Skills Review for The Limit Laws

Learning Outcomes

Factor the greatest common factor (monomial) of a polynomial
Factor a trinomial with a leading coefficient of 1
Rewrite a trinomial as a four term polynomial and factor by grouping terms
Factor difference of squares
Factor sum or difference of cubes
Simplify a rational expression
Remove radicals from a multiple term denominator
Simplify complex rational expressions

The Limit Laws section will introduce yet another way to calculate a limit, using limit laws. Basically, calculating limits of functions algebraically will be the topic of focus in the section. Several techniques that are used to simplify functions are sometimes needed to successfully calculate a function’s limit. Here we will review factoring, simplifying rational expressions, rationalizing radical expressions, and simplifying complex rational expressions.

Factor Polynomials

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

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.

A General Note: Greatest Common Factor

The greatest common factor (GCF) of a polynomial is the largest polynomial that divides evenly into each term of the polynomial.

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.

Distributive Property Forward and Backward

Forward: We distribute $a$ over $b+c$ .

$a\left(b+c\right)=ab+ac$ .

Backward: We factor $a$ out of $ab+ac$ .

$ab+ac=a\left(b+c\right)$ .

We have seen that we can distribute a factor over a sum or difference. Now we see that we can “undo” the distributive property with factoring.

Example: Factoring The Greatest Common Factor

Factor $25b^{3}+10b^{2}$ .

Show Solution

Find the GCF.

$\begin{array}{l}\,\,25b^{3}=5\cdot5\cdot{b}\cdot{b}\cdot{b}\\\,\,10b^{2}=5\cdot2\cdot{b}\cdot{b}\\\text{GCF}=5\cdot{b}\cdot{b}=5b^{2}\end{array}$

Rewrite each term with the GCF as one factor.

$\begin{array}{l}25b^{3} = 5b^{2}\cdot5b\\10b^{2}=5b^{2}\cdot2\end{array}$

Rewrite the polynomial using the factored terms in place of the original terms.

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

Factor out the $5b^{2}$ .

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

Answer

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

Analysis

The factored form of the polynomial $25b^{3}+10b^{2}$ is $5b^{2}\left(5b+2\right)$ . You can check this by doing the multiplication. $5b^{2}\left(5b+2\right)=25b^{3}+10b^{2}$ .

Note that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.

For example:

$\begin{array}{l}25b^{3}+10b^{2}=5\left(5b^{3}+2b^{2}\right)\,\,\,\,\,\,\,\,\,\,\,\text{Factor out }5.\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=5b^{2}\left(5b+2\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Factor out }b^{2}.\end{array}$

Notice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.

How To: Given a polynomial expression, factor out the greatest common factor

Identify the GCF of the coefficients.
Identify the GCF of the variables.
Combine to find the GCF of the expression.
Determine what the GCF needs to be multiplied by to obtain each term in the expression.
Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.

Example: Factoring the Greatest Common Factor

Factor $6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy$ .

Show Solution

First find the GCF of the expression. The GCF of $6,45$ , and $21$ is $3$ . The GCF of ${x}^{3},{x}^{2}$ , and $x$ is $x$ . (Note that the GCF of a set of expressions of the form ${x}^{n}$ will always be the lowest exponent.) The GCF of ${y}^{3},{y}^{2}$ , and $y$ is $y$ . Combine these to find the GCF of the polynomial, $3xy$ .

Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that $3xy\left(2{x}^{2}{y}^{2}\right)=6{x}^{3}{y}^{3}, 3xy\left(15xy\right)=45{x}^{2}{y}^{2}$ , and $3xy\left(7\right)=21xy$ .

Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.

(3 x y) (2 x^{2} y^{2} + 15 x y + 7)

$\left(3xy\right)\left(2{x}^{2}{y}^{2}+15xy+7\right)$

Analysis of the Solution

After factoring, we can check our work by multiplying. Use the distributive property to confirm that $\left(3xy\right)\left(2{x}^{2}{y}^{2}+15xy+7\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy$ .

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 ${x}^{2}+5x+6$ has a GCF of 1, but it can be written as the product of the factors $\left(x+2\right)$ and $\left(x+3\right)$ .

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

A General Note: Factoring a Trinomial with Leading Coefficient 1

A trinomial of the form ${x}^{2}+bx+c$ can be written in factored form as $\left(x+p\right)\left(x+q\right)$ where $pq=c$ and $p+q=b$ .

Q & A

Can every trinomial be factored as a product of binomials?

No. Some polynomials cannot be factored. These polynomials are said to be prime.

How To: Given a trinomial in the form ${x}^{2}+bx+c$ , factor it

List factors of $c$ .
Find $p$ and $q$ , a pair of factors of $c$ with a sum of $b$ .
Write the factored expression $\left(x+p\right)\left(x+q\right)$ .

Example: Factoring a Trinomial with Leading Coefficient 1

Factor ${x}^{2}+2x - 15$ .

Show Solution

We have a trinomial with leading coefficient $1,b=2$ , and $c=-15$ . We need to find two numbers with a product of $-15$ and a sum of $2$ . In the table, we list factors until we find a pair with the desired sum.

Factors of $-15$	Sum of Factors
$1,-15$	$-14$
$-1,15$	$14$
$3,-5$	$-2$
$-3,5$	$2$

Now that we have identified $p$ and $q$ as $-3$ and $5$ , write the factored form as $\left(x - 3\right)\left(x+5\right)$ .

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that $\left(x - 3\right)\left(x+5\right)={x}^{2}+2x - 15$ .

Q & A

Does the order of the factors matter?

No. Multiplication is commutative, so the order of the factors does not matter.

Try It

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 $2{x}^{2}+5x+3$ can be rewritten as $\left(2x+3\right)\left(x+1\right)$ using this process. We begin by rewriting the original expression as $2{x}^{2}+2x+3x+3$ and then factor each portion of the expression to obtain $2x\left(x+1\right)+3\left(x+1\right)$ . We then pull out the GCF of $\left(x+1\right)$ to find the factored expression.

A General Note: Factor by Grouping

To factor a trinomial of the form $a{x}^{2}+bx+c$ by grouping, we find two numbers with a product of $ac$ and a sum of $b$ . We use these numbers to divide the $x$ term into the sum of two terms and factor each portion of the expression separately then factor out the GCF of the entire expression.

How To: Given a trinomial in the form $a{x}^{2}+bx+c$ , factor by grouping

List factors of $ac$ .
Find $p$ and $q$ , a pair of factors of $ac$ with a sum of $b$ .
Rewrite the original expression as $a{x}^{2}+px+qx+c$ .
Pull out the GCF of $a{x}^{2}+px$ .
Pull out the GCF of $qx+c$ .
Factor out the GCF of the expression.

Example: Factoring a Trinomial by Grouping

Factor $5{x}^{2}+7x - 6$ by grouping.

Show Solution

We have a trinomial with $a=5,b=7$ , and $c=-6$ . First, determine $ac=-30$ . We need to find two numbers with a product of $-30$ and a sum of $7$ . In the table, we list factors until we find a pair with the desired sum.

Factors of $-30$	Sum of Factors
$1,-30$	$-29$
$-1,30$	$29$
$2,-15$	$-13$
$-2,15$	$13$
$3,-10$	$-7$
$-3,10$	$7$

So $p=-3$ and $q=10$ .

\begin{array}{cc} 5 x^{2} - 3 x + 10 x - 6 & Rewrite the original expression as a x^{2} + p x + q x + c . \\ x (5 x - 3) + 2 (5 x - 3) & Factor out the GCF of each part . \\ (5 x - 3) (x + 2) & Factor out the GCF of the expression . \end{array}

$\begin{array}{cc}5{x}^{2}-3x+10x - 6 \hfill & \text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\hfill \\ x\left(5x - 3\right)+2\left(5x - 3\right)\hfill & \text{Factor out the GCF of each part}.\hfill \\ \left(5x - 3\right)\left(x+2\right)\hfill & \text{Factor out the GCF}\text{ }\text{ of the expression}.\hfill \end{array}$

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that $\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6$ .

Try It

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.

a^{2} - b^{2} = (a + b) (a - b)

${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

$\\$

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

A General Note: Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.

a^{2} - b^{2} = (a + b) (a - b)

${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

How To: Given a difference of squares, factor it into binomials

Confirm that the first and last term are perfect squares.
Write the factored form as $\left(a+b\right)\left(a-b\right)$ .

Example: Factoring a Difference of Squares

Factor $9{x}^{2}-25$ .

Show Solution

Notice that $9{x}^{2}$ and $25$ are perfect squares because $9{x}^{2}={\left(3x\right)}^{2}$ and $25={5}^{2}$ . The polynomial represents a difference of squares and can be rewritten as $\left(3x+5\right)\left(3x - 5\right)$ .

Try It

Q & A

Is there a formula to factor the sum of squares?

No. A sum of squares cannot be factored.

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.

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$

$\\$

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

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$

$\\$

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.

x^{3} - 2^{3} = (x - 2) (x^{2} + 2 x + 4)

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

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

A General Note: Sum and Difference of Cubes

We can factor the sum of two cubes as

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$

We can factor the difference of two cubes as

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$

How To: Given a sum of cubes or difference of cubes, factor it

Confirm that the first and last term are cubes, ${a}^{3}+{b}^{3}$ or ${a}^{3}-{b}^{3}$ .
For a sum of cubes, write the factored form as $\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$ . For a difference of cubes, write the factored form as $\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$ .

Example: Factoring a Sum of Cubes

Factor ${x}^{3}+512$ .

Show Solution

Notice that ${x}^{3}$ and $512$ are cubes because ${8}^{3}=512$ . Rewrite the sum of cubes as $\left(x+8\right)\left({x}^{2}-8x+64\right)$ .

Analysis of the Solution

After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.

Try It

Factor the sum of cubes $216{a}^{3}+{b}^{3}$ .

Show Solution

$\left(6a+b\right)\left(36{a}^{2}-6ab+{b}^{2}\right)$

Example: Factoring a Difference of Cubes

Factor $8{x}^{3}-125$ .

Show Solution

Notice that $8{x}^{3}$ and $125$ are cubes because $8{x}^{3}={\left(2x\right)}^{3}$ and $125={5}^{3}$ . Write the difference of cubes as $\left(2x - 5\right)\left(4{x}^{2}+10x+25\right)$ .

Analysis of the Solution

Just as with the sum of cubes, we will not be able to further factor the trinomial portion.

Try It

Factor the difference of cubes: $1,000{x}^{3}-1$ .

Show Solution

$\left(10x - 1\right)\left(100{x}^{2}+10x+1\right)$

Try It

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).

Simplify Rational Expressions

The quotient of two polynomial expressions is called a rational expression. We can apply the properties of fractions to rational expressions such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let’s start with the rational expression shown.

$\frac{{x}^{2}+8x+16}{{x}^{2}+11x+28}$

$\\$ We can factor the numerator and denominator to rewrite the expression as $\frac{{\left(x+4\right)}^{2}}{\left(x+4\right)\left(x+7\right)}$

$\\$

Then we can simplify the expression by canceling the common factor

(x + 4)

$\left(x+4\right)$ to get

\frac{x + 4}{x + 7}

$\frac{x+4}{x+7}$

$\\$

How To: Given a rational expression, simplify it

Factor the numerator and denominator.
Cancel any common factors.

Example: Simplifying Rational Expressions

Simplify $\frac{{x}^{2}-9}{{x}^{2}+4x+3}$ .

Show Solution

$\begin{array}{lllllllll}\frac{\left(x+3\right)\left(x - 3\right)}{\left(x+3\right)\left(x+1\right)}\hfill & \hfill & \hfill & \hfill & \text{Factor the numerator and the denominator}.\hfill \\ \frac{x - 3}{x+1}\hfill & \hfill & \hfill & \hfill & \text{Cancel common factor }\left(x+3\right).\hfill \end{array}$

Analysis of the Solution

We can cancel the common factor because any expression divided by itself is equal to 1.

Q & A

Can the ${x}^{2}$ term be cancelled in the above example?

No. A factor is an expression that is multiplied by another expression. The ${x}^{2}$ term is not a factor of the numerator or the denominator.

Try It

Simplify $\frac{x - 6}{{x}^{2}-36}$ .

Show Solution

$\frac{1}{x+6}$

Try It

Rationalize Radical Expressions

When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator.

We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by a form of 1 that will eliminate the radical.

For a denominator containing a binomial where at least one of the terms is a square root, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign in the middle of the binomial. If the denominator is $a+b\sqrt{c}$ , then the conjugate is $a-b\sqrt{c}$ .

How To: Given an expression with a Binomial containing a square root in the denominator, rationalize the denominator

Find the conjugate of the denominator.
Multiply the numerator and denominator by the conjugate.
Use the distributive property.
Simplify.

Example: Rationalizing a Denominator with a binomial Containing a square root

Rationalize $\dfrac{4}{1+\sqrt{5}}$ .

Show Solution

Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of $1+\sqrt{5}$ is $1-\sqrt{5}$ . Then multiply the fraction by $\dfrac{1-\sqrt{5}}{1-\sqrt{5}}$ .

\begin{aligned} \frac{4}{1 + \sqrt{5}} \cdot \frac{1 - \sqrt{5}}{1 - \sqrt{5}} & = \frac{4 - 4 \sqrt{5}}{- 4} & Use the distributive property . \\ = \sqrt{5} - 1 & Simplify . \end{aligned}

$\begin{align}\frac{4}{1+\sqrt{5}}\cdot \frac{1-\sqrt{5}}{1-\sqrt{5}} &= \frac{4 - 4\sqrt{5}}{-4} && \text{Use the distributive property}. \\ &=\sqrt{5}-1 && \text{Simplify}. \end{align}$

Try It

Write $\dfrac{7}{2+\sqrt{3}}$ in simplest form.

Show Solution

$14 - 7\sqrt{3}$

Try It

You can view the transcript for “Ex: Rationalize the Denominator of a Radical Expression – Conjugate” here (opens in new window).

Simplify Complex Rational Expressions

A complex rational expression is a rational expression that contains additional rational expressions in the numerator, the denominator, or both. We can simplify complex rational expressions by rewriting the numerator and denominator as single rational expressions and dividing. The complex rational expression $\dfrac{a}{\dfrac{1}{b}+c}$ can be simplified by rewriting the numerator as the fraction $\dfrac{a}{1}$ and combining the expressions in the denominator as $\dfrac{1+bc}{b}$ . We can then rewrite the expression as a multiplication problem using the reciprocal of the denominator. We get $\dfrac{a}{1}\cdot \dfrac{b}{1+bc}$ which is equal to $\dfrac{ab}{1+bc}$ .

How To: Given a complex rational expression, simplify it

Combine the expressions in the numerator into a single rational expression by adding or subtracting.
Combine the expressions in the denominator into a single rational expression by adding or subtracting.
Rewrite as the numerator divided by the denominator.
Rewrite as multiplication.
Multiply.
Simplify.

Example: Simplifying Complex Rational Expressions

Simplify: $\dfrac{y+\dfrac{1}{x}}{\dfrac{x}{y}}$ .

Show Solution

Begin by combining the expressions in the numerator into one expression.

\begin{array}{cc} y \cdot \frac{x}{x} + \frac{1}{x} & Multiply by \frac{x}{x} to get LCD as denominator . \\ \frac{x y}{x} + \frac{1}{x} \\ \frac{x y + 1}{x} & Add numerators . \end{array}

$\begin{array}{cc}y\cdot \dfrac{x}{x}+\dfrac{1}{x}\hfill & \text{Multiply by }\dfrac{x}{x}\text{to get LCD as denominator}.\hfill \\ \dfrac{xy}{x}+\dfrac{1}{x}\hfill & \\ \dfrac{xy+1}{x}\hfill & \text{Add numerators}.\hfill \end{array}$

Now the numerator is a single rational expression and the denominator is a single rational expression.

\frac{\frac{x y + 1}{x}}{\frac{x}{y}}

$\dfrac{\dfrac{xy+1}{x}}{\dfrac{x}{y}}$

We can rewrite this as division and then multiplication.

\begin{array}{cc} \frac{x y + 1}{x} \div \frac{x}{y} \\ \frac{x y + 1}{x} \cdot \frac{y}{x} & Rewrite as multiplication . \\ \frac{y (x y + 1)}{x^{2}} & Multiply . \end{array}

$\begin{array}{cc}\dfrac{xy+1}{x}\div \dfrac{x}{y}\hfill & \\ \dfrac{xy+1}{x}\cdot \dfrac{y}{x}\hfill & \text{Rewrite as multiplication}\text{.}\hfill \\ \dfrac{y\left(xy+1\right)}{{x}^{2}}\hfill & \text{Multiply}\text{.}\hfill \end{array}$

Limits: Prerequisite Material