# 6.5: Use the Zero Product Property - Mathematics

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## Algebra (need help quick)

19. A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y=0.006x^2+10.1x+5, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?
A.168.83 m
B.5.00 m
C.84.17 m
D.168.34 m

20. How many real-number solutions does the equation have? -7x^2+6x+3=0
A. one solution
B. two solutions
C. no solutions
D. infinitely many solutions
If you have the rest of the Unit 8 lesson 2 Semester B exam for Algebra on connexus please let me know

17.
well if x = 0, that works
and if 3x +1 = 0, that works

18. well I already know that x = 0 and x = -1/3 works but
18 x^2 + 6 x + 0 = 0
x = [ -6 +/- sqrt (36 - 0) ] / 36
x = [ -6 +/-6 ] /36
x = 0/36 or x = -12/36
but we knew that

19. just learned how to solve quadratic. So what is x when y = 0? (the big one, not the one from before the launch.)

if 0, one solution (parabola vertex just touches x axis)
if negative, complex solutions, none real
if positive, two real

Thanks Damon for the help! Can you list all the answers in one message so im not confused and put the wrong answer?

## PCC SLC Math Resources

When the product of two numbers is zero, at least one of the numbers in the product must be zero. This is a property unique to zero e.g., when two numbers multiply to, say, ten, not only is it not the case that one of the two numbers need be ten, the only restriction on the two numbers at all is that neither be zero.

When solving an equation of form:

the next step in the process is to state:

We then solve the two new equations separately and state our solutions, or solution set. For example:

The solutions are (-frac<7><4>) and (frac<9><2> ext<.>) The solution set is (<-frac<7><4>, frac<9><2>> ext<.>)

We frequently have to perform some preliminary steps before invoking the zero-product property. Specifically:

1. Completely expand both sides of the equation.
2. Add and/or subtract to/from both sides of the equation so that one side of the equation is zero. The next step will be easier if make sure that the second degree term ((ax^2)) has a positive coefficient.
3. Factor the non-zero side of the equation.
4. We are now set to invoke the zero-product property.
###### Example 12.3.1 .

Use the zero-product property to solve (4x^2=4x+15 ext<.>)

The solutions are (frac<5><2>) and (-frac<3><2> ext<.>) The solution set is (left<2>, -frac<3><2> ight> ext<.>)

###### Example 12.3.2 .

Use the zero-product property to solve ((x+6)(x-2)=-16 ext<.>)

The solution is (-2 ext<.>) The solution set is (<-2> ext<.>)

###### Example 12.3.3 .

Use the zero-product property to solve ((x+4)(4x-5)=(7x+10)(3x-2) ext<.>)

The solutions are (0) and (-frac<5><17> ext<.>) The solution set is (left<0, -frac<5><17> ight> ext<.>)

###### Example 12.3.4 .

Use the zero-product property to solve (2-x^2=(2-x)^2 ext<.>)

The solution is (1 ext<.>) The solution set is (<1> ext<.>)

### Exercises Exercises

Use the zero-product property to solve each quadratic equation. State the solutions to each equation as well as the solution set to each equation.

We begin by setting each of the factors from the left side of the equation equal to zero.

The solutions are (-frac<8><3>) and (frac<7><5> ext<.>)

The solution set is (left<-frac<8><3>, frac<7><5> ight> ext<.>)

We begin by setting each of the factors from the left side of the equation equal to zero.

The solutions are (0) and (6 ext<.>)

We begin by making the right side of the equation zero and then factoring the left side of the equation.

The solutions are (5) and (-2 ext<.>)

We begin by making the right side of the equation zero and then factoring the left side of the equation.

We begin by expanding the left side of the equation, making the right side of the equation zero, and then factoring the left side of the equation.

The solutions are (-frac<5><4>) and (frac<1><8> ext<.>)

The solution set is (left<-frac<5><4>, frac<1><8> ight> ext<.>)

We begin by making the right side of the equation zero and then factoring the left side of the equation.

The solutions are (0) and (frac<1><2> ext<.>)

The solution set is (left<0, frac<1><2> ight> ext<.>)

We begin by expanding the left side of the equation, making the right side of the equation zero, and then factoring the left side of the equation.

## 5.3: Back and Forth (15 minutes)

### Activity

This activity builds toward completing the square. Students rewrite an equation of the form ((x-h)^2 + (y-k)^2 = r^2) , putting it in the form (x^2 + y^2 + ax +by + c = 0) . Then, they’re presented with an equation for a circle in which the squared binomials have been expanded. Students rewrite 2 perfect square trinomials in factored form, then identify the center and radius of the circle.

1. Here is the equation of a circle: ((x-2)^2+(y+7)^2=10^2)
1. What are the center and radius of the circle?
2. Apply the distributive property to the squared binomials and rearrange the equation so that one side is 0. This is the form in which many circle equations are written.
1. How can you rewrite this equation to find the center and radius of the circle?
2. What are the center and radius of the circle?

### Student Response

#### Are you ready for more?

In three-dimensional space, there are 3 coordinate axes, called the (x) -axis, the (y) -axis, and the (z) -axis. Write an equation for a sphere with center ((a,b,c)) and radius (r) .

### Anticipated Misconceptions

Students may struggle to decide whether the coordinates of the circles’ centers are positive or negative. Encourage them to rewrite the equation in the form ((x-h)^2+(y-k)^2=r^2) . Remind them that we subtract the coordinates of the center from the given point ((x,y)) to get the distance between the center and the point.

### Activity Synthesis

Ask students to rearrange the circle equation from the second problem so that there is a 0 on one side of the equation: (x^2+6x+y^2-10y-30=0) . Display these 3 forms of this equation for all to see, emphasizing that these are all equivalent equations and therefore represent the same circle:

The purpose of the discussion is to make connections between different forms of the equation in preparation for completing the square. Ask students:

• “In which form is it easiest to find the center and radius of the circle?” (The first one.)
• “Compare and contrast the second and third forms.” (Each form contains the terms (x^2) , (6x) , (y^2) , and ( ext-10y) , but the constant terms are different.)
• “How can you go from the first form to the second one?” (Distribute the two sets of squared binomials.)
• “How can you go from the second form to the third one?” (Combine like terms.)
• “How can you go from the second form to the first one?” (Rewrite the perfect square trinomials as squared binomials, and rewrite 64 as 8 2 .)

## The Importance of zero in Mathematics

The concept of zero seems to have originated around 520 AD with the Indian Aryabhata who used a symbol he called “kha” as a place holder. Brahmagupta, another Indian mathematician who lived in the 5th century, is credited for developing the Hindu-Arabic number system which included zero as an actual number in the system. Other mathematicians like al-Khwarizmi and Leonardo Fibonacci expanded the use of zero. By the Middle Ages, around the early 1200’s, this concept had come to Western society.

How important is zero? It is the number around which the negative numbers to its left stretch into infinity and the positive numbers to the right do likewise. It is neither positive nor negative. For that reason, zero is a pivotal point on thermometers and is the origin point for bathroom scales and the coordinate axis.

Zero is also important when you think of sets. An empty or null set is one which has no items in it.

Zero is so important that each of the mathematical operations has special rules called properties governing its use with other whole numbers.

The addition property of zero states that whenever zero is added to a whole number or vice versa the sum will be the whole number. Example: Ben had 3 apples and Sara had none. If they combined their apples, how many would there be in all?

The subtraction property of zero says that whenever zero is subtracted from a whole number, the difference will be the whole number, and whenever a whole number is subtracted from itself, the difference will be zero.

The multiplication property of zero is a little like the addition property in that it does not matter in what order you do the operation to the whole number. Thus, a whole number multiplied by zero equals zero, and vice versa.

The division property of zero is interesting. If zero is divided by a whole number, the quotient will be zero. This is the same as saying “divide zero into x number of groups and how many items will be in each group?” The answer, of course, is zero. You can not, however, divide a whole number by zero, because you can not come up with an inverse statement that makes sense. Can you divide x number of coins into groups of zero? Impossible! That is why mathematicians have a special term for x/0 they call it infinity.

Zero is very important for its place-holding value. If you have a number like two hundred four, how do you write it so that you understand that there are no tens in the number? You can not write it as 24 because that is a totally different number.

One neat thing when dealing with powers of ten: 10 squared=100. Notice the exponent 2 shows how many zeros will be in the written out form of the number. When you multiply two numbers which are powers of ten, the number of zeros in the answer equals the sum of the zeros in the factors. For example, 2000 multiplied by 300 equals 600,000, or 6 with 5 zeros after it.

When you round numbers like 6934 to the nearest ten, you place a zero in the ones place. 6934 rounded to the nearest ten equals 6930.

If you are writing a number with a decimal, you do not need to continue placing zeros to the right of the decimal. The decimal .033 (thirty-three thousandths) is still thirty-three thousandths if you write it .03300000. Why write all those extra zeros? But the zero in the tenths place is extremely important since it holds’ the tenths place by showing there are no tenths in the decimal.

Whether you call it zero, naught, or nil, zero has an important place in the field of mathematics.

## 6.5 Polynomial Equations

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

### Polynomial Equation

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

We have already solved polynomial equations of degree one . Polynomial equations of degree one are linear equations are of the form a x + b = c . a x + b = c .

We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:

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 . n 2 + n .

An equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 is called a quadratic equation.

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

### Use 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, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

### Zero Product Property

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

### Example 6.44

#### How to Solve a Quadratic Equation Using the Zero Product Property

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

#### Solution

Solve: ( 3 m − 2 ) ( 2 m + 1 ) = 0 . ( 3 m − 2 ) ( 2 m + 1 ) = 0 .

Solve: ( 4 p + 3 ) ( 4 p − 3 ) = 0 . ( 4 p + 3 ) ( 4 p − 3 ) = 0 .

### How To

#### Use the Zero Product Property.

1. Step 1. Set each factor equal to zero.
2. Step 2. Solve the linear equations.
3. Step 3. Check.

### Solve Quadratic Equations by Factoring

The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we must be sure to start with the quadratic equation in standard form , a x 2 + b x + c = 0 . a x 2 + b x + c = 0 . Then we must factor the expression on the left.

### How To

#### Solve a quadratic equation by factoring.

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

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?

### Example 6.46

#### Solution

We leave the check up to you.

In the next example, the left side of the equation 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.

### Example 6.47

Solve: ( 3 x − 8 ) ( x − 1 ) = 3 x . ( 3 x − 8 ) ( x − 1 ) = 3 x .

#### Solution

Solve: ( 2 m + 1 ) ( m + 3 ) = 1 2 m . ( 2 m + 1 ) ( m + 3 ) = 1 2 m .

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

### Example 6.48

#### Solution

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 greater than two by using the Zero Product Property, just like we solved quadratic equations.

### Example 6.49

Solve: 9 m 3 + 100 m = 60 m 2 . 9 m 3 + 100 m = 60 m 2 .

#### Solution

Solve: 8 x 3 = 24 x 2 − 18 x . 8 x 3 = 24 x 2 − 18 x .

Solve: 16 y 2 = 32 y 3 + 2 y . 16 y 2 = 32 y 3 + 2 y .

### Solve Equations with Polynomial Functions

As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.

### Example 6.50

For the function f ( x ) = x 2 + 2 x − 2 , f ( x ) = x 2 + 2 x − 2 ,

#### Solution

For the function f ( x ) = x 2 − 2 x − 8 , f ( x ) = x 2 − 2 x − 8 ,

For the function f ( x ) = x 2 − 8 x + 3 , f ( x ) = x 2 − 8 x + 3 ,

The Zero Product Property also helps us determine where the function is zero. A value of x where the function is 0, is called a zero of the function .

### Example 6.51

For the function f ( x ) = 3 x 2 + 10 x − 8 , f ( x ) = 3 x 2 + 10 x − 8 , find

ⓐ the zeros of the function, ⓑ any x-intercepts of the graph of the function, ⓒ any y-intercepts of the graph of the function

#### Solution

ⓐ To find the zeros of the function, we need to find when the function value is 0.

For the function f ( x ) = 2 x 2 − 7 x + 5 , f ( x ) = 2 x 2 − 7 x + 5 , find

ⓐ the zeros of the function, ⓑ any x-intercepts of the graph of the function, ⓒ any y-intercepts of the graph of the function.

For the function f ( x ) = 6 x 2 + 13 x − 15 , f ( x ) = 6 x 2 + 13 x − 15 , find

ⓐ the zeros of the function, ⓑ any x-intercepts of the graph of the function, ⓒ any y-intercepts of the graph of the function.

### Solve Applications Modeled by Polynomial 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 polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.

### How To

#### Use a problem solving strategy to solve word problems.

1. Step 1. Read the problem. Make sure all the words and ideas are understood.
2. Step 2. Identify what we are looking for.
3. Step 3. Name what we are looking for. Choose a variable to represent that quantity.
4. Step 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 algebraic equation.
5. Step 5. Solve the equation using appropriate algebra techniques.
6. Step 6. Check the answer in the problem and make sure it makes sense.
7. Step 7. Answer the question with a complete sentence.

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

### Example 6.52

The product of two consecutive odd integers is 323. Find the integers.

#### Solution

 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. Let n = the first integer. n = the first integer. n + 2 = next consecutive odd integer n + 2 = next consecutive odd integer Step 4. Translate into an equation. Restate the problem in a sentence. The product of the two consecutive odd integers is 323. n ( n + 2 ) = 323 n ( n + 2 ) = 323 Step 5. Solve the equation. n 2 + 2 n = 323 n 2 + 2 n = 323 Bring all the terms to one side. n 2 + 2 n − 323 = 0 n 2 + 2 n − 323 = 0 Factor the trinomial. ( n − 17 ) ( n + 19 ) = 0 ( n − 17 ) ( n + 19 ) = 0 Use the Zero Product Property. Solve the equations. n − 17 = 0 n + 19 = 0 n = 17 n = −19 n − 17 = 0 n + 19 = 0 n = 17 n = −19 There are two values for n that are solutions to this problem. So there are two sets of consecutive odd integers that will work. If the first integer is n = 17 n = 17 If the first integer is n = −19 n = −19 then the next odd integer is then the next odd integer is n + 2 n + 2 n + 2 n + 2 17 + 2 17 + 2 − 19 + 2 − 19 + 2 19 19 − 17 − 17 17 , 19 17 , 19 − 17 , −19 − 17 , −19 Step 6. Check the answer. The results are consecutive odd integers 17 , 19 and − 19 , −17 . 17 , 19 and − 19 , −17 . 17 · 19 = 323 ✓ − 19 ( − 17 ) = 323 ✓ 17 · 19 = 323 ✓ − 19 ( − 17 ) = 323 ✓ Both pairs of consecutive integers are solutions. Step 7. Answer the question The consecutive integers are 17, 19 and − 19 , −17 . − 19 , −17 .

The product of two consecutive odd integers is 255. Find the integers.

The product of two consecutive odd integers is 483 Find the integers.

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 positive results.

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

### Example 6.53

A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.

#### 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. Let w = w = the width of the bedroom. The length is four feet more than the width. w + 4 = w + 4 = the length of the garden Step 4. Translate into an equation. Restate the important information in a sentence. The area of the bedroom is 117 square feet. Use the formula for the area of a rectangle. A = l · w A = l · w Substitute in the variables. 117 = ( w + 4 ) w 117 = ( w + 4 ) w Step 5. Solve the equation Distribute first. 117 = w 2 + 4 w 117 = w 2 + 4 w Get zero on one side. 117 = w 2 + 4 w 117 = w 2 + 4 w Factor the trinomial. 0 = w 2 + 4 w − 117 0 = w 2 + 4 w − 117 Use the Zero Product Property. 0 = ( w 2 + 13 ) ( w − 9 ) 0 = ( w 2 + 13 ) ( w − 9 ) Solve each equation. 0 = w + 13 0 = w − 9 0 = w + 13 0 = w − 9 Since w is the width of the bedroom, it does not make sense for it to be negative. We eliminate that value for w. −13 = w 9 = w −13 = w 9 = w w = 9 w = 9 Width is 9 feet. Find the value of the length. w + 4 w + 4 9 + 4 9 + 4 13 Length is 13 feet. Step 6. Check the answer. Does the answer make sense? Yes, this makes sense. Step 7. Answer the question. The width of the bedroom is 9 feet and the length is 13 feet.

A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.

A rectangular patio has an area of 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.

We will use this formula to in the next example.

### Example 6.54

A boat’s sail is in the shape of a right triangle as shown. 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 sail.

#### 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 sail. Step 3. Name what you are looking for. One side is 7 less than the other. Let x = x = length of a side of the sail. x − 7 = 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 a 2 + b 2 = c 2 Substitute in the variables. x 2 + ( x − 7 ) 2 = 17 2 x 2 + ( x − 7 ) 2 = 17 2 Step 5. Solve the equation Simplify. x 2 + x 2 − 14 x + 49 = 289 x 2 + x 2 − 14 x + 49 = 289 2 x 2 − 14 x + 49 = 289 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 2 x 2 − 14 x − 240 = 0 Factor the greatest common factor. 2 ( x 2 − 7 x − 120 ) = 0 2 ( x 2 − 7 x − 120 ) = 0 Factor the trinomial. 2 ( x − 15 ) ( x + 8 ) = 0 2 ( x − 15 ) ( x + 8 ) = 0 Use the Zero Product Property. 2 ≠ 0 x − 15 = 0 x + 8 = 0 2 ≠ 0 x − 15 = 0 x + 8 = 0 Solve. 2 ≠ 0 x = 15 x = −8 2 ≠ 0 x = 15 x = −8 Since x is a side of the triangle, x = −8 x = −8 does not make sense. 2 ≠ 0 x = 15 x = −8 2 ≠ 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 in the problem Do these numbers make sense? Step 7. Answer the question The sides of the sail are 8, 15 and 17 feet.

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.

A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg.

The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.

### Example 6.55

Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from 80 feet above the ground, the function h ( t ) = −16 t 2 + 64 t + 80 h ( t ) = −16 t 2 + 64 t + 80 models the height, h, of the ball above the ground as a function of time, t. Find:

ⓐ the zeros of this function which tell us when the ball hits the ground, ⓑ when the ball will be 80 feet above the ground, ⓒ the height of the ball at t = 2 t = 2 seconds.

#### Solution

ⓐ The zeros of this function are found by solving h ( t ) = 0 . h ( t ) = 0 . This will tell us when the ball will hit the ground.

ⓑ The ball will be 80 feet above the ground when h ( t ) = 80 . h ( t ) = 80 .

Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from 160 feet above the ocean, the function h ( t ) = −16 t 2 + 48 t + 160 h ( t ) = −16 t 2 + 48 t + 160 models the height, h, of the rock above the ocean as a function of time, t. Find:

ⓐ the zeros of this function which tell us when the rock will hit the ocean, ⓑ when the rock will be 160 feet above the ocean, ⓒ the height of the rock at t = 1.5 t = 1.5 seconds.

Calib is going to throw his lucky penny from his balcony on a cruise ship. When he throws the penny upward from 128 feet above the ground, the function h ( t ) = −16 t 2 + 32 t + 128 h ( t ) = −16 t 2 + 32 t + 128 models the height, h, of the penny above the ocean as a function of time, t. Find:

ⓐ the zeros of this function which is when the penny will hit the ocean, ⓑ when the penny will be 128 feet above the ocean, ⓒ the height the penny will be at t = 1 t = 1 seconds which is when the penny will be at its highest point.

### Media

Access this online resource for additional instruction and practice with quadratic equations.

### Section 6.5 Exercises

#### Practice Makes Perfect

Use the Zero Product Property

In the following exercises, solve.

In the following exercises, solve.

Solve Equations with Polynomial Functions

In the following exercises, solve.

In the following exercises, for each function, find: ⓐ the zeros of the function ⓑ the x-intercepts of the graph of the function ⓒ the y-intercept of the graph of the function.

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve.

The product of two consecutive odd integers is 143. Find the integers.

The product of two consecutive odd integers is 195. Find the integers.

The product of two consecutive even integers is 168. Find the integers.

The product of two consecutive even integers is 288. 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.

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.

The area of a bulletin board is 55 square feet. The length is four feet less than three times the width. Find the length and the width of the a bulletin board.

A rectangular carport has area 150 square feet. The width of the carport is five feet less than twice its length. Find the width and the length of the carport.

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.

A stained glass window is shaped like a right triangle. The hypotenuse is 15 feet. One leg is three more than the other. Find the lengths of the legs.

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.

A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other leg is 4 feet more than the leg against the barn. The hypotenuse is 8 feet more than the leg along the barn. Find the three sides of the goat enclosure.

Juli is going to launch a model rocket in her back yard. When she launches the rocket, the function h ( t ) = −16 t 2 + 32 t h ( t ) = −16 t 2 + 32 t models the height, h, of the rocket above the ground as a function of time, t. Find:

ⓐ the zeros of this function, which tell us when the rocket will be on the ground. ⓑ the time the rocket will be 16 feet above the ground.

Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48 feet above the ground, the function h ( t ) = −16 t 2 + 32 t + 48 h ( t ) = −16 t 2 + 32 t + 48 models the height, h, of the ball above the ground as a function of time, t. Find:

ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 48 feet above the ground. ⓒ the height the ball will be at t = 1 t = 1 seconds which is when the ball will be at its highest point.

#### Writing Exercises

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

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.

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

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## 6.5: Use the Zero Product Property - Mathematics

This method of solving quadratic equations may be familiar to you. Recall that before trying to solve any quadratic equation, we must put it in standard form :

Once the equation is in standard form try to factor it. If you are unsure about factoring, review the Factoring Notes . After factoring, we will apply the zero product property.

The Zero Product Property

If AB = 0, then A = 0 or B = 0. Meaning that if we multiply two things together and get 0, one of them must be 0.

To apply the zero product property, we set each factor equal to zero and solve for the variable.

Here is an example: x 2 - 2x - 15 = 0

Since the equation is already in standard form, the first step is to factor:

Now apply the zero product property. Setting x - 5 = 0 and x + 3 = 0 produces the answers x = 5 and x = -3.

Solve each of the following by factoring and using the zero product property.

Since the equation is already in standard form, the first step is to factor. Note that we have a difference of squares :

Now apply the zero product property and set each of the factors equal to zero.

We now have two linear equations to solve.

Since the equation is not in standard form , the first step is to put it in standard form. Subtract 43x and 40 from both sides of this equation. Doing this, we get:

Now that the equation is in standard form, factor the trinomial.

Next apply the zero product property and set each of the factors equal to zero.

## Example

### Solution This image shows the steps for solving 3 p (10 p + 7) = 0. The first step is using the zero product property to set each factor equal to 0, 3p = 0 or 10 p + 7 = 0. The next step is solving both equations, p = 0 or p = negative 7/10. Finally, check the solutions by substituting the answers into the original equation.

It may appear that there is only one factor in the next example. Remember, however, that (^<2>) means (left(y-8 ight)left(y-8 ight)) .

## Commutative Property - Definition with Examples

The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer. The property holds for Addition and Multiplication, but not for subtraction and division.

 Addition Subtraction Multiplication Division The above examples clearly show that we can apply the commutative property on addition and multiplication. However, we cannot apply commutative property on subtraction and division. If you move the position of numbers in subtraction or division, it changes the entire problem.

Therefore, if a and b are two non-zero numbers, then:

The commutative property of addition is:

The commutative property of multiplication is:

In short, in commutative property, the numbers can be added or multiplied to each other in any order without changing the answer.

Let us see some examples to understand commutative property.

Example 1: Commutative property with addition

Myra has 5 marbles, and Rick has 3 marbles. How many marbles they have in total? Hence, we can see whether we add 5 + 3 or 3 + 5, the answer is always 8.

Example 2: Commutative property with subtraction.

Alvin has 12 apples. He gives 8 apples to his sister. How many apples are left with Alvin?

Here, we subtract 8 from 12 and get the answer as 4 apples. However, we cannot subtract 12 from 8 and get 8 as the answer. Example 3: Commutative property with multiplication.

Sara buys 3 packs of buns. Each pack has 4 buns. How many buns did she buy?

Here, if we multiply 3 by 4 or 4 by 3, in both cases we get the answer as 12 buns.

So, the commutative property holds for multiplication. So, commutative property holds true for multiplication.

Example 4: Commutative property with division.

If you have to divide 25 strawberries to 5 kids, each kid will receive 5 strawberries. However, if you have to divide 5 strawberries amongst 25 children, every kid will get a tiny fraction of the strawberry. Therefore, we cannot apply the commutative property with the division.

## 6.5: Use the Zero Product Property - Mathematics

Here are the steps required for Solving Quadratics by Factoring:

 Step 1: Write the equation in the correct form. To be in the correct form, you must remove all parentheses from each side of the equation by distributing, combine all like terms, and finally set the equation equal to zero with the terms written in descending order. Step 2: Use a factoring strategies to factor the problem. Step 3: Use the Zero Product Property and set each factor containing a variable equal to zero. Step 4: Solve each factor that was set equal to zero by getting the x on one side and the answer on the other side.

Example 1 &ndash Solve: x 2 + 16 = 10x

Example 2 &ndash Solve: 18x 2 – 3x = 6

Example 3 &ndash Solve: 50x 2 = 72

Example 4 &ndash Solve: x(2x – 1) = 3

Example 5 &ndash Solve: (x + 3)(x – 5) = 𔃅