# Partial Fractions Decomposition – Examples and Practice Problems

There are four main types of partial fractions: denominator with linear factors, denominator with irreducible quadratic factor, denominator with a repeated factor, and improper fractions.

Here, we will look at some examples of partial fractions decomposition, where we will apply the four types of partial fractions mentioned. In addition, we will look at some practice problems to apply the concepts.

##### ALGEBRA

Relevant for

Solving examples of partial fractions decomposition.

See examples

##### ALGEBRA

Relevant for

Solving examples of partial fractions decomposition.

See examples

## 10 Examples with answers of partial fractions decomposition

You can make a revision of the four types of partial fractions decomposition in this article.

### EXAMPLE 1

What are the partial fractions of the following fraction?

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

This fraction corresponds to the first type of partial fractions because we have only linear factors in its denominator, $latex (x+4)$ and $latex (x-3)$.

Then, the partial fractions of this fraction will have the following form:

$$\frac{5x+6}{(x+4)(x-3)}=\frac{A}{x+4}+\frac{B}{x-3}$$

To find the values of A and B, we start by multiplying the whole expression by $latex (x+4)(x-3)$ and we have:

$$5x+6=A(x-3)+B(x+4)$$

Now, we can substitute the value $latex x=3$ to remove the constant A and find the value of B:

$$5(3)+6=A(3-3)+B(3+4)$$

$latex 21=7B$

$latex B=3$

Then, we substitute $latex x=-4$ to find the value of A:

$$5(-4)+6=A(-4-3)+B(-4+4)$$

$latex -14=-7A$

$latex A=2$

Therefore, we have

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

### EXAMPLE 2

Find the partial fractions of the following fraction:

$$\frac{7x+8}{(x+4)(x-6)}$$

The denominator of the fraction has two linear factors, $latex (x+4)$ and $latex (x-6)$. Then, its decomposition into partial fractions has the form:

$$\frac{7x+8}{(x+4)(x-6)}=\frac{A}{x+4}+\frac{B}{x-6}$$

When we multiply the whole expression by $latex (x+4)(x-6)$, we get the following:

$$7x+8=A(x-6)+B(x+4)$$

Now, we can find the value of B by using the value $latex x=6$ to remove the constant A:

$$7(6)+8=A(6-6)+B(6+4)$$

$latex 50=10B$

$latex B=5$

Then, we substitute $latex x=-4$ to find the value of A:

$$7(-4)+8=A(-4-6)+B(-4+4)$$

$latex -20=-10A$

$latex A=2$

Therefore, we have:

$$\frac{7x+8}{(x+4)(x-6)}=\frac{2}{x+4}+\frac{5}{x-6}$$

### EXAMPLE 3

Find the partial fractions of the following fraction:

$$\frac{9x^2+34x+14}{(x+2)(x^2-x-12)}$$

At first glance, the fraction appears to have a quadratic factor in the denominator. However, we can factor this expression as follows:

$$x^2-x-12=(x+3)(x-4)$$

Then, the expression has only linear factors:

$$\frac{9x^2+34x+14}{(x+2)(x^2-x-12)}=\frac{9x^2+34x+14}{(x+2)(x+3)(x-4)}$$

Since we have only linear factors, the partial fractions have the following form:

$$\frac{9x^2+34x+14}{(x+2)(x+3)(x-4)}=\frac{A}{x+2}+\frac{B}{x+3}+\frac{C}{x-4}$$

Now, let’s multiply the whole expression by $latex (x+2)(x+3)(x-4)$:

$$9x^2+34x+14=A(x+3)(x-4)+B(x+2)(x-4)+C(x+2)(x+3)$$

We can find the value of A by using $latex x=-2$:

$$9(-2)^2+34(-2)+14=A(-2+3)(-2-4)+B(-2+2)(-2-4)+C(-2+2)(-2+3)$$

$latex -18=-6A$

$latex A=3$

We can find the value of B by using $latex x=-3$:

$$9(-3)^2+34(-3)+14=A(-3+3)(-3-4)+B(-3+2)(-3-4)+C(-3+2)(-3+3)$$

$latex -7=7B$

$latex B=-1$

We can find the value of C by using $latex x=4$:

$$9(4)^2+34(4)+14=A(4+3)(4-4)+B(4+2)(4-4)+C(4+2)(4+3)$$

$latex 294=42C$

$latex C=7$

Therefore, the partial fractions are:

$$\frac{9x^2+34x+14}{(x+2)(x+3)(x-4)}=\frac{3}{x+2}-\frac{1}{x+3}+\frac{7}{x-4}$$

### EXAMPLE 4

Express the following fraction as partial fractions:

$$\frac{6x+7}{(x^2+2)(x+3)}$$

The quadratic factor $latex (x^2+2)$ in the denominator cannot be factored. Then, the partial fractions have the form:

$$\frac{6x+7}{(x^2+2)(x+3)}=\frac{Ax+B}{x^2+2}+\frac{C}{x+3}$$

Multiplying both sides of the expression by $latex (x^2+2)(x+3)$, we have:

$$6x+7=(Ax+B)(x+3)+C(x^2+2)$$

We can find the value of C by substituting $latex x=-3$, so that the term $latex (Ax+B)$ equals zero:

$$6(-3)+7=(A(-3)+B)(-3+3)+C((-3)^2+2)$$

$latex -11=11C$

$latex C=-1$

Now, we can find the value of A by comparing the coefficients of the terms with $latex x^2$. On the left-hand side, we have none and on the right-hand side we have $latex Ax^2+Cx^2$. Then:

$latex 0=A+C$

Substituting $latex C=-1$ in this equation, we obtain the value $latex A=1$.

We can find the value of B by comparing the coefficients of the constant terms:

$latex 7=3B+2C$

Substituting $latex C=-1$ in this equation, we obtain the value $latex B=3$.

Therefore, the partial fractions are:

$$\frac{6x+7}{(x^2+2)(x+3)}=\frac{x+3}{x^2+2}-\frac{1}{x+3}$$

### EXAMPLE 5

What are the partial fractions of the following fraction?

$$\frac{7x^2+2x-28}{(x-6)(x^2+3x+5)}$$

The quadratic expression $latex x^2+3x+5$ in the denominator cannot be factored. Then, the partial fractions will have the following form:

$$\frac{7x^2+2x-28}{(x-6)(x^2+3x+5)}=\frac{A}{x-6}+\frac{Bx+C}{x^2+3x+5}$$

To find the values of A, B and C, we start by multiplying the whole expression by $latex (x-6)(x^2+3x+5)$ and we have:

$$7x^2+2x-28=A(x^2+3x+5)+(Bx+C)(x-6)$$

Now, we use the value $latex x=6$ to find the value of A:

$$7(6)^2+2(6)-28=A((6)^2+3(6)+5)+(B(6)+C)(6-6)$$

$latex 236=59A$

$latex A=4$

Then, we find the value of B by comparing the coefficients of the terms with $latex x^2$:

$latex 7=A+B$

When we substitute $latex A=4$ in this equation, we obtain the value $latex B=3$.

Finally, to find the value of C, we compare the constant terms:

$latex -28=5A-6C$

When we substitute $latex A=4$ in this equation, we obtain the value $latex C=8$.

Therefore, the partial fractions are:

$$\frac{7x^2+2x-28}{(x-6)(x^2+3x+5)}=\frac{4}{x-6}+\frac{3x+8}{x^2+3x+5}$$

### EXAMPLE 6

Express the following fraction as partial fractions:

$$\frac{5x+7}{(x+1)^2(x+2)}$$

The denominator of the fraction has a linear factor $latex (x+1)$ and a repeated factor $latex (x+2)^2$. In this case, the partial fractions have the following form:

$$\frac{5x+7}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}$$

To find the values of A, B and C, let’s multiply the whole expression by $latex (x+1)^2(x+2)$:

$$5x+7=A(x+1)(x+2)+B(x+2)+C(x+1)^2$$

Using the value $latex x=-1$, we have:

$$5(-1)+7=A(-1+1)(-1+2)+B(-1+2)+C(-1+1)^2$$

$latex 2=B$

$latex B=2$

Using the value $latex x=-2$, we have:

$$5(-2)+7=A(-2+1)(-2+2)+B(-2+2)+C(-2+1)^2$$

$latex -3=C$

$latex C=-3$

Finally, we find the value of A by comparing the coefficients with the term $latex x^2$. We have no terms on the left-hand side, and we have: $latex Ax^2+Cx^2$ on the right-hand side:

$latex 0=A+C$

Using the value $latex C=-3$ in this equation, we find the value of $latex A=3$. Therefore, we have:

$$\frac{5x+7}{(x+1)^2(x+2)}=\frac{3}{x+1}+\frac{2}{(x+1)^2}-\frac{3}{x+2}$$

### EXAMPLE 7

Write the following fraction using partial fractions:

$$\frac{2x^2+29x-1}{(2x+1)(x-2)^2}$$

The denominator of this fraction has a linear factor $latex (2x+1)$ and a repeated factor $latex (x-2)^2$. Then, its partial fractions have the following form:

$$\frac{2x^2+29x-1}{(2x+1)(x-2)^2}=\frac{A}{2x+1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$

To find the values of the constants, we multiply the whole expression by $latex (2x+1)(x-2)^2$:

$$2x^2+29x-11=A(x-2)^2+B(2x+1)(x-2)+C(2x+1)$$

Now, we use the value $latex x=-\frac{1}{2}$ to find the value of A:

$$2(-\frac{1}{2})^2+29(-\frac{1}{2})-11=A(-\frac{1}{2}-2)^2+B(2(-\frac{1}{2})+1)(-\frac{1}{2}-2)+C(2(-\frac{1}{2})+1)$$

$$-25=\frac{25}{4}A$$

$latex A=-4$

Then, we use the value $latex x=2$ to find the value of C:

$$2(2)^2+29(2)-11=A(2-2)^2+B(2(2)+1)(2-2)+C(2(2)+1)$$

$latex 55=5C$

$latex C=11$

If we compare the coefficients of the terms with $latex x^2$, we have:

$latex 2=A+2B$

Substituting $latex A=-4$ in this equation, we find the value of $latex B=3$. Therefore, we have:

$$\frac{2x^2+29x-1}{(2x+1)(x-2)^2}=-\frac{4}{2x+1}+\frac{3}{x-2}+\frac{11}{(x-2)^2}$$

### EXAMPLE 8

Find the partial fractions decomposition of the following fraction:

$$\frac{5x^2-71}{(x+5)(x-4)}$$

We can observe that the degree of the polynomial of the numerator is equal to 2 and the degree of the denominator is also equal to 2.

In this case, we have an improper fraction and its partial fractions have the following form:

$$\frac{5x^2-71}{(x+5)(x-4)}=A+\frac{B}{x+5}+\frac{C}{x-4}$$

Now, let’s multiply the whole expression by $latex (x+5)(x-4)$ and we have:

$$5x^2-71=A(x+5)(x-4)+B(x-4)+C(x+5)$$

Then, we can find the value of A by comparing the coefficients of the terms with $latex x^2$ and conclude that $latex A=5$.

The value of B can be found by substituting $latex x=-5$ in the expression:

$$5(-5)^2-71=B(-5-4)$$

$latex 54=-9B$

$latex B=-6$

We can find the value of C by substituting $latex x=4$ into the expression:

$$5(4)^2-71=C(4+5)$$

$latex 9=9C$

$latex C=1$

Therefore, we have:

$$\frac{5x^2-71}{(x+5)(x-4)}=5-\frac{6}{x+5}+\frac{1}{x-4}$$

### EXAMPLE 9

What are the partial fractions of the following fraction?

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

This example is similar to the previous one because the polynomials of the numerator and the denominator are of the same degree. Then, we have:

$$\frac{2x^2+x-5}{(x+2)(x+1)}=A+\frac{B}{x+2}+\frac{C}{x+1}$$

Now, let’s multiply the whole expression by $latex (x+2)(x+1)$ and we have:

$$2x^2+x-5=A(x+2)(x+1)+B(x+1)+C(x+2)$$

The value of A is found by comparing the coefficients of the terms with $latex x^2$, and we find that $latex A=2$.

To find the value of B, we substitute $latex x=-2$, and we have:

$$2(-2)^2-2-5=A(-2+2)(-2+1)+B(-2+1)+C(-2+2)$$

$latex 1=-B$

$latex B=-1$

To find the value of C, we substitute $latex x=-1$, and we have:

$$2(-1)^2-1-5=A(-1+2)(-1+1)+B(-1+1)+C(-1+2)$$

$latex -4=C$

$latex C=-4$

Therefore, we have:

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

### EXAMPLE 10

Write the following fraction using partial fractions

$$\frac{3x^4+7x^3+8x^2+53x-186}{(x+4)(x^2+9)}$$

The degree of the polynomial in the numerator is 4 and the degree of the polynomial in the denominator is 3. Then, the quotient will be a polynomial of degree 4-3=1.

Therefore, the quotient is a linear expression and its partial fractions have the following form:

$$\frac{3x^4+7x^3+8x^2+53x-186}{(x+4)(x^2+9)}=Ax+B+\frac{C}{x+4}+\frac{Dx+E}{x^2+9}$$

When we multiply the whole expression by $latex (x+4)(x^2+9)$, we have:

$$3x^4+7x^3+8x^2+53x-186=(Ax+B)(x+4)(x^2+9)+C(x^2+9)+(Dx+E)(x+4)$$

The value of A is found by comparing the coefficients of the terms with $latex x^4$, and we find that $latex A=3$.

Comparing the coefficients of the $latex x^3$ terms, we have:

$latex 7=4A+B$

Substituting $latex A=3$, we find that $latex B=-5$.

When we use the value $latex x=-4$, we have:

$$3(-4)^4+7(-4)^3+8(-4)^2+53(-4)-186=C((-4)^2+9)$$

$latex 50=25C$

$latex C=2$

Comparing the coefficients of the $latex x^2$ terms, we have:

$$8=9A+4B+C+D$$

Substituting the values $latex A=4$, $latex B=-5$ and $latex C=2$, we have $latex D=-1$.

Finally, comparing the constant terms, we have:

$latex -186=36B+9C+4E$

Substituting the values $latex B=-5$ and $latex C=2$, we have $latex E=-6$.

Therefore, we have:

$$\frac{3x^4+7x^3+8x^2+53x-186}{(x+4)(x^2+9)}=3x-5+\frac{2}{x+4}+\frac{-x-6}{x^2+9}$$

## Partial fractions decomposition – Practice problems

Partial fractions quiz
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#### The partial fractions decomposition of $latex \frac{x^4+x^3-19x^2-44x-21}{(x+3)(x+2)(x+1)}$ has two terms of the form $latex Ax+B$.

Write these terms in the input box.

$latex ~~=$