### SubsectionQuotients of Powers

We can reduce a fraction by dividing numerator and denominator by any common factors. For example,

\begin{equation*} \dfrac{10}{15} = \dfrac{2 \cdot 5}{3 \cdot 5} = \dfrac{2}{3} \end{equation*}

We first factored the numerator and denominator of the fraction and then canceled the 5's by dividing. We can apply the same technique to quotients of powers.

###### Example7.20.

Simplify.

1. $\dfrac{a^5}{a^3}$
2. $\dfrac{a^4}{a^8}$
Solution
1. We first write the numerator and denominator in factored form. Then we divide any common factors from the numerator and denominator.

\begin{equation*} \dfrac{a^5}{a^3} = \dfrac{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot a\cdot a}{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}} = \dfrac{a^2}{1} = a^2 \end{equation*}

You may observe that the exponent of the quotient can be obtained by subtracting the exponent of the denominator from the exponent of the numerator. In other words,

\begin{equation*} \dfrac{a^5}{a^3}= a^{5-3} = a^2 \end{equation*}
2. In this quotient, the larger power occurs in the denominator.

\begin{equation*} \dfrac{a^4}{a^8} = \dfrac{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot \cancel{a}}{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot a\cdot a\cdot a \cdot a} = \dfrac{1}{a^4} \end{equation*}

We see that we can subtract the exponent of the numerator from the exponent of the denominator. That is,

\begin{equation*} \dfrac{a^4}{a^8}=\dfrac{1} {a^{8-4}} = \dfrac{1}{a^4} \end{equation*}

These examples suggest the following property.

###### Second Law of Exponents.

To divide two powers with the same base, we subtract the smaller exponent from the larger one, and keep the same base.

1. If the larger exponent occurs in the numerator, put the power in the numerator.
2. If the larger exponent occurs in the denominator, put the power in the denominator.

In symbols,

1. $\blert{\dfrac{a^m}{a^n} = a^{m-n}~~~~~~\text{if}~~~~n \lt m}$
2. $\blert{\dfrac{a^m}{a^n} = \dfrac{1}{a^{n-m}}~~~~~~\text{if}~~~~n \gt m}$

###### 1.

In a quotient of powers, how do we know whether the new power appears in the numerator or the denominator?

### SubsectionQuotients of Monomials

To divide one monomial by another, we apply the second law of exponents to the powers of each variable.

###### Example7.21.

Divide $~~\dfrac{3x^2y^4}{6x^3y}$

Solution

We consider the numerical coefficients and the powers of each base separately. We use the second law of exponents to simplify each quotient of powers.

\begin{align*} \dfrac{3x^2y^4}{6x^3y} \amp = \dfrac{3}{6} \cdot \dfrac{x^2}{x^3} \cdot \dfrac{y^4}{y} \amp \amp \blert{\text{Subtract exponents on each base.}}\\ \amp = \dfrac{1}{2} \cdot \dfrac{1}{x^{3-2}} \cdot y^{4-1}\\ \amp = \dfrac{1}{2} \cdot \dfrac{1}{x} \cdot \dfrac{y^3}{1}=\dfrac{y^3}{2x} \amp \amp \blert{\text{Multiply factors.}} \end{align*}

### SubsectionGreatest Common Factors

Now we consider several techniques for factoring polynomials. The first of these is factoring out the greatest common factor (GCF).

The greatest common factor (GCF) is the largest factor that divides evenly into each term of the polynomial: the largest numerical factor and the highest power of each variable.

###### Example7.22.

Find the greatest common factor for $~~4a^3b^2+6ab^3-18a^2b^4$

Solution

The largest integer that divides evenly into all three coefficients is 2. The highest power of $a$ is $a^1\text{,}$ and the highest power of $b$ is $b^2\text{.}$ Thus, the GCF is $2ab^2\text{.}$ Note that the exponent on each variable of $2ab^2$ is the smallest exponent that appears on that variable among the terms of the polynomial.

###### 2.

The exponent on each variable of the GCF is the exponent that appears on that variable among the terms of the polynomial.

Once we have found the greatest common factor for the polynomial, we can write each term as a product of the GCF and another factor. For example, the GCF of $8x^2-6x$ is $2x\text{,}$ and we can write

\begin{equation*} 8x^2-6x = \blert{2x} \cdot 4x - \blert{2x} \cdot 3 \end{equation*}

We can then use the distributive law to write the expression on the right side as a product.

\begin{equation*} \blert{2x} \cdot 4x - \blert{2x} \cdot 3 = \blert{2x} (4x-3) \end{equation*}

We say that we have factored out the greatest common factor from the polynomial.

For more complicated polynomials, we can divide the GCF into each term to find the remaining factors.

###### Example7.23.

Factor $~~4a^3b^2+6ab^3-18a^2b^4$

Solution

The GCF for this polynomial is $2ab^2\text{,}$ as we saw in Example 7.22. We factor out the GCF from each term and write the polynomial as a product,

\begin{equation*} 2ab^2(\hphantom{0000000000}) \end{equation*}

To find the factor inside parentheses, we divide each term of the polynomial by the GCF.

\begin{equation*} \dfrac{4a^3b^2}{2ab^2}=\blert{2a^2},~~~~\dfrac{6ab^3}{2ab^2}=\blert{3b},~~~\dfrac{-18a^2b^4}{2ab^2}=\blert{-9ab^2} \end{equation*}

We apply the distributive law to factor $2ab^2$ from each term.

\begin{align*} 4a^3b^2+6ab^3-18a^2b^4 \amp =2ab^2\cdot \blert{2a^2}+2ab^2 \cdot \blert{3b-}2ab^2 \cdot \blert{9ab^2}\\ \amp = 2ab^2(2a^2+3b-9ab^2) \end{align*}
###### Look Closer.

Sometimes the greatest common factor for a polynomial is not a monomial, but may instead have two or more terms.

###### Example7.24.

Factor $x^2(2x+1)-3(2x+1)$

Solution

The given expression has two terms, and $(2x+1)$ is a factor of each. We factor out the entire binomial $(2x+1)\text{.}$

\begin{equation*} x^2\blert{(2x+1)}-3\blert{(2x+1)} = \blert{(2x+1)} (x^2-3) \end{equation*}

### SubsectionFactoring Quadratic Trinomials by Guess-and-Check

So far, we can factor quadratic trinomials of the form $x^2+bx+c=0\text{,}$ where $a=1\text{.}$ Recall that the factored form of $x^2+bx+c=0$ looks like

\begin{equation*} (x+p)(x+q)~~~~~~ \text{where}~~~pq=c ~~~ \text{and} ~~~p+q=b \end{equation*}

You may want to review the FOIL method discussed in Lesson 6.3.

What if the coefficient of $a$ is not 1? Sometimes, if the coefficients are small, we can factor a quadratic expression by the guess-and-check method.

###### Example7.25.

Factor $~2x^2-7x+3~$ into a product of two binomials.

Solution

We begin by factoring the quadratic term, $2x^2\text{,}$ which can only be factored as $x$ times $2x\text{,}$ so we can fill in the First terms in each binomial.

\begin{equation*} 2x^2-7x+3 = (2x\underline{\hspace{2.727272727272727em}})(x\underline{\hspace{2.727272727272727em}}) \end{equation*}

Next, we factor the constant term, $3\text{,}$ which can only be factored as 3 times 1. Because the linear term is negative, we make both factors negative: $3=-3(-1)\text{.}$ Finally, we have to decide which number appears as the Last term in each binomial. We check $O+I$ for each possibility.

\begin{align*} (2x-3)(x-1)~~~~O+I \amp = -2x-3x=-5x\\ (2x-1)(x-3)~~~~\blert{O+I} \amp \blert{= -6x-x=-7x} \end{align*}

The second choice gives the correct middle term, so the factorization is

\begin{equation*} 2x^2-7x+3 = (2x-1)(x-3) \end{equation*}

###### 3.

When we factor a quadratic trinomial by the guess-and-check method, how do we check for the correct middle term?

Remember that you can always check your factorization by multiplying the factors together.

### SubsectionQuadratic Trinomials in Two Variables

So far, we have factored quadratic trinomials in one variable, that is, polynomials of the form

\begin{equation*} ax^2+bx+c \end{equation*}

The method we learned can also be used to factor trinomials in two variables of the form

\begin{equation*} ax^2+bxy+cy^2 \end{equation*}

In this expression, the first and last terms are quadratic terms, while the middle term is a cross-term consisting of the product of the two variables.

###### Example7.26.

Factor $~x^2+5xy+6y^2~$

Solution

As usual, we begin by factoring the first term into $x$ times $x\text{.}$

\begin{equation*} x^2+5xy+6y^2 = (x + \underline{\hspace{2.727272727272727em}})(x + \underline{\hspace{2.727272727272727em}}) \end{equation*}

Next we look for factors of the last term, $6y^2\text{.}$ In order to obtain the $xy$-term in the middle, we need a $y$ in each factor. Thus the possibilities are

\begin{equation*} y~~\text{and}~~6y~~~~~~~~\text{or}~~~~~~~~2y~~\text{and}~~3y \end{equation*}

We'll check the sum $O+I$ for each possibility.

\begin{align*} (x+y)(x+6y)~~~~O+I \amp = 6xy+xy=7xy\\ (x+2y)(x+3y)~~~~\blert{O+I} \amp \blert{= 3xy+2xy=5xy} \end{align*}

The second possibility gives the correct middle term, so the factorization is

\begin{equation*} x^2+5xy+6y^2 = (x+2y)(x+3y) \end{equation*}

### SubsectionCombining Factoring Techniques

We should always begin factoring by checking to see if there is a common factor that can be factored out.

###### Example7.27.

Factor completely $~2b^3+8b^2+6b~$

Solution

We begin by factoring out the greatest common factor, $2b\text{.}$

\begin{equation*} 2b^3+8b^2+6b = 2b(b^2+4b+3) \end{equation*}

The remaining factor, $b^2+4b+3\text{,}$ is a quadratic trinomial that can be factored. We look for two numbers $p$ and $q$ so that $pq=3$ and $p+q=4\text{.}$ You can check that $p=3$ and $q=1$ will work.

Thus, $~~b^2+4b+3 = (b+3)(b+1)\text{,}$ and

\begin{equation*} 2b^3+8b^2+6b = 2b(b+3)(b+1) \end{equation*}

###### 4.

What should always be the first step in factoring a polynomial?

### SubsectionSkills Warm-Up

#### ExercisesExercises

Mental exercise: Find the other factor of each quadratic trinomial without using pencil, paper, or calculator.

###### 1.
$b^2+8b-240 = (b-12)(\underline{\hspace{2.727272727272727em}})$
###### 2.
$n^2-97n-300 = (n-100)(\underline{\hspace{2.727272727272727em}})$
###### 3.
$3u^2-17u-6 = (u-6)(\underline{\hspace{2.727272727272727em}})$
###### 4.
$2t^2-21t+54 = (t-6)(\underline{\hspace{2.727272727272727em}})$

### ExercisesHomework 7.3

###### 1.

Find each quotient by using the second law of exponents.

1. $\dfrac{a^6}{a^3}$
2. $\dfrac{3^9}{3^4}$
3. $\dfrac{z^6}{z^9}$
###### 2.

Choose a value for the variable and evaluate to show that the following pairs of expressions are not equivalent.

1. $t^2 \cdot t^3,~~t^6$
2. $\dfrac{v^8}{v^2},~~v^4$
3. $\dfrac{n^3}{n^5},~~n^2$

For Problems 3–5, divide.

###### 3.

$\dfrac{2x^3y}{8x^4y^5}$

###### 4.

$\dfrac{-12bx^4}{8bx^2}$

###### 5.

$\dfrac{-15x^3y^2}{-3x^3y^4}$

For Problems 6–7, factor out a negative monomial.

###### 6.

$-b^2-bc-ab$

###### 7.

$-4k^4+4k^2-2k$

For Problems 8–11, factor out the greatest common factor.

###### 8.

$2x^4-4x^2+8$

###### 9.

$16a^3b^3-12a^2b+8ab^2$

###### 10.

$9x^2-12x^5+3x^3$

###### 11.

$14x^3y-35x^2y^2+21xy^3$

For Problems 12–13, factor out the common factor.

###### 12.

$2x(x+6)-3(2x+6)$

###### 13.

$3x^2(2x+3)-(2x+3)$

For Problems 14–16, factor by the guess-and-check method.

###### 14.

$2x^2+11x+5$

###### 15.

$5t^2+7t+2$

###### 16.

$3x^2-8x+5$

For Problems 17–28, factor completely.

###### 17.

$2x^2+10x+12$

###### 18.

$4a^2b+12ab-7b$

###### 19.

$4z^3+10z^2+6z$

###### 20.

$18a^2b-9ab-27b$

###### 21.

$x^2-5xy+6y^2$

###### 22.

$x^2+4xy+4y^2$

###### 23.

$x^2+4ax-77a^2$

###### 24.

$4x^3+12x^2y+8xy^2$

###### 25.

$9a^3b+9a^2b^2-18ab^3$

###### 26.

$2t^2-5st-3s^2$

###### 27.

$4b^2y^2+5by+1$

###### 28.

$12ab^2+15a^2b+3a^3$

For Problems 29–34, solve the equation by factoring.

###### 29.

$3n^2-n=4$

###### 30.

$11t=6t^2+3$

###### 31.

$1=4y-4y^2$

###### 32.

$12z^2+26z=10$

###### 33.

$y(3y+4)=4$

###### 34.

$(2x-1)(2x-1)=-1$

###### 35.

The cost $C$ of producing a wool rug depends on the number of hours $t$ it takes to weave it, where

\begin{equation*} C=3t^2-4t+100 \end{equation*}

How many hours did it take to weave a rug that costs \$120?

###### 36.

Steve's boat locker is 2 feet longer than twice its width. Find the dimensions of the locker if the 13-foot mast of Steve's boat will just fit diagonally across the floor of the locker.