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Partial Fractions

Decompose

Partial Fractions

  • Find the least common denominator of rational expressions (IA 7.2.3)
  • Solve a system of equations by elimination (IA 4.1.4)

A rational expression is an expression of the form \(\frac{p}{q}\) where p and q are polynomials and \(q\ne 0\) .

\(\frac{2}{7},\frac{5y}{7x{z}^{2}},\frac{x+1}{x+2},and\frac{2{x}^{2}+5x-7}{{x}^{2}-9}\) are examples of rational expressions.

Example

Find the least common denominator of the following rationals:

Try it.

\(\frac{2}{3},\frac{5}{12},and\frac{1}{18}\)

Solution

To find the LCD of the fractions, we factored 3, 12 and 18 into primes, lining up any common primes in columns. Then we “brought down” one prime from each column. Finally, we multiplied the factors to find the LCD.
\(\begin{array}{llllll}3 & = & & & 3 & \\ 18 & = & 2* & & 3* & 3 \\ 12 & = & 2* & 2* & 3 & \\ LCD & = & 2* & 2* & 3* & 3 \\ LCD & = & 36 & & & \end{array}\)

Find the least common denominator of the following rationals:

Try it.

\(\frac{1}{5}\), \(\frac{2}{7}\), and \(\frac{2}{75}\)

Example

Find the least common denominator of the following rational expressions:

Try it.

\(\frac{x+1}{x+3}\) and \(\frac{2}{{x}^{2}-9}\)

Solution
Step 1. List the factors of each denominator. Match factors vertically when possible \(\begin{array}{l}x+3=(x+3) \\ {x}^{2}-9=(x+3)(x-3)\end{array}\)
Step 2. Bring down the columns by including all factors, but do not include common factors twice. \(\frac{\begin{array}{l}x+3=(x+3) \\ {x}^{2}-9=(x+3)(x-3)\end{array}}{\text{LCD}=(x+3)(x-3)}\)
Step 3. Write the LCD as the product of the factors.The LCD is \((x+3)(x-3)\)

Condensed — the full section is in OpenStax College Algebra 2e.

Decomposing

Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fraction.

For example, suppose we add the following fractions:

\[\frac{2}{x-3}+\frac{-1}{x+2}\]

We would first need to find a common denominator, \((x+2)(x-3).\)

Next, we would write each expression with this common denominator and find the sum of the terms.

\[\begin{array}{l}\frac{2}{x-3}(\frac{x+2}{x+2})+\frac{-1}{x+2}(\frac{x-3}{x-3})= \\ \ \frac{2x+4-x+3}{(x+2)(x-3)}=\frac{x+7}{{x}^{2}-x-6}\end{array}\]

Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.

\[\underset{\begin{array}{l} \\ \text{Simplified}\ \text{sum}\end{array}}{\frac{x+7}{{x}^{2}-x-6}}\ \ \ =\underset{\begin{array}{l} \\ \text{Partial}\ \text{fraction}\ \text{decomposition}\end{array}}{\frac{2}{x-3}+\frac{-1}{x+2}}\]

We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.

When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of \({x}^{2}-x-6\) are \((x-3)(x+2),\) the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.

Condensed — the full section is in OpenStax College Algebra 2e.

Decomposing

Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.

Condensed — the full section is in OpenStax College Algebra 2e.

Decomposing

So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators \(\ A,B,\\) or \(\ C\\) representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as \(\ Ax+B,Bx+C,\\) etc.

Condensed — the full section is in OpenStax College Algebra 2e.

Decomposing

Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

  • Decompose \(\frac{P(x)}{Q(x)}\) by writing the partial fractions as \(\frac{A}{{a}_{1}x+{b}_{1}}+\frac{B}{{a}_{2}x+{b}_{2}}.\) 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 .
  • The decomposition of \(\frac{P(x)}{Q(x)}\) with repeated linear factors must account for the factors of the denominator in increasing powers. See .
  • The decomposition of \(\frac{P(x)}{Q(x)}\) with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in \(\frac{A}{x}+\frac{Bx+C}{(a{x}^{2}+bx+c)}.\) See .
  • In the decomposition of \(\frac{P(x)}{Q(x)},\) where \(Q(x)\) 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
    \[\frac{Ax+B}{(a{x}^{2}+bx+c)}+\frac{{A}_{2}x+{B}_{2}}{{(a{x}^{2}+bx+c)}^{2}}+\cdots \text{+}\frac{{A}_{n}x+{B}_{n}}{{(a{x}^{2}+bx+c)}^{n}}.\] See .

Partial Fractions

  • Find the least common denominator of rational expressions (IA 7.2.3)
  • Solve a system of equations by elimination (IA 4.1.4)

A rational expression is an expression of the form \(\frac{p}{q}\) where p and q are polynomials and \(q\ne 0\) .

\(\frac{2}{7},\frac{5y}{7x{z}^{2}},\frac{x+1}{x+2},and\frac{2{x}^{2}+5x-7}{{x}^{2}-9}\) are examples of rational expressions.

Example

Find the least common denominator of the following rationals:

Try it.

\(\frac{2}{3},\frac{5}{12},and\frac{1}{18}\)

Solution

To find the LCD of the fractions, we factored 3, 12 and 18 into primes, lining up any common primes in columns. Then we “brought down” one prime from each column. Finally, we multiplied the factors to find the LCD.
\(\begin{array}{llllll}3 & = & & & 3 & \\ 18 & = & 2* & & 3* & 3 \\ 12 & = & 2* & 2* & 3 & \\ LCD & = & 2* & 2* & 3* & 3 \\ LCD & = & 36 & & & \end{array}\)

Find the least common denominator of the following rationals:

Try it.

\(\frac{1}{5}\), \(\frac{2}{7}\), and \(\frac{2}{75}\)

Example

Find the least common denominator of the following rational expressions:

Try it.

\(\frac{x+1}{x+3}\) and \(\frac{2}{{x}^{2}-9}\)

Solution
Step 1. List the factors of each denominator. Match factors vertically when possible \(\begin{array}{l}x+3=(x+3) \\ {x}^{2}-9=(x+3)(x-3)\end{array}\)
Step 2. Bring down the columns by including all factors, but do not include common factors twice. \(\frac{\begin{array}{l}x+3=(x+3) \\ {x}^{2}-9=(x+3)(x-3)\end{array}}{\text{LCD}=(x+3)(x-3)}\)
Step 3. Write the LCD as the product of the factors.The LCD is \((x+3)(x-3)\)

Condensed — the full section is in OpenStax Precalculus 2e.

Decomposing

Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fraction.

For example, suppose we add the following fractions:

\[\frac{2}{x-3}+\frac{-1}{x+2}\]

We would first need to find a common denominator, \((x+2)(x-3).\)

Next, we would write each expression with this common denominator and find the sum of the terms.

\[\begin{array}{l}\frac{2}{x-3}(\frac{x+2}{x+2})+\frac{-1}{x+2}(\frac{x-3}{x-3})= \\ \ \frac{2x+4-x+3}{(x+2)(x-3)}=\frac{x+7}{{x}^{2}-x-6}\end{array}\]

Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.

\[\underset{\begin{array}{l} \\ \text{Simplified}\ \text{sum}\end{array}}{\frac{x+7}{{x}^{2}-x-6}}\ \ \ =\underset{\begin{array}{l} \\ \text{Partial}\ \text{fraction}\ \text{decomposition}\end{array}}{\frac{2}{x-3}+\frac{-1}{x+2}}\]

We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.

When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of \({x}^{2}-x-6\) are \((x-3)(x+2),\) the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.

Condensed — the full section is in OpenStax Precalculus 2e.

Decomposing

Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.

Condensed — the full section is in OpenStax Precalculus 2e.

Decomposing

So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators \(\ A,B,\\) or \(\ C\\) representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as \(\ Ax+B,Bx+C,\\) etc.

Condensed — the full section is in OpenStax Precalculus 2e.

Decomposing

Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.

Condensed — the full section is in OpenStax Precalculus 2e.

Key Concepts

  • Decompose \(\frac{P(x)}{Q(x)}\) by writing the partial fractions as \(\frac{A}{{a}_{1}x+{b}_{1}}+\frac{B}{{a}_{2}x+{b}_{2}}.\) 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 .
  • The decomposition of \(\frac{P(x)}{Q(x)}\) with repeated linear factors must account for the factors of the denominator in increasing powers. See .
  • The decomposition of \(\frac{P(x)}{Q(x)}\) with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in \(\frac{A}{x}+\frac{Bx+C}{(a{x}^{2}+bx+c)}.\) See .
  • In the decomposition of \(\frac{P(x)}{Q(x)},\) where \(Q(x)\) 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
    \[\frac{Ax+B}{(a{x}^{2}+bx+c)}+\frac{{A}_{2}x+{B}_{2}}{{(a{x}^{2}+bx+c)}^{2}}+\cdots \text{+}\frac{{A}_{n}x+{B}_{n}}{{(a{x}^{2}+bx+c)}^{n}}.\] See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. \(\frac{2}{3},\frac{5}{12},and\frac{1}{18}\)

    Afslør svaret

    To find the LCD of the fractions, we factored 3, 12 and 18 into primes, lining up any common primes in columns. Then we “brought down” one prime from each column. Finally, we multiplied the factors to find the LCD.
    \(\begin{array}{llllll}3 & = & & & 3 & \\ 18 & = & 2* & & 3* & 3 \\ 12 & = & 2* & 2* & 3 & \\ LCD & = & 2* & 2* & 3* & 3 \\ LCD & = & 36 & & & \end{array}\)

  2. \(\frac{1}{5}\), \(\frac{2}{7}\), and \(\frac{2}{75}\)

  3. \(\frac{x+1}{x+3}\) and \(\frac{2}{{x}^{2}-9}\)

    Afslør svaret
    Step 1. List the factors of each denominator. Match factors vertically when possible \(\begin{array}{l}x+3=(x+3) \\ {x}^{2}-9=(x+3)(x-3)\end{array}\)
    Step 2. Bring down the columns by including all factors, but do not include common factors twice. \(\frac{\begin{array}{l}x+3=(x+3) \\ {x}^{2}-9=(x+3)(x-3)\end{array}}{\text{LCD}=(x+3)(x-3)}\)
    Step 3. Write the LCD as the product of the factors.The LCD is \((x+3)(x-3)\)

  4. \(\left\{\begin{array}{l}3x+y=5 \\ 2x-3y=7\end{array}\right\)

    Afslør svaret
    Step 1Write the equations in standard form. If any coefficients are fractions, clear them.
    \(\left\{\begin{array}{l}3x+y=5 \\ 2x-3y=7\end{array}\right\)
    Step 2Let’s eliminate y
    \(\left\{\begin{array}{l}3x+y=5 \\ 2x-3y=7\end{array}\right\begin{array}{l}\underset{\text{Multiply by 3}}{\to } \\ \end{array}\left\{\begin{array}{l}9x+3y=15 \\ 2x-3y=7\end{array}\right\)
    Step 3 \(\frac{\begin{array}{l}9x+3y=15 \\ 2x-3y=7\end{array}}{11x=22}\)
    Step 4 \(\begin{array}{l}11x=22 \\ x=2\end{array}\)
    Step 5Use the value of the variable found in Step 2 to find the second variable.
    Let’s substitute \(x=2\) into \(3x+y=5\)
    \(\begin{array}{lll}3(2)+y & = & 5 \\ 6+y & = & 5 \\ y & = & -1\end{array}\)
    Step 6Write the solution as an ordered pair: (2, -1)
    Step 7Check the solution into the original equations.
    \(\begin{array}{llllll}3x+y & = & 5 & 2x-3y & = & 7 \\ 3(2)-1 & = & 5 & (2)-3(-1) & = & 7 \\ 6-1 & = & 5 & 4+3 & = & 7 \\ 5 & = & 5 & 7 & = & 7\end{array}\)

  5. Decompose the given rational expression with distinct linear factors.

    \[\frac{3x}{(x+2)(x-1)}\]
    Afslør svaret

    We will separate the denominator factors and give each numerator a symbolic label, like \(A,B\text{,}\) or \(C.\)

    \[\frac{3x}{(x+2)(x-1)}=\frac{A}{(x+2)}+\frac{B}{(x-1)}\]

    Multiply both sides of the equation by the common denominator to eliminate the fractions:

    \[(x+2)(x-1)[\frac{3x}{(x+2)(x-1)}]=(x+2)(x-1)[\frac{A}{(x+2)}]+(x+2)(x-1)[\frac{B}{(x-1)}]\]

    The resulting equation is

    \[3x=A(x-1)+B(x+2)\]

    Expand the right side of the equation and collect like terms.

    \[\begin{array}{l}3x=Ax-A+Bx+2B \\ 3x=(A+B)x-A+2B\end{array}\]

    Set up a system of equations associating corresponding coefficients.

    \[\begin{array}{l}3=\ \ A+B \\ 0=-A+2B\end{array}\]

    Add the two equations and solve for \(B.\)

    \[\begin{array}{l}\underset{\bar }{\begin{array}{l}3=\ \ \ A+B \\ 0=-A+2B\end{array}} \\ 3\ =\ \ \ 0\ +3B \\ 1=B\end{array}\]

    Substitute \(B=1\) into one of the original equations in the system.

    \[\begin{array}{l}3=A+1 \\ 2=A\end{array}\]

    Thus, the partial fraction decomposition is

    \[\frac{3x}{(x+2)(x-1)}=\frac{2}{(x+2)}+\frac{1}{(x-1)}\]

    Another method to use to solve for \(A\) or \(B\) is by considering the equation that resulted from eliminating the fractions and substituting a value for \(x\) that will make either the A- or B-term equal 0. If we let \(x=1,\) the
    \(A-\) term becomes 0 and we can simply solve for \(B.\)

    \[\begin{array}{l}\ 3x=A(x-1)+B(x+2) \\ 3(1)=A[(1)-1]+B[(1)+2] \\ \ 3=0+3B \\ \ 1=B\end{array}\]

    Next, either substitute \(B=1\) into the equation and solve for \(A,\) or make the B-term 0 by substituting \(x=-2\) into the equation.

    \[\begin{array}{l}\ 3x=A(x-1)+B(x+2) \\ \ \ 3(-2)=A[(-2)-1]+B[(-2)+2] \\ \ -6=-3A+0 \\ \ \frac{-6}{-3}=A \\ \text{ 2}=A\end{array}\]

    We obtain the same values for \(A\) and \(B\) using either method, so the decompositions are the same using either method.

    \[\frac{3x}{(x+2)(x-1)}=\frac{2}{(x+2)}+\frac{1}{(x-1)}\]

    Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the Heaviside method, named after Charles Heaviside, a pioneer in the study of electronics.

  6. Find the partial fraction decomposition of the following expression.

    \[\frac{x}{(x-3)(x-2)}\]
    Afslør svaret

    \(\frac{3}{x-3}-\frac{2}{x-2}\)

  7. Decompose the given rational expression with repeated linear factors.

    \[\frac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}\]
    Afslør svaret

    The denominator factors are \(x{(x-2)}^{2}.\) To allow for the repeated factor of \((x-2),\) the decomposition will include three denominators: \(x,(x-2),\) and \({(x-2)}^{2}.\) Thus,

    \[\frac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\frac{A}{x}+\frac{B}{(x-2)}+\frac{C}{{(x-2)}^{2}}\]

    Next, we multiply both sides by the common denominator.

    \[\begin{array}{l}x{(x-2)}^{2}[\frac{-{x}^{2}+2x+4}{x{(x-2)}^{2}}]=[\frac{A}{x}+\frac{B}{(x-2)}+\frac{C}{{(x-2)}^{2}}]x{(x-2)}^{2} \\ \ -{x}^{2}+2x+4=A{(x-2)}^{2}+Bx(x-2)+Cx\end{array}\]

    On the right side of the equation, we expand and collect like terms.

    \[\begin{array}{l}-{x}^{2}+2x+4=A({x}^{2}-4x+4)+B({x}^{2}-2x)+Cx \\ \ =A{x}^{2}-4Ax+4A+B{x}^{2}-2Bx+Cx \\ \ =(A+B){x}^{2}+(-4A-2B+C)x+4A\end{array}\]

    Next, we compare the coefficients of both sides. This will give the system of equations in three variables:

    \[-{x}^{2}+2x+4=(A+B){x}^{2}+(-4A-2B+C)x+4A\]\[\begin{array}{ll}A+B=-1 & \text{(1)} \\ -4A-2B+C=2\ \ & \text{(2)} \\ 4A=4\ \ & \text{(3)}\end{array}\]

    Solving for \(A\) , we have

    \[\begin{array}{l}4A=4 \\ \ A=1\end{array}\]

    Substitute \(A=1\) into equation (1).

    \[\begin{array}{l}\ A+B=-1 \\ (1)+B=-1 \\ \ B=-2\end{array}\]

    Then, to solve for \(C,\) substitute the values for \(A\) and \(B\) into equation (2).

    \[\begin{array}{l}\ -4A-2B+C=2 \\ -4(1)-2(-2)+C=2 \\ \ -4+4+C=2 \\ \ C=2\end{array}\]

    Thus,

    \[\frac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\frac{1}{x}-\frac{2}{(x-2)}+\frac{2}{{(x-2)}^{2}}\]

  8. Find the partial fraction decomposition of the expression with repeated linear factors.

    \[\frac{6x-11}{{(x-1)}^{2}}\]
    Afslør svaret

    \(\frac{6}{x-1}-\frac{5}{{(x-1)}^{2}}\)

  9. Find a partial fraction decomposition of the given expression.

    \[\frac{8{x}^{2}+12x-20}{(x+3)({x}^{2}+x+2)}\]
    Afslør svaret

    We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus,

    \[\frac{8{x}^{2}+12x-20}{(x+3)({x}^{2}+x+2)}=\frac{A}{(x+3)}+\frac{Bx+C}{({x}^{2}+x+2)}\]

    We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.

    \[\begin{array}{l}(x+3)({x}^{2}+x+2)[\frac{8{x}^{2}+12x-20}{(x+3)({x}^{2}+x+2)}]=[\frac{A}{(x+3)}+\frac{Bx+C}{({x}^{2}+x+2)}](x+3)({x}^{2}+x+2) \\ \ 8{x}^{2}+12x-20=A({x}^{2}+x+2)+(Bx+C)(x+3)\end{array}\]

    Notice we could easily solve for \(\ A\\) by choosing a value for \(\ x\\) that will make the \(\ Bx+C\\) term equal 0. Let \(x=-3\\) and substitute it into the equation.

    \[\begin{array}{l}\ \ \ \ 8{x}^{2}+12x-20=A({x}^{2}+x+2)+(Bx+C)(x+3) \\ \ 8{(-3)}^{2}+12(-3)-20=A({(-3)}^{2}+(-3)+2)+(B(-3)+C)((-3)+3) \\ \ \ \ \ 16=8A \\ \ \ \ \ A=2\end{array}\]

    Now that we know the value of \(\ A,\\) substitute it back into the equation. Then expand the right side and collect like terms.

    \[\begin{array}{l} \\ 8{x}^{2}+12x-20=2({x}^{2}+x+2)+(Bx+C)(x+3) \\ 8{x}^{2}+12x-20=2{x}^{2}+2x+4+B{x}^{2}+3B+Cx+3C \\ 8{x}^{2}+12x-20=(2+B){x}^{2}+(2+3B+C)x+(4+3C)\end{array}\]

    Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.

    \[\begin{array}{ll}\ 2+B=8 & \text{(1)} \\ 2+3B+C=12 & \text{(2)} \\ \ 4+3C=-20 & \text{(3)}\end{array}\]

    Solve for \(\ B\\) using equation (1) and solve for \(\ C\\) using equation (3).

    \[\begin{array}{ll}\ 2+B=8 & \text{(1)} \\ \ B=6 & \\ & \\ 4+3C=-20 & \text{(3)} \\ \ 3C=-24 & \\ \ C=-8 & \end{array}\]

    Thus, the partial fraction decomposition of the expression is

    \[\frac{8{x}^{2}+12x-20}{(x+3)({x}^{2}+x+2)}=\frac{2}{(x+3)}+\frac{6x-8}{({x}^{2}+x+2)}\]

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

    \[\frac{5{x}^{2}-6x+7}{(x-1)({x}^{2}+1)}\]
    Afslør svaret

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

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

    \[\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{({x}^{2}+1)}^{2}}\]
    Afslør svaret

    The factors of the denominator are \(\ x,({x}^{2}+1),\\) and \(\ {({x}^{2}+1)}^{2}.\\) 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 \(\ Ax+B.\\) So, let’s begin the decomposition.

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

    We eliminate the denominators by multiplying each term by \(\ x{({x}^{2}+1)}^{2}.\\) Thus,

    \[{x}^{4}+{x}^{3}+{x}^{2}-x+1=A{({x}^{2}+1)}^{2}+(Bx+C)(x)({x}^{2}+1)+(Dx+E)(x)\]

    Expand the right side.

    \[\begin{array}{l}\ {x}^{4}+{x}^{3}+{x}^{2}-x+1=A({x}^{4}+2{x}^{2}+1)+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex \\ \ \ =A{x}^{4}+2A{x}^{2}+A+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex\end{array}\]

    Now we will collect like terms.

    \[{x}^{4}+{x}^{3}+{x}^{2}-x+1=(A+B){x}^{4}+(C){x}^{3}+(2A+B+D){x}^{2}+(C+E)x+A\]

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

    \[\begin{array}{l}\ A+B=1 \\ \ C=1 \\ 2A+B+D=1 \\ \ C+E=-1 \\ \ A=1\end{array}\]

    We can use substitution from this point. Substitute \(A=1\) into the first equation.

    \[\begin{array}{l}1+B=1 \\ \ B=0\end{array}\]

    Substitute \(A=1\) and \(B=0\) into the third equation.

    \[\begin{array}{l}2(1)+0+D=1 \\ \ D=-1\end{array}\]

    Substitute \(C=1\) into the fourth equation.

    \[\begin{array}{l}1+E=-1 \\ \ E=-2\end{array}\]

    Now we have solved for all of the unknowns on the right side of the equal sign. We have \(A=1,\) \(B=0,\) \(C=1,\) \(D=-1,\) and \(E=-2.\) We can write the decomposition as follows:

    \[\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{({x}^{2}+1)}^{2}}=\frac{1}{x}+\frac{1}{({x}^{2}+1)}-\frac{x+2}{{({x}^{2}+1)}^{2}}\]

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

    \[\frac{{x}^{3}-4{x}^{2}+9x-5}{{({x}^{2}-2x+3)}^{2}}\]
    Afslør svaret

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

  13. 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

    Afslør svaret

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

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

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

    Afslør svaret

    Graph both sides and ensure they are equal.

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

  17. 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 \(\frac{7x+13}{3{x}^{2}+8x+15}=\frac{A}{x+1}+\frac{B}{3x+5}\), we eventually simplify to \(7x+13=A(3x+5)+B(x+1).\) Explain how you could intelligently choose an \(x\) -value that will eliminate either \(A\) or \(B\) and solve for \(A\) and \(B.\)

    Afslør svaret

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

  18. \(\frac{5x+16}{{x}^{2}+10x+24}\)

  19. \(\frac{3x-79}{{x}^{2}-5x-24}\)

    Afslør svaret

    \(\frac{8}{x+3}-\frac{5}{x-8}\)

  20. \(\frac{-x-24}{{x}^{2}-2x-24}\)

  21. \(\frac{10x+47}{{x}^{2}+7x+10}\)

    Afslør svaret

    \(\frac{1}{x+5}+\frac{9}{x+2}\)

  22. \(\frac{x}{6{x}^{2}+25x+25}\)

  23. \(\frac{32x-11}{20{x}^{2}-13x+2}\)

    Afslør svaret

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

  24. \(\frac{x+1}{{x}^{2}+7x+10}\)

  25. \(\frac{5x}{{x}^{2}-9}\)

    Afslør svaret

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

  26. \(\frac{10x}{{x}^{2}-25}\)

  27. \(\frac{6x}{{x}^{2}-4}\)

    Afslør svaret

    \(\frac{3}{x+2}+\frac{3}{x-2}\)

  28. \(\frac{2x-3}{{x}^{2}-6x+5}\)

  29. \(\frac{4x-1}{{x}^{2}-x-6}\)

    Afslør svaret

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

  30. \(\frac{4x+3}{{x}^{2}+8x+15}\)

  31. \(\frac{3x-1}{{x}^{2}-5x+6}\)

    Afslør svaret

    \(\frac{8}{x-3}-\frac{5}{x-2}\)

  32. \(\frac{-5x-19}{{(x+4)}^{2}}\)

  33. \(\frac{x}{{(x-2)}^{2}}\)

    Afslør svaret

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

  34. \(\frac{7x+14}{{(x+3)}^{2}}\)

  35. \(\frac{-24x-27}{{(4x+5)}^{2}}\)

    Afslør svaret

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

  36. \(\frac{-24x-27}{{(6x-7)}^{2}}\)

  37. \(\frac{5-x}{{(x-7)}^{2}}\)

    Afslør svaret

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

  38. \(\frac{5x+14}{2{x}^{2}+12x+18}\)

  39. \(\frac{5{x}^{2}+20x+8}{2x{(x+1)}^{2}}\)

    Afslør svaret

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

  40. \(\frac{4{x}^{2}+55x+25}{5x{(3x+5)}^{2}}\)

Symbols used here

P(A),\ P(A \mid B)
probability, conditional probability
Chance of A; chance of A given that B happened.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\neq
not equal
The two sides are different.
\mathbf{v},\ \vec{v}
vector
A quantity with magnitude and direction; a column of numbers.
A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
matrix
A rectangular array of numbers; a linear map.
\det A,\ |A|
determinant
Scaling factor of area/volume under A; zero means singular.
A^{-1},\ A^{T}
inverse, transpose
The matrix that undoes A; A with rows and columns swapped.
\lambda
lambda (eigenvalue)
The factor by which an eigenvector is stretched: Av = λv.
\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|
dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Partial Fractions

  1. Decompose
  2. Decompose
  3. Decompose
  4. Decompose
  5. Find the least common denominator of rational expressions (IA 7.2.3)
  6. Solve a system of equations by elimination (IA 4.1.4)
  7. List the factors of each denominator. Match factors vertically when possible.
  8. Bring down the columns by including all factors, but do not include common factors twice.

Questions people ask

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn — it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Mere i Linear Algebra

DeterminantsMatrix inverseEigenvalues and eigenvectorsMatrix multiplicationRow reductionVector spaces, span and linear independenceOrthogonality, projections and least squaresDiagonalisation and matrix powersLinear transformations and change of basis