Quadratic Equations

We have already solved linear equations, equations of the form $a x + b y = c$ . In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. Listed below are some examples of quadratic equations:

\begin{matrix} x^{2} + 5 x + 6 = 0 & 3 y^{2} + 4 y = 10 & 64 u^{2} - 81 = 0 & n (n + 1) = 42 \end{matrix}

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get $n^{2} + n$ .

The general form of a quadratic equation is $a x^{2} + b x + c = 0, with a \neq 0$ .

To solve quadratic equations we need methods different than the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

Solve Quadratic Equations Using the Zero Product Property

We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

We will now use the Zero Product Property, to solve a quadratic equation.

How to Use the Zero Product Property to Solve a Quadratic Equation

Solve: $(x + 1) (x - 4) = 0$ .

Solution

This table gives the steps for solving (x + 1)(x – 4) = 0. The first step is to set each factor equal to 0. Since it is a product equal to 0, at least one factor must equal 0. x + 1 = 0 or x – 4 = 0.

The last step is to check both answers by substituting the values for x into the original equation.

We usually will do a little more work than we did in this last example to solve the linear equations that result from using the Zero Product Property.

Solve: $(5 n - 2) (6 n - 1) = 0$ .

Solution

	$(5 n - 2) (6 n - 1) = 0$
Use the Zero Product Property to set each factor to 0.	$5 n - 2 = 0$	$6 n - 1 = 0$
Solve the equations.	$n = \frac{2}{5}$	$n = \frac{1}{6}$
Check your answers.

Notice when we checked the solutions that each of them made just one factor equal to zero. But the product was zero for both solutions.

Solve: $3 p (10 p + 7) = 0$ .

Solution

	$3 p (10 p + 7) = 0$
Use the Zero Product Property to set each factor to 0.	$3 p = 0$	$10 p + 7 = 0$
Solve the equations.	$p = 0$	$10 p = −7$
		$p = - \frac{7}{10}$
Check your answers.

It may appear that there is only one factor in the next example. Remember, however, that ${(y - 8)}^{2}$ means $(y - 8) (y - 8)$ .

Solve: ${(y - 8)}^{2} = 0$ .

Solution

	${(y - 8)}^{2} = 0$
Rewrite the left side as a product.	$(y - 8) (y - 8) = 0$
Use the Zero Product Property and set each factor to 0.	$y - 8 = 0$	$y - 8 = 0$
Solve the equations.	$y = 8$	$y = 8$
When a solution repeats, we call it a double root.
Check your answer.

Solve Quadratic Equations by Factoring

Each of the equations we have solved in this section so far had one side in factored form. In order to use the Zero Product Property, the quadratic equation must be factored, with zero on one side. So we must be sure to start with the quadratic equation in standard form, $a x^{2} + b x + c = 0$ . Then we can factor the expression on the left.

How to Solve a Quadratic Equation by Factoring

Solve: $x^{2} + 2 x - 8 = 0$ .

Solution

This table gives the steps for solving the equation x squared + 2 x – 8 = 0. The first step is writing the equation in standard quadratic form, which it is.

The second step is factoring the quadratic expression x squared + 2 x – 8. The factors are (x + 4), (x – 2).

The next step is using the zero product property and set each factor equal to 0, x + 4 = 0 and x – 2 = 0.

The next step is solving both linear equations, x = −4 or x = 2.

The last step is checking both solutions by substituting them into the original equation.

Before we factor, we must make sure the quadratic equation is in standard form.

Solve: $2 y^{2} = 13 y + 45$ .

Solution

	$2 y^{2} = 13 y + 45$
Write the quadratic equation in standard form.	$2 y^{2} - 13 y - 45 = 0$
Factor the quadratic expression.	$(2 y + 5) (y - 9) = 0$
Use the Zero Product Property to set each factor to 0.	$2 y + 5 = 0$	$y - 9 = 0$
Solve each equation.	$y = - \frac{5}{2}$	$y = 9$
Check your answers.

Solve: $5 x^{2} - 13 x = 7 x$ .

Solution

	$5 x^{2} - 13 x = 7 x$
Write the quadratic equation in standard form.	$5 x^{2} - 20 x = 0$
Factor the left side of the equation.	$5 x (x - 4) = 0$
Use the Zero Product Property to set each factor to 0.	$5 x = 0$	$x - 4 = 0$
Solve each equation.	$x = 0$	$x = 4$
Check your answers.

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

Solve: $144 q^{2} = 25$ .

Solution


Write the quadratic equation in standard form.
Factor. It is a difference of squares.
Use the Zero Product Property to set each factor to 0.	$\begin{matrix} 12 q - 5 = 0 \end{matrix}$	$\begin{matrix} 12 q + 5 = 0 \end{matrix}$
Solve each equation.	$\begin{matrix} 12 q & = & 5 \\ q & = & \frac{5}{12} \end{matrix}$	$\begin{matrix} 12 q & = & –5 \\ q & = & - \frac{5}{12} \end{matrix}$
Check your answers.

The left side in the next example is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Solve: $(3 x - 8) (x - 1) = 3 x$ .

Solution

Step-by-step solution of a quadratic equation by factoring, demonstrating transformations to find the roots.
	$\begin{matrix} (3 x - 8) (x - 1) & = & 3 x \end{matrix}$
Multiply the binomials.	$\begin{matrix} 3 x^{2} - 11 x + 8 & = & 3 x \end{matrix}$
Write the quadratic equation in standard form.	$\begin{matrix} 3 x^{2} - 14 x + 8 & = & 0 \end{matrix}$
Factor the trinomial.	$\begin{matrix} (3 x - 2) (x - 4) & = & 0 \end{matrix}$
Use the Zero Product Property to set each factor to 0. Solve each equation.	$\begin{matrix} 3 x - 2 & = & 0 & x - 4 & = & 0 \\ 3 x & = & 2 & x & = & 4 \\ x & = & \frac{2}{3} \end{matrix}$
Check your answers.	The check is left to you!

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree more than two by using the Zero Product Property, just like we solved quadratic equations.

Solve: $9 m^{3} + 100 m = 60 m^{2}$ .

Solution

Steps for solving algebraic equations by factoring, showing each action and its mathematical representation.

Bring all the terms to one side so that the other side is zero.
Factor the greatest common factor first.
Factor the trinomial.
Use the Zero Product Property to set each factor to 0.
Solve each equation.	$Three mathematical expressions are shown: m = 0, m = 10/3, and m = 10/3. The letter 'm' is equal to zero in the first expression and to the fraction 10 over 3 in the other two.$
Check your answers.	The check is left to you.

When we factor the quadratic equation in the next example we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

Solve: $4 x^{2} = 16 x + 84$ .

Solution

Steps demonstrating how to solve a quadratic equation by factoring, from the initial equation to the final solutions.
	$\begin{matrix} 4 x^{2} & = & 16 x + 84 \end{matrix}$
Write the quadratic equation in standard form.	$\begin{matrix} 4 x^{2} - 16 x - 84 & = & 0 \end{matrix}$
Factor the greatest common factor first.	$\begin{matrix} 4 (x^{2} - 4 x - 21) & = & 0 \end{matrix}$
Factor the trinomial.	$\begin{matrix} 4 (x - 7) (x + 3) & = & 0 \end{matrix}$
Use the Zero Product Property to set each factor to 0. Solve each equation.	$\begin{matrix} 4 & \neq & 0 & x - 7 & = & 0 & x + 3 & = & 0 \\ 4 & \neq & 0 & x & = & 7 & x & = & −3 \end{matrix}$
Check your answers.	The check is left to you.

Solve Applications Modeled by Quadratic Equations

The problem solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to quadratic equations. We will copy the problem solving strategy here so we can use it for reference.

We will start with a number problem to get practice translating words into a quadratic equation.

The product of two consecutive integers is 132. Find the integers.

Solution

This table outlines a 7-step process for solving a word problem to find consecutive integers whose product is 132, from reading the problem to checking the final answer.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for two consecutive integers.
Step 3. Name what we are looking for.	$\begin{matrix} Let n = the first integer \\ n + 1 = the next consecutive integer \end{matrix}$
Step 4. Translate into an equation. Restate the problem in a sentence.	The product of the two consecutive integers is 132.
	The first integer times the next integer is 132.
Translate to an equation.	$\begin{matrix} n (n + 1) & = & 132 \end{matrix}$
Step 5. Solve the equation.	$\begin{matrix} n^{2} + n & = & 132 \end{matrix}$
Bring all the terms to one side.	$\begin{matrix} n^{2} + n - 132 & = & 0 \end{matrix}$
Factor the trinomial.	$\begin{matrix} (n - 11) (n + 12) & = & 0 \end{matrix}$
Use the zero product property. Solve the equations.	$\begin{matrix} n - 11 & = & 0 & n + 12 & = & 0 \\ n & = & 11 & n & = & −12 \end{matrix}$
There are two values for $n$ that are solutions to this problem. So there are two sets of consecutive integers that will work.
$\begin{matrix} If the first integer is n = 11 & If the first integer is n = −12 \\ then the next integer is n + 1 & then the next integer is n + 1 \\ 11 + 1 & −12 + 1 \\ 12 & −11 \end{matrix}$
Step 6. Check the answer.
The consecutive integers are $11, 12$ and $−11, −12$ . The product $11 \cdot 12 = 132$ and the product $−11 (−12) = 132$ . Both pairs of consecutive integers are solutions.
Step 7. Answer the question. The consecutive integers are $11, 12$ and $−11, −12$ .

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give 132.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

A rectangular garden has an area $15$ square feet. The length of the garden is two feet more than the width. Find the length and width of the garden.

Solution

Step 1. Read the problem. In problems involving geometric figures, a sketch can help you visualize the situation.
Step 2. Identify what you are looking for.	We are looking for the length and width.
Step 3. Name what you are looking for. The length is two feet more than width.	Let W = the width of the garden. W + 2 = the length of the garden
Step 4. Translate into an equation. Restate the important information in a sentence.	The area of the rectangular garden is 15 square feet.
Use the formula for the area of a rectangle.	$A = L \cdot W$
Substitute in the variables.	$15 = (W + 2) W$
Step 5. Solve the equation. Distribute first.	$15 = W^{2} + 2 W$
Get zero on one side.	$0 = W^{2} + 2 W - 15$
Factor the trinomial.	$0 = (W + 5) (W - 3)$
Use the Zero Product Property.	$0 = W + 5$	$0 = W - 3$
Solve each equation.	$−5 = W$	$3 = W$
Since W is the width of the garden, it does not make sense for it to be negative. We eliminate that value for W.	$−5 = W$ $W = 3$	$3 = W$ Width is 3 feet.
Find the value of the length.	$W + 2 = length$
	$3 + 2$
	$5$	Length is 5 feet.
Step 6. Check the answer. Does the answer make sense?

	Yes, this makes sense.
Step 7. Answer the question.	The width of the garden is 3 feet and the length is 5 feet.

In an earlier chapter, we used the Pythagorean Theorem $(a^{2} + b^{2} = c^{2})$ . It gave the relation between the legs and the hypotenuse of a right triangle.

We will use this formula in the next example.

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle, as shown below. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the deck.

This figure is a right triangle. The vertical leg is labeled “x – 7”. the horizontal leg, the base, is labeled “x”. The hypotenuse is labeled “17”.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	We are looking for the lengths of the sides of the deck.
Step 3. Name what you are looking for. One side is 7 less than the other.	Let x = length of a side of the deck x − 7 = length of other side
Step 4. Translate into an equation. Since this is a right triangle we can use the Pythagorean Theorem.	$a^{2} + b^{2} = c^{2}$
Substitute in the variables.	$x^{2} + {(x - 7)}^{2} = 17^{2}$
Step 5. Solve the equation.	$x^{2} + x^{2} - 14 x + 49 = 289$
Simplify.	$2 x^{2} - 14 x + 49 = 289$
It is a quadratic equation, so get zero on one side.	$2 x^{2} - 14 x - 240 = 0$
Factor the greatest common factor.	$2 (x^{2} - 7 x - 120) = 0$
Factor the trinomial.	$2 (x - 15) (x + 8) = 0$
Use the Zero Product Property.	$2 \neq 0$	$x - 15 = 0$	$x + 8 = 0$
Solve.	$2 \neq 0$	$x = 15$	$x = −8$
Since x is a side of the triangle, $x = −8$ does not make sense.	$2 \neq 0$	$x = 15$	$x = −8$
Find the length of the other side.
If the length of one side is
then the length of the other side is

	8 is the length of the other side.
Step 6. Check the answer. Do these numbers make sense?

Step 7. Answer the question.	The sides of the deck are 8, 15, and 17 feet.

Key Concepts

Zero Product Property If $a \cdot b = 0$ , then either $a = 0$ or $b = 0$ or both. See Example 1.
Solve a quadratic equation by factoring To solve a quadratic equation by factoring: See Example 5.
1. Write the quadratic equation in standard form, $a x^{2} + b x + c = 0$ .
2. Factor the quadratic expression.
3. Use the Zero Product Property.
4. Solve the linear equations.
5. Check.
Use a problem solving strategy to solve word problems See Example 12.
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Section Exercises

Practice Makes Perfect

Use the Zero Product Property

In the following exercises, solve.

$(x - 3) (x + 7) = 0$

Solution

$x = 3, x = −7$

$(y - 11) (y + 1) = 0$

$(3 a - 10) (2 a - 7) = 0$

Solution

$a = 10 / 3, a = 7 / 2$

$(5 b + 1) (6 b + 1) = 0$

$6 m (12 m - 5) = 0$

Solution

$m = 0, m = 5 / 12$

$2 x (6 x - 3) = 0$

${(y - 3)}^{2} = 0$

Solution

$y = 3$

${(b + 10)}^{2} = 0$

${(2 x - 1)}^{2} = 0$

Solution

$x = 1 / 2$

${(3 y + 5)}^{2} = 0$

Solve Quadratic Equations by Factoring

In the following exercises, solve.

$x^{2} + 7 x + 12 = 0$

Solution

$x = −3, x = −4$

$y^{2} - 8 y + 15 = 0$

$5 a^{2} - 26 a = 24$

Solution

$a = −4 / 5, a = 6$

$4 b^{2} + 7 b = −3$

$4 m^{2} = 17 m - 15$

Solution

$m = 5 / 4, m = 3$

$n^{2} = 5 n - 6$

$7 a^{2} + 14 a = 7 a$

Solution

$a = −1, a = 0$

$12 b^{2} - 15 b = −9 b$

$49 m^{2} = 144$

Solution

$m = 12 / 7, m = −12 / 7$

$625 = x^{2}$

$(y - 3) (y + 2) = 4 y$

Solution

$y = −1, y = 6$

$(p - 5) (p + 3) = −7$

$(2 x + 1) (x - 3) = −4 x$

Solution

$x = 3 / 2, x = −1$

$(x + 6) (x - 3) = −8$

$16 p^{3} = 24 p^{2} - 9 p$

Solution

$p = 0, p = ¾$

$m^{3} - 2 m^{2} = − m$

$20 x^{2} - 60 x = −45$

Solution

$x = 3 / 2$

$3 y^{2} - 18 y = −27$

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve.

The product of two consecutive integers is 56. Find the integers.

Solution

$7 and 8; - 8 and −7$

The product of two consecutive integers is 42. Find the integers.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

Solution

$4 feet and 7 feet$

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

Solution

$6 feet and 8 feet$

A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool.

Mixed Practice

In the following exercises, solve.

$(x + 8) (x - 3) = 0$

Solution

$x = −8, x = 3$

$(3 y - 5) (y + 7) = 0$

$p^{2} + 12 p + 11 = 0$

Solution

$p = −1, p = −11$

$q^{2} - 12 q - 13 = 0$

$m^{2} = 6 m + 16$

Solution

$m = −2, m = 8$

$4 n^{2} + 19 n = 5$

$a^{3} - a^{2} - 42 a = 0$

Solution

$a = 0, a = −6, a = 7$

$4 b^{2} - 60 b + 224 = 0$

The product of two consecutive integers is 110. Find the integers.

Solution

$10 and 11; - 11 and −10$

The length of one leg of a right triangle is three feet more than the other leg. If the hypotenuse is 15 feet, find the lengths of the two legs.

Everyday Math

Area of a patio If each side of a square patio is increased by 4 feet, the area of the patio would be 196 square feet. Solve the equation ${(s + 4)}^{2} = 196$ for s to find the length of a side of the patio.

Solution

10 feet

Watermelon drop A watermelon is dropped from the tenth story of a building. Solve the equation $−16 t^{2} + 144 = 0$ for $t$ to find the number of seconds it takes the watermelon to reach the ground.

Writing Exercises

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

Solution

Answers may vary.

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has the following statements all to be preceded by “I can…”. The first row is “solve quadratic equations by using the zero product property”. The second row is “solve quadratic equations by factoring”. The third row is “solve applications modeled by quadratic equations”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter 7 Review Exercises

7.1 Greatest Common Factor and Factor by Grouping

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

$42, 60$

Solution

$450, 420$

$90, 150, 105$

Solution

$15$

$60, 294, 630$

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

$24 x - 42$

Solution

$6 (4 x - 7)$

$35 y + 84$

$15 m^{4} + 6 m^{2} n$

Solution

$3 m^{2} (5 m^{2} + 2 n)$

$24 p t^{4} + 16 t^{7}$

Factor by Grouping

In the following exercises, factor by grouping.

$a x - a y + b x - b y$

Solution

$(a + b) (x - y)$

$x^{2} y - x y^{2} + 2 x - 2 y$

$x^{2} + 7 x - 3 x - 21$

Solution

$(x - 3) (x + 7)$

$4 x^{2} - 16 x + 3 x - 12$

$m^{3} + m^{2} + m + 1$

Solution

$(m^{2} + 1) (m + 1)$

$5 x - 5 y - y + x$

7.2 Factor Trinomials of the form $x^{2} + b x + c$

Factor Trinomials of the Form $x^{2} + b x + c$

In the following exercises, factor each trinomial of the form $x^{2} + b x + c$ .

$u^{2} + 17 u + 72$

Solution

$(u + 8) (u + 9)$

$a^{2} + 14 a + 33$

$k^{2} - 16 k + 60$

Solution

$(k - 6) (k - 10)$

$r^{2} - 11 r + 28$

$y^{2} + 6 y - 7$

Solution

$(y + 7) (y - 1)$

$m^{2} + 3 m - 54$

$s^{2} - 2 s - 8$

Solution

$(s - 4) (s + 2)$

$x^{2} - 3 x - 10$

Factor Trinomials of the Form $x^{2} + b x y + c y^{2}$

In the following examples, factor each trinomial of the form $x^{2} + b x y + c y^{2}$ .

$x^{2} + 12 x y + 35 y^{2}$

Solution

$(x + 5 y) (x + 7 y)$

$u^{2} + 14 u v + 48 v^{2}$

$a^{2} + 4 a b - 21 b^{2}$

Solution

$(a + 7 b) (a - 3 b)$

$p^{2} - 5 p q - 36 q^{2}$

7.3 Factoring Trinomials of the form $a x^{2} + b x + c$

Recognize a Preliminary Strategy to Factor Polynomials Completely

In the following exercises, identify the best method to use to factor each polynomial.

$y^{2} - 17 y + 42$

Solution

Undo FOIL

$12 r^{2} + 32 r + 5$

$8 a^{3} + 72 a$

Solution

Factor the GCF

$4 m - m n - 3 n + 12$

Factor Trinomials of the Form $a x^{2} + b x + c$ with a GCF

In the following exercises, factor completely.

$6 x^{2} + 42 x + 60$

Solution

$6 (x + 2) (x + 5)$

$8 a^{2} + 32 a + 24$

$3 n^{4} - 12 n^{3} - 96 n^{2}$

Solution

$3 n^{2} (n - 8) (n + 4)$

$5 y^{3} + 25 y^{2} - 70 y$

Factor Trinomials Using the “ac” Method

In the following exercises, factor.

$2 x^{2} + 9 x + 4$

Solution

$(x + 4) (2 x + 1)$

$3 y^{2} + 17 y + 10$

$18 a^{2} - 9 a + 1$

Solution

$(3 a - 1) (6 a - 1)$

$8 u^{2} - 14 u + 3$

$15 p^{2} + 2 p - 8$

Solution

$(5 p + 4) (3 p - 2)$

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

$40 s^{2} - s - 6$

Solution

$(5 s - 2) (8 s + 3)$

$20 n^{2} - 7 n - 3$

Factor Trinomials with a GCF Using the “ac” Method

In the following exercises, factor.

$3 x^{2} + 3 x - 36$

Solution

$3 (x + 4) (x - 3)$

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

$60 y^{2} - 85 y - 25$

Solution

$5 (4 y + 1) (3 y - 5)$

$18 a^{2} - 57 a - 21$

7.4 Factoring Special Products

Factor Perfect Square Trinomials

In the following exercises, factor.

$25 x^{2} + 30 x + 9$

Solution

${(5 x + 3)}^{2}$

$16 y^{2} + 72 y + 81$

$36 a^{2} - 84 a b + 49 b^{2}$

Solution

${(6 a - 7 b)}^{2}$

$64 r^{2} - 176 r s + 121 s^{2}$

$40 x^{2} + 360 x + 810$

Solution

$10 {(2 x + 9)}^{2}$

$75 u^{2} + 180 u + 108$

$2 y^{3} - 16 y^{2} + 32 y$

Solution

$2 y {(y - 4)}^{2}$

$5 k^{3} - 70 k^{2} + 245 k$

Factor Differences of Squares

In the following exercises, factor.

$81 r^{2} - 25$

Solution

$(9 r - 5) (9 r + 5)$

$49 a^{2} - 144$

$169 m^{2} - n^{2}$

Solution

$(13 m + n) (13 m - n)$

$64 x^{2} - y^{2}$

$25 p^{2} - 1$

Solution

$(5 p - 1) (5 p + 1)$

$1 - 16 s^{2}$

$9 - 121 y^{2}$

Solution

$(3 + 11 y) (3 - 11 y)$

$100 k^{2} - 81$

$20 x^{2} - 125$

Solution

$5 (2 x - 5) (2 x + 5)$

$18 y^{2} - 98$

$49 u^{3} - 9 u$

Solution

$u (7 u + 3) (7 u - 3)$

$169 n^{3} - n$

Factor Sums and Differences of Cubes

In the following exercises, factor.

$a^{3} - 125$

Solution

$(a - 5) (a^{2} + 5 a + 25)$

$b^{3} - 216$

$2 m^{3} + 54$

Solution

$2 (m + 3) (m^{2} - 3 m + 9)$

$81 x^{3} + 3$

7.5 General Strategy for Factoring Polynomials

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

In the following exercises, factor completely.

$24 x^{3} + 44 x^{2}$

Solution

$4 x^{2} (6 x + 11)$

$24 a^{4} - 9 a^{3}$

$16 n^{2} - 56 m n + 49 m^{2}$

Solution

${(4 n - 7 m)}^{2}$

$6 a^{2} - 25 a - 9$

$5 r^{2} + 22 r - 48$

Solution

$(r + 6) (5 r - 8)$

$5 u^{4} - 45 u^{2}$

$n^{4} - 81$

Solution

$(n^{2} + 9) (n + 3) (n - 3)$

$64 j^{2} + 225$

$5 x^{2} + 5 x - 60$

Solution

$5 (x - 3) (x + 4)$

$b^{3} - 64$

$m^{3} + 125$

Solution

$(m + 5) (m^{2} - 5 m + 25)$

$2 b^{2} - 2 b c + 5 c b - 5 c^{2}$

7.6 Quadratic Equations

Use the Zero Product Property

In the following exercises, solve.

$(a - 3) (a + 7) = 0$

Solution

$a = 3, a = −7$

$(b - 3) (b + 10) = 0$

$3 m (2 m - 5) (m + 6) = 0$

Solution

$m = 0 m = –6 m = \frac{5}{2}$

$7 n (3 n + 8) (n - 5) = 0$

Solve Quadratic Equations by Factoring

In the following exercises, solve.

$x^{2} + 9 x + 20 = 0$

Solution

$x = −4, x = −5$

$y^{2} - y - 72 = 0$

$2 p^{2} - 11 p = 40$

Solution

$p = - \frac{5}{2}, p = 8$

$q^{3} + 3 q^{2} + 2 q = 0$

$144 m^{2} - 25 = 0$

Solution

$m = \frac{5}{12}, m = - \frac{5}{12}$

$4 n^{2} = 36$

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve.

The product of two consecutive numbers is $462$ . Find the numbers.

Solution

$−21 and −22; 21 and 22$

The area of a rectangular shaped patio $400$ square feet. The length of the patio is $9$ feet more than its width. Find the length and width.

Practice Test

In the following exercises, find the Greatest Common Factor in each expression.

$14 y - 42$

Solution

$14 (y - 3)$

$−6 x^{2} - 30 x$

$80 a^{2} + 120 a^{3}$

Solution

$40 a^{2} (2 + 3 a)$

$5 m (m - 1) + 3 (m - 1)$

In the following exercises, factor completely.

$x^{2} + 13 x + 42$

Solution

$(x + 7) (x + 6)$

$p^{2} + p q - 12 q^{2}$

$3 a^{3} - 6 a^{2} - 72 a$

Solution

$3 a (a - 6) (a + 4)$

$s^{2} - 25 s + 84$

$5 n^{2} + 30 n + 45$

Solution

$5 (n + 3)^{2}$

$64 y^{2} - 49$

$x y - 8 y + 7 x - 56$

Solution

$(x - 8) (y + 7)$

$40 r^{2} + 810$

$9 s^{2} - 12 s + 4$

Solution

${(3 s - 2)}^{2}$

$n^{2} + 12 n + 36$

$100 - a^{2}$

Solution

$(10 - a) (10 + a)$

$6 x^{2} - 11 x - 10$

$3 x^{2} - 75 y^{2}$

Solution

$3 (x + 5 y) (x - 5 y)$

$c^{3} - 1000 d^{3}$

$a b - 3 b - 2 a + 6$

Solution

$(a - 3) (b - 2)$

$6 u^{2} + 3 u - 18$

$8 m^{2} + 22 m + 5$

Solution

$(4 m + 1) (2 m + 5)$

In the following exercises, solve.

$x^{2} + 9 x + 20 = 0$

$y^{2} = y + 132$

Solution

$y = −11, y = 12$

$5 a^{2} + 26 a = 24$

$9 b^{2} - 9 = 0$

Solution

$b = 1, b = −1$

$16 - m^{2} = 0$

$4 n^{2} + 19 n + 21 = 0$

Solution

$n = - \frac{7}{4}, n = −3$

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

The product of two consecutive integers is 156. Find the integers.

Solution

$12 and 13; - 13 and −12$

The area of a rectangular place mat is $168$ square inches. Its length is two inches longer than the width. Find the length and width of the place mat.

quadratic equations

are equations in which the variable is squared.

Zero Product Property

The Zero Product Property states that, if the product of two quantities is zero, at least one of the quantities is zero.

Quadratic Equations

Solve Quadratic Equations Using the Zero Product Property

How to Use the Zero Product Property to Solve a Quadratic Equation

Solution

Solution

Solution

Solution

Solve Quadratic Equations by Factoring

How to Solve a Quadratic Equation by Factoring

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solve Applications Modeled by Quadratic Equations

Solution

Solution

Solution

Key Concepts

Section Exercises

Practice Makes Perfect

Everyday Math

Writing Exercises

Self Check

Chapter 7 Review Exercises

7.1 Greatest Common Factor and Factor by Grouping

7.2 Factor Trinomials of the form x2+bx+c

7.3 Factoring Trinomials of the form ax2+bx+c

7.4 Factoring Special Products

7.5 General Strategy for Factoring Polynomials

7.6 Quadratic Equations

Practice Test

7.2 Factor Trinomials of the form $x^{2} + b x + c$

7.3 Factoring Trinomials of the form $a x^{2} + b x + c$