}$
$\frac{3}{x+2}+\frac{3}{x\mathrm{-2}}$
$\frac{2x\mathrm{-3}}{{x}^{2}\mathrm{-6}x+5}$
$\frac{4x\mathrm{-1}}{{x}^{2}-x\mathrm{-6}}$
$\frac{9}{5\left(x+2\right)}+\frac{11}{5\left(x\mathrm{-3}\right)}$
$\frac{4x+3}{{x}^{2}+8x+15}$
$\frac{3x\mathrm{-1}}{{x}^{2}\mathrm{-5}x+6}$
$\frac{8}{x\mathrm{-3}}-\frac{5}{x\mathrm{-2}}$
For the following exercises, find the decomposition of the partial fraction for the repeating linear factors.
$\frac{\mathrm{-5}x\mathrm{-19}}{{\left(x+4\right)}^{2}}$
$\frac{x}{{\left(x\mathrm{-2}\right)}^{2}}$
$\frac{1}{x\mathrm{-2}}+\frac{2}{{\left(x\mathrm{-2}\right)}^{2}}$
$\frac{7x+14}{{\left(x+3\right)}^{2}}$
$\frac{\mathrm{-24}x\mathrm{-27}}{{\left(4x+5\right)}^{2}}$
$-\frac{6}{4x+5}+\frac{3}{{\left(4x+5\right)}^{2}}$
$\frac{\mathrm{-24}x\mathrm{-27}}{{\left(6x\mathrm{-7}\right)}^{2}}$
$\frac{5-x}{{\left(x\mathrm{-7}\right)}^{2}}$
$-\frac{1}{x\mathrm{-7}}-\frac{2}{{\left(x\mathrm{-7}\right)}^{2}}$
$\frac{5x+14}{2{x}^{2}+12x+18}$
$\frac{5{x}^{2}+20x+8}{2x{\left(x+1\right)}^{2}}$
$\frac{4}{x}-\frac{3}{2\left(x+1\right)}+\frac{7}{2{\left(x+1\right)}^{2}}$
$\frac{4{x}^{2}+55x+25}{5x{\left(3x+5\right)}^{2}}$
$\frac{54{x}^{3}+127{x}^{2}+80x+16}{2{x}^{2}{\left(3x+2\right)}^{2}}$
$\frac{4}{x}+\frac{2}{{x}^{2}}-\frac{3}{3x+2}+\frac{7}{2{\left(3x+2\right)}^{2}}$
$\frac{{x}^{3}\mathrm{-5}{x}^{2}+12x+144}{{x}^{2}\left({x}^{2}+12x+36\right)}$
For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor.
$\frac{4{x}^{2}+6x+11}{\left(x+2\right)\left({x}^{2}+x+3\right)}$
$\frac{x+1}{{x}^{2}+x+3}+\frac{3}{x+2}$
$\frac{4{x}^{2}+9x+23}{\left(x\mathrm{-1}\right)\left({x}^{2}+6x+11\right)}$
$\frac{\mathrm{-2}{x}^{2}+10x+4}{\left(x\mathrm{-1}\right)\left({x}^{2}+3x+8\right)}$
$\frac{4\mathrm{-3}x}{{x}^{2}+3x+8}+\frac{1}{x\mathrm{-1}}$
$\frac{{x}^{2}+3x+1}{\left(x+1\right)\left({x}^{2}+5x\mathrm{-2}\right)}$
$\frac{4{x}^{2}+17x\mathrm{-1}}{\left(x+3\right)\left({x}^{2}+6x+1\right)}$
$\frac{2x\mathrm{-1}}{{x}^{2}+6x+1}+\frac{2}{x+3}$
$\frac{4{x}^{2}}{\left(x+5\right)\left({x}^{2}+7x\mathrm{-5}\right)}$
$\frac{4{x}^{2}+5x+3}{{x}^{3}\mathrm{-1}}$
$\frac{1}{{x}^{2}+x+1}+\frac{4}{x\mathrm{-1}}$
$\frac{\mathrm{-5}{x}^{2}+18x\mathrm{-4}}{{x}^{3}+8}$
$\frac{3{x}^{2}\mathrm{-7}x+33}{{x}^{3}+27}$
$\frac{2}{{x}^{2}\mathrm{-3}x+9}+\frac{3}{x+3}$
$\frac{{x}^{2}+2x+40}{{x}^{3}\mathrm{-125}}$
$\frac{4{x}^{2}+4x+12}{8{x}^{3}\mathrm{-27}}$
$-\frac{1}{4{x}^{2}+6x+9}+\frac{1}{2x\mathrm{-3}}$
$\frac{\mathrm{-50}{x}^{2}+5x\mathrm{-3}}{125{x}^{3}\mathrm{-1}}$
$\frac{\mathrm{-2}{x}^{3}\mathrm{-30}{x}^{2}+36x+216}{{x}^{4}+216x}$
$\frac{1}{x}+\frac{1}{x+6}-\frac{4x}{{x}^{2}\mathrm{-6}x+36}$
For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor.
$\frac{3{x}^{3}+2{x}^{2}+14x+15}{{\left({x}^{2}+4\right)}^{2}}$
$\frac{{x}^{3}+6{x}^{2}+5x+9}{{\left({x}^{2}+1\right)}^{2}}$
$\frac{x+6}{{x}^{2}+1}+\frac{4x+3}{{\left({x}^{2}+1\right)}^{2}}$
$\frac{{x}^{3}-{x}^{2}+x\mathrm{-1}}{{\left({x}^{2}\mathrm{-3}\right)}^{2}}$
$\frac{{x}^{2}+5x+5}{{\left(x+2\right)}^{2}}$
$\frac{x+1}{x+2}+\frac{2x+3}{{\left(x+2\right)}^{2}}$
$\frac{{x}^{3}+2{x}^{2}+4x}{{\left({x}^{2}+2x+9\right)}^{2}}$
$\frac{{x}^{2}+25}{{\left({x}^{2}+3x+25\right)}^{2}}$
$\frac{1}{{x}^{2}+3x+25}-\frac{3x}{{\left({x}^{2}+3x+25\right)}^{2}}$
$\frac{2{x}^{3}+11x+7x+70}{{\left(2{x}^{2}+x+14\right)}^{2}}$
$\frac{5x+2}{x{\left({x}^{2}+4\right)}^{2}}$
$\frac{1}{8x}-\frac{x}{8\left({x}^{2}+4\right)}+\frac{10-x}{2{\left({x}^{2}+4\right)}^{2}}$
$\frac{{x}^{4}+{x}^{3}+8{x}^{2}+6x+36}{x{\left({x}^{2}+6\right)}^{2}}$
$\frac{2x\mathrm{-9}}{{\left({x}^{2}-x\right)}^{2}}$
$-\frac{16}{x}-\frac{9}{{x}^{2}}+\frac{16}{x\mathrm{-1}}-\frac{7}{{\left(x\mathrm{-1}\right)}^{2}}$
$\frac{5{x}^{3}\mathrm{-2}x+1}{{\left({x}^{2}+2x\right)}^{2}}$
For the following exercises, find the partial fraction expansion.
$\frac{{x}^{2}+4}{{\left(x+1\right)}^{3}}$
$\frac{1}{x+1}-\frac{2}{{\left(x+1\right)}^{2}}+\frac{5}{{\left(x+1\right)}^{3}}$
$\frac{{x}^{3}\mathrm{-4}{x}^{2}+5x+4}{{\left(x\mathrm{-2}\right)}^{3}}$
For the following exercises, perform the operation and then find the partial fraction decomposition.
$\frac{7}{x+8}+\frac{5}{x\mathrm{-2}}-\frac{x\mathrm{-1}}{{x}^{2}\mathrm{-6}x\mathrm{-16}}$
$\frac{5}{x\mathrm{-2}}-\frac{3}{10\left(x+2\right)}+\frac{7}{x+8}-\frac{7}{10\left(x\mathrm{-8}\right)}$
$\frac{1}{x\mathrm{-4}}-\frac{3}{x+6}-\frac{2x+7}{{x}^{2}+2x\mathrm{-24}}$
$\frac{2x}{{x}^{2}\mathrm{-16}}-\frac{1\mathrm{-2}x}{{x}^{2}+6x+8}-\frac{x\mathrm{-5}}{{x}^{2}\mathrm{-4}x}$
$-\frac{5}{4x}-\frac{5}{2\left(x+2\right)}+\frac{11}{2\left(x+4\right)}+\frac{5}{4\left(x+4\right)}$
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