In the Partial Fractions section, we will learn how to split fractions that have multiple terms in the denominator. This will then allow us to evaluate the resulting “split” integral. Here we will review polynomial long division, factoring polynomials, simplifying rational equations, and solving systems of equations.

Perform Polynomial Long Division

We are familiar with the long division algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division.

Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.

[latex]\begin{array}{l}\left(\text{divisor }\cdot \text{ quotient}\right)\text{ + remainder}\text{ = dividend}\hfill \\ \left(3\cdot 59\right)+1 = 177+1 = 178\hfill \end{array}[/latex]

We call this the Division Algorithm and will discuss it more formally after looking at an example.

Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide [latex]2{x}^{3}-3{x}^{2}+4x+5[/latex] by [latex]x+2[/latex] using the long division algorithm, it would look like this:

We have found

[latex]\frac{2{x}^{3}-3{x}^{2}+4x+5}{x+2}=2{x}^{2}-7x+18-\frac{31}{x+2}[/latex]

or

[latex]2{x}^{3}-3{x}^{2}+4x+5=\left(x+2\right)\left(2{x}^{2}-7x+18\right)-31[/latex]

We can identify the dividend, divisor, quotient, and remainder.

Writing the result in this manner illustrates the Division Algorithm.

Example: Using Long Division to Divide a Third-Degree Polynomial

Divide [latex]6{x}^{3}+11{x}^{2}-31x+15[/latex] by [latex]3x - 2[/latex].

Show Solution

There is a remainder of 1. We can express the result as:

[latex]\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x - 2}=2{x}^{2}+5x - 7+\frac{1}{3x - 2}[/latex]

Analysis of the Solution

We can check our work by using the Division Algorithm to rewrite the solution then multiplying.

[latex]\left(3x - 2\right)\left(2{x}^{2}+5x - 7\right)+1=6{x}^{3}+11{x}^{2}-31x+15[/latex]

Notice, as we write our result,

• the dividend is [latex]6{x}^{3}+11{x}^{2}-31x+15[/latex]
• the divisor is [latex]3x - 2[/latex]
• the quotient is [latex]2{x}^{2}+5x - 7[/latex]
• the remainder is 1

Factor Polynomials

Recall that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For example, [latex]4[/latex] is the GCF of [latex]16[/latex] and [latex]20[/latex] because it is the largest number that divides evenly into both [latex]16[/latex] and [latex]20[/latex]. The GCF of polynomials works the same way: [latex]4x[/latex] is the GCF of [latex]16x[/latex] and [latex]20{x}^{2}[/latex] because it is the largest polynomial that divides evenly into both [latex]16x[/latex] and [latex]20{x}^{2}[/latex].

Finding and factoring out a GCF from a polynomial is the first skill involved in factoring polynomials.

Factor a GCF out of a Polynomial

When factoring a polynomial expression, our first step is to check to see if each term contains a common factor. If so, we factor out the greatest amount we can from each term. To make it less challenging to find this GCF of the polynomial terms, first look for the GCF of the coefficients, and then look for the GCF of the variables.

To factor out a GCF from a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property “backwards” to rewrite the polynomial in a factored form. Recall that the distributive property of multiplication over addition states that a product of a number and a sum is the same as the sum of the products.

Watch this video to see more examples of how to factor the GCF from a trinomial.

You can view the transcript for “Ex 2: Identify GCF and Factor a Trinomial” here (opens in new window).

Factor a Trinomial with Leading Coefficient 1

Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]{x}^{2}+5x+6[/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\left(x+2\right)[/latex] and [latex]\left(x+3\right)[/latex].

Trinomials of the form [latex]{x}^{2}+bx+c[/latex] can be factored by finding two numbers with a product of [latex]c[/latex] and a sum of [latex]b[/latex]. The trinomial [latex]{x}^{2}+10x+16[/latex], for example, can be factored using the numbers [latex]2[/latex] and [latex]8[/latex] because the product of these numbers is [latex]16[/latex] and their sum is [latex]10[/latex]. The trinomial can be rewritten as the product of [latex]\left(x+2\right)[/latex] and [latex]\left(x+8\right)[/latex].

Factor by Grouping

Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[/latex] can be rewritten as [latex]\left(2x+3\right)\left(x+1\right)[/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[/latex] and then factor each portion of the expression to obtain [latex]2x\left(x+1\right)+3\left(x+1\right)[/latex]. We then pull out the GCF of [latex]\left(x+1\right)[/latex] to find the factored expression.

In the next video, we see another example of how to factor a trinomial by grouping.

You can view the transcript for “Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique” here (opens in new window).

Factor a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.

[latex]\\[/latex]

We can use this equation to factor any differences of squares.

Watch this video to see another example of how to factor a difference of squares.

You can view the transcript for “Ex: Factor a Difference of Squares” here (opens in new window).

Factor the Sum and Difference of Cubes

Now we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.

[latex]\\[/latex]

Similarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.

[latex]\\[/latex]

We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example.

The sign of the first 2 is the same as the sign between [latex]{x}^{3}-{2}^{3}[/latex]. The sign of the [latex]2x[/latex] term is opposite the sign between [latex]{x}^{3}-{2}^{3}[/latex]. And the sign of the last term, 4, is always positive.

In the following two video examples, we show more binomials that can be factored as a sum or difference of cubes.

You can view the transcript for “Ex 1: Factor a Sum or Difference of Cubes” here (opens in new window).

You can view the transcript for “Ex 3: Factor a Sum or Difference of Cubes” here (opens in new window).

Solve Rational Equations

Equations that contain rational expressions are called rational equations. For example, [latex] \frac{2x+1}{4}=\frac{x}{3}[/latex] is a rational equation. One of the most straightforward ways to solve a rational equation is to eliminate denominators with the common denominator and then use properties of equality to isolate the variable. This method is often used to solve linear equations that involve fractions as in the following example:

Solve  [latex]\frac{1}{2}x-3=2-\frac{3}{4}x[/latex] by clearing the fractions in the equation first.

Multiply both sides of the equation by [latex]4[/latex], the common denominator of the fractional coefficients.

[latex]\begin{array}{c}\frac{1}{2}x-3=2-\frac{3}{4}x\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\\\text{}\\\,\,\,\,4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)+4\left(-\frac{3}{4}x\right)\\2x-12=8-3x\\\underline{+3x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+3x}\\5x-12=8\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\underline{+12}\,\,\,\,\underline{+12} \\5x=20\\x=4\end{array}[/latex]

We could have found a common denominator and worked with fractions, but that often leads to more mistakes. We can apply the same idea to solving rational equations.  The difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions. This means that clearing the denominator may sometimes mean multiplying the whole rational equation by a polynomial. In the next example, we will clear the denominators of a rational equation with a term that has a polynomial in the numerator.

In the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.

Solve Systems of Equations

A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.

We will first look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.

The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.

Two of the most common ways to solve a system of linear equations are the substitution method and the addition (elimination) method.

The Substitution Method

The Addition (Elimination) Method

Systems of Equations Containing Three Variables