homework 4 9 writing polynomials

1.4 Polynomials

Learning objectives.

In this section, you will:

• Identify the degree and leading coefficient of polynomials.
• Add and subtract polynomials.
• Multiply polynomials.
• Use FOIL to multiply binomials.
• Perform operations with polynomials of several variables.

Maahi is building a little free library (a small house-shaped book repository), whose front is in the shape of a square topped with a triangle. There will be a rectangular door through which people can take and donate books. Maahi wants to find the area of the front of the library so that they can purchase the correct amount of paint. Using the measurements of the front of the house, shown in Figure 1 , we can create an expression that combines several variable terms, allowing us to solve this problem and others like it.

First find the area of the square in square feet.

Then find the area of the triangle in square feet.

Next find the area of the rectangular door in square feet.

The area of the front of the library can be found by adding the areas of the square and the triangle, and then subtracting the area of the rectangle. When we do this, we get 4 x 2 + 3 2 x − x ft 2 , 4 x 2 + 3 2 x − x ft 2 , or 4 x 2 + 1 2 x 4 x 2 + 1 2 x ft 2 .

In this section, we will examine expressions such as this one, which combine several variable terms.

Identifying the Degree and Leading Coefficient of Polynomials

The formula just found is an example of a polynomial , which is a sum of or difference of terms, each consisting of a variable raised to a nonnegative integer power. A number multiplied by a variable raised to an exponent, such as 384 π , 384 π , is known as a coefficient . Coefficients can be positive, negative, or zero, and can be whole numbers, decimals, or fractions. Each product a i x i , a i x i , such as 384 π w , 384 π w , is a term of a polynomial . If a term does not contain a variable, it is called a constant .

A polynomial containing only one term, such as 5 x 4 , 5 x 4 , is called a monomial . A polynomial containing two terms, such as 2 x − 9 , 2 x − 9 , is called a binomial . A polynomial containing three terms, such as −3 x 2 + 8 x − 7 , −3 x 2 + 8 x − 7 , is called a trinomial .

We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. The term with the highest degree is called the leading term because it is usually written first. The coefficient of the leading term is called the leading coefficient . When a polynomial is written so that the powers are descending, we say that it is in standard form.

Polynomials

A polynomial is an expression that can be written in the form

Each real number a i is called a coefficient . The number a 0 a 0 that is not multiplied by a variable is called a constant . Each product a i x i a i x i is a term of a polynomial . The highest power of the variable that occurs in the polynomial is called the degree of a polynomial. The leading term is the term with the highest power, and its coefficient is called the leading coefficient .

Given a polynomial expression, identify the degree and leading coefficient .

• Find the highest power of x to determine the degree.
• Identify the term containing the highest power of x to find the leading term.
• Identify the coefficient of the leading term.

Identifying the Degree and Leading Coefficient of a Polynomial

For the following polynomials, identify the degree, the leading term, and the leading coefficient.

• ⓐ 3 + 2 x 2 − 4 x 3 3 + 2 x 2 − 4 x 3
• ⓑ 5 t 5 − 2 t 3 + 7 t 5 t 5 − 2 t 3 + 7 t
• ⓒ 6 p − p 3 − 2 6 p − p 3 − 2
• ⓐ The highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, −4 x 3 . −4 x 3 . The leading coefficient is the coefficient of that term, −4. −4.
• ⓑ The highest power of t is 5 , 5 , so the degree is 5. 5. The leading term is the term containing that degree, 5 t 5 . 5 t 5 . The leading coefficient is the coefficient of that term, 5. 5.
• ⓒ The highest power of p is 3 , 3 , so the degree is 3. 3. The leading term is the term containing that degree, − p 3 , − p 3 , The leading coefficient is the coefficient of that term, −1. −1.

Identify the degree, leading term, and leading coefficient of the polynomial 4 x 2 − x 6 + 2 x − 6. 4 x 2 − x 6 + 2 x − 6.

• Adding and Subtracting Polynomials

We can add and subtract polynomials by combining like terms, which are terms that contain the same variables raised to the same exponents. For example, 5 x 2 5 x 2 and −2 x 2 −2 x 2 are like terms, and can be added to get 3 x 2 , 3 x 2 , but 3 x 3 x and 3 x 2 3 x 2 are not like terms, and therefore cannot be added.

Given multiple polynomials, add or subtract them to simplify the expressions.

• Combine like terms.
• Simplify and write in standard form.

Adding Polynomials

Find the sum.

( 12 x 2 + 9 x − 21 ) + ( 4 x 3 + 8 x 2 − 5 x + 20 ) ( 12 x 2 + 9 x − 21 ) + ( 4 x 3 + 8 x 2 − 5 x + 20 )

4 x 3 + ( 12 x 2 + 8 x 2 ) + ( 9 x − 5 x ) + ( −21 + 20 )      Combine like terms . 4 x 3 + 20 x 2 + 4 x − 1    Simplify . 4 x 3 + ( 12 x 2 + 8 x 2 ) + ( 9 x − 5 x ) + ( −21 + 20 )      Combine like terms . 4 x 3 + 20 x 2 + 4 x − 1    Simplify .

We can check our answers to these types of problems using a graphing calculator. To check, graph the problem as given along with the simplified answer. The two graphs should be equivalent. Be sure to use the same window to compare the graphs. Using different windows can make the expressions seem equivalent when they are not.

( 2 x 3 + 5 x 2 − x + 1 ) + ( 2 x 2 − 3 x − 4 ) ( 2 x 3 + 5 x 2 − x + 1 ) + ( 2 x 2 − 3 x − 4 )

Subtracting Polynomials

Find the difference.

( 7 x 4 − x 2 + 6 x + 1 ) − ( 5 x 3 − 2 x 2 + 3 x + 2 ) ( 7 x 4 − x 2 + 6 x + 1 ) − ( 5 x 3 − 2 x 2 + 3 x + 2 )

7 x 4 − x 2 + 6 x + 1 - 5 x 3 + 2 x 3 - 3 x − 2 Distribute negative sign. 7 x 4 − 5 x 3 + x 2 + 6 x − 3 x + 1 - 2 Group like terms. 7 x 4 − 5 x 3 + x 2 + 3 x - 1 Combine/simplify. 7 x 4 − x 2 + 6 x + 1 - 5 x 3 + 2 x 3 - 3 x − 2 Distribute negative sign. 7 x 4 − 5 x 3 + x 2 + 6 x − 3 x + 1 - 2 Group like terms. 7 x 4 − 5 x 3 + x 2 + 3 x - 1 Combine/simplify.

Note that finding the difference between two polynomials is the same as adding the opposite of the second polynomial to the first.

( −7 x 3 − 7 x 2 + 6 x − 2 ) − ( 4 x 3 − 6 x 2 − x + 7 ) ( −7 x 3 − 7 x 2 + 6 x − 2 ) − ( 4 x 3 − 6 x 2 − x + 7 )

• Multiplying Polynomials

Multiplying polynomials is a bit more challenging than adding and subtracting polynomials. We must use the distributive property to multiply each term in the first polynomial by each term in the second polynomial. We then combine like terms. We can also use a shortcut called the FOIL method when multiplying binomials. Certain special products follow patterns that we can memorize and use instead of multiplying the polynomials by hand each time. We will look at a variety of ways to multiply polynomials.

Multiplying Polynomials Using the Distributive Property

To multiply a number by a polynomial, we use the distributive property. The number must be distributed to each term of the polynomial. We can distribute the 2 2 in 2 ( x + 7 ) 2 ( x + 7 ) to obtain the equivalent expression 2 x + 14. 2 x + 14. When multiplying polynomials, the distributive property allows us to multiply each term of the first polynomial by each term of the second. We then add the products together and combine like terms to simplify.

Given the multiplication of two polynomials, use the distributive property to simplify the expression.

• Multiply each term of the first polynomial by each term of the second.

Find the product.

( 2 x + 1 ) ( 3 x 2 − x + 4 ) ( 2 x + 1 ) ( 3 x 2 − x + 4 )

2 x ( 3 x 2 − x + 4 ) + 1 ( 3 x 2 − x + 4 )      Use the distributive property . ( 6 x 3 − 2 x 2 + 8 x ) + ( 3 x 2 − x + 4 )    Multiply . 6 x 3 + ( −2 x 2 + 3 x 2 ) + ( 8 x − x ) + 4    Combine like terms . 6 x 3 + x 2 + 7 x + 4      Simplify . 2 x ( 3 x 2 − x + 4 ) + 1 ( 3 x 2 − x + 4 )      Use the distributive property . ( 6 x 3 − 2 x 2 + 8 x ) + ( 3 x 2 − x + 4 )    Multiply . 6 x 3 + ( −2 x 2 + 3 x 2 ) + ( 8 x − x ) + 4    Combine like terms . 6 x 3 + x 2 + 7 x + 4      Simplify .

We can use a table to keep track of our work, as shown in Table 1 . Write one polynomial across the top and the other down the side. For each box in the table, multiply the term for that row by the term for that column. Then add all of the terms together, combine like terms, and simplify.

( 3 x + 2 ) ( x 3 − 4 x 2 + 7 ) ( 3 x + 2 ) ( x 3 − 4 x 2 + 7 )

Using FOIL to Multiply Binomials

A shortcut called FOIL is sometimes used to find the product of two binomials. It is called FOIL because we multiply the f irst terms, the o uter terms, the i nner terms, and then the l ast terms of each binomial.

The FOIL method arises out of the distributive property. We are simply multiplying each term of the first binomial by each term of the second binomial, and then combining like terms.

Given two binomials, use FOIL to simplify the expression.

• Multiply the first terms of each binomial.
• Multiply the outer terms of the binomials.
• Multiply the inner terms of the binomials.
• Multiply the last terms of each binomial.
• Add the products.
• Combine like terms and simplify.

Use FOIL to find the product.

( 2 x - 18 ) ( 3 x + 3 ) ( 2 x - 18 ) ( 3 x + 3 )

Find the product of the first terms.

Find the product of the outer terms.

Find the product of the inner terms.

Find the product of the last terms.

6 x 2 + 6 x − 54 x − 54 Add the products . 6 x 2 + ( 6 x − 54 x ) − 54 Combine like terms . 6 x 2 − 48 x − 54 Simplify . 6 x 2 + 6 x − 54 x − 54 Add the products . 6 x 2 + ( 6 x − 54 x ) − 54 Combine like terms . 6 x 2 − 48 x − 54 Simplify .

( x + 7 ) ( 3 x − 5 ) ( x + 7 ) ( 3 x − 5 )

Perfect Square Trinomials

Certain binomial products have special forms. When a binomial is squared, the result is called a perfect square trinomial . We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier and faster. Let’s look at a few perfect square trinomials to familiarize ourselves with the form.

Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial.

When a binomial is squared, the result is the first term squared added to double the product of both terms and the last term squared.

Given a binomial, square it using the formula for perfect square trinomials.

• Square the first term of the binomial.
• Square the last term of the binomial.
• For the middle term of the trinomial, double the product of the two terms.
• Add and simplify.

Expanding Perfect Squares

Expand ( 3 x − 8 ) 2 . ( 3 x − 8 ) 2 .

Begin by squaring the first term and the last term. For the middle term of the trinomial, double the product of the two terms.

Expand ( 4 x − 1 ) 2 . ( 4 x − 1 ) 2 .

Difference of Squares

Another special product is called the difference of squares , which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let’s see what happens when we multiply ( x + 1 ) ( x − 1 ) ( x + 1 ) ( x − 1 ) using the FOIL method.

The middle term drops out, resulting in a difference of squares. Just as we did with the perfect squares, let’s look at a few examples.

Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.

Is there a special form for the sum of squares?

No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares.

When a binomial is multiplied by a binomial with the same terms separated by the opposite sign, the result is the square of the first term minus the square of the last term.

Given a binomial multiplied by a binomial with the same terms but the opposite sign, find the difference of squares.

• Square the first term of the binomials.
• Square the last term of the binomials.
• Subtract the square of the last term from the square of the first term.

Multiplying Binomials Resulting in a Difference of Squares

Multiply ( 9 x + 4 ) ( 9 x − 4 ) . ( 9 x + 4 ) ( 9 x − 4 ) .

Square the first term to get ( 9 x ) 2 = 81 x 2 . ( 9 x ) 2 = 81 x 2 . Square the last term to get 4 2 = 16. 4 2 = 16. Subtract the square of the last term from the square of the first term to find the product of 81 x 2 − 16. 81 x 2 − 16.

Multiply ( 2 x + 7 ) ( 2 x − 7 ) . ( 2 x + 7 ) ( 2 x − 7 ) .

Performing Operations with Polynomials of Several Variables

We have looked at polynomials containing only one variable. However, a polynomial can contain several variables. All of the same rules apply when working with polynomials containing several variables. Consider an example:

Multiplying Polynomials Containing Several Variables

Multiply ( x + 4 ) ( 3 x − 2 y + 5 ) . ( x + 4 ) ( 3 x − 2 y + 5 ) .

Follow the same steps that we used to multiply polynomials containing only one variable.

Multiply ( 3 x − 1 ) ( 2 x + 7 y − 9 ) . ( 3 x − 1 ) ( 2 x + 7 y − 9 ) .

Access these online resources for additional instruction and practice with polynomials.

• Special Products of Polynomials

1.4 Section Exercises

Evaluate the following statement: The degree of a polynomial in standard form is the exponent of the leading term. Explain why the statement is true or false.

Many times, multiplying two binomials with two variables results in a trinomial. This is not the case when there is a difference of two squares. Explain why the product in this case is also a binomial.

You can multiply polynomials with any number of terms and any number of variables using four basic steps over and over until you reach the expanded polynomial. What are the four steps?

State whether the following statement is true and explain why or why not: A trinomial is always a higher degree than a monomial.

For the following exercises, identify the degree of the polynomial.

7 x − 2 x 2 + 13 7 x − 2 x 2 + 13

14 m 3 + m 2 − 16 m + 8 14 m 3 + m 2 − 16 m + 8

−625 a 8 + 16 b 4 −625 a 8 + 16 b 4

200 p − 30 p 2 m + 40 m 3 200 p − 30 p 2 m + 40 m 3

x 2 + 4 x + 4 x 2 + 4 x + 4

6 y 4 − y 5 + 3 y − 4 6 y 4 − y 5 + 3 y − 4

For the following exercises, find the sum or difference.

( 12 x 2 + 3 x ) − ( 8 x 2 −19 ) ( 12 x 2 + 3 x ) − ( 8 x 2 −19 )

( 4 z 3 + 8 z 2 − z ) + ( −2 z 2 + z + 6 ) ( 4 z 3 + 8 z 2 − z ) + ( −2 z 2 + z + 6 )

( 6 w 2 + 24 w + 24 ) − ( 3 w − 2 6 w + 3 ) ( 6 w 2 + 24 w + 24 ) − ( 3 w − 2 6 w + 3 )

( 7 a 3 + 6 a 2 − 4 a − 13 ) + ( − 3 a 3 − 4 a 2 + 6 a + 17 ) ( 7 a 3 + 6 a 2 − 4 a − 13 ) + ( − 3 a 3 − 4 a 2 + 6 a + 17 )

( 11 b 4 − 6 b 3 + 18 b 2 − 4 b + 8 ) − ( 3 b 3 + 6 b 2 + 3 b ) ( 11 b 4 − 6 b 3 + 18 b 2 − 4 b + 8 ) − ( 3 b 3 + 6 b 2 + 3 b )

( 49 p 2 − 25 ) + ( 16 p 4 − 32 p 2 + 16 ) ( 49 p 2 − 25 ) + ( 16 p 4 − 32 p 2 + 16 )

For the following exercises, find the product.

( 4 x + 2 ) ( 6 x − 4 ) ( 4 x + 2 ) ( 6 x − 4 )

( 14 c 2 + 4 c ) ( 2 c 2 − 3 c ) ( 14 c 2 + 4 c ) ( 2 c 2 − 3 c )

( 6 b 2 − 6 ) ( 4 b 2 − 4 ) ( 6 b 2 − 6 ) ( 4 b 2 − 4 )

( 3 d − 5 ) ( 2 d + 9 ) ( 3 d − 5 ) ( 2 d + 9 )

( 9 v − 11 ) ( 11 v − 9 ) ( 9 v − 11 ) ( 11 v − 9 )

( 4 t 2 + 7 t ) ( −3 t 2 + 4 ) ( 4 t 2 + 7 t ) ( −3 t 2 + 4 )

( 8 n − 4 ) ( n 2 + 9 ) ( 8 n − 4 ) ( n 2 + 9 )

For the following exercises, expand the binomial.

( 4 x + 5 ) 2 ( 4 x + 5 ) 2

( 3 y − 7 ) 2 ( 3 y − 7 ) 2

( 12 − 4 x ) 2 ( 12 − 4 x ) 2

( 4 p + 9 ) 2 ( 4 p + 9 ) 2

( 2 m − 3 ) 2 ( 2 m − 3 ) 2

( 3 y − 6 ) 2 ( 3 y − 6 ) 2

( 9 b + 1 ) 2 ( 9 b + 1 ) 2

For the following exercises, multiply the binomials.

( 4 c + 1 ) ( 4 c − 1 ) ( 4 c + 1 ) ( 4 c − 1 )

( 9 a − 4 ) ( 9 a + 4 ) ( 9 a − 4 ) ( 9 a + 4 )

( 15 n − 6 ) ( 15 n + 6 ) ( 15 n − 6 ) ( 15 n + 6 )

( 25 b + 2 ) ( 25 b − 2 ) ( 25 b + 2 ) ( 25 b − 2 )

( 4 + 4 m ) ( 4 − 4 m ) ( 4 + 4 m ) ( 4 − 4 m )

( 14 p + 7 ) ( 14 p − 7 ) ( 14 p + 7 ) ( 14 p − 7 )

( 11 q − 10 ) ( 11 q + 10 ) ( 11 q − 10 ) ( 11 q + 10 )

For the following exercises, multiply the polynomials.

( 2 x 2 + 2 x + 1 ) ( 4 x − 1 ) ( 2 x 2 + 2 x + 1 ) ( 4 x − 1 )

( 4 t 2 + t − 7 ) ( 4 t 2 − 1 ) ( 4 t 2 + t − 7 ) ( 4 t 2 − 1 )

( x − 1 ) ( x 2 − 2 x + 1 ) ( x − 1 ) ( x 2 − 2 x + 1 )

( y − 2 ) ( y 2 − 4 y − 9 ) ( y − 2 ) ( y 2 − 4 y − 9 )

( 6 k − 5 ) ( 6 k 2 + 5 k − 1 ) ( 6 k − 5 ) ( 6 k 2 + 5 k − 1 )

( 3 p 2 + 2 p − 10 ) ( p − 1 ) ( 3 p 2 + 2 p − 10 ) ( p − 1 )

( 4 m − 13 ) ( 2 m 2 − 7 m + 9 ) ( 4 m − 13 ) ( 2 m 2 − 7 m + 9 )

( a + b ) ( a − b ) ( a + b ) ( a − b )

( 4 x − 6 y ) ( 6 x − 4 y ) ( 4 x − 6 y ) ( 6 x − 4 y )

( 4 t − 5 u ) 2 ( 4 t − 5 u ) 2

( 9 m + 4 n − 1 ) ( 2 m + 8 ) ( 9 m + 4 n − 1 ) ( 2 m + 8 )

( 4 t − x ) ( t − x + 1 ) ( 4 t − x ) ( t − x + 1 )

( b 2 − 1 ) ( a 2 + 2 a b + b 2 ) ( b 2 − 1 ) ( a 2 + 2 a b + b 2 )

( 4 r − d ) ( 6 r + 7 d ) ( 4 r − d ) ( 6 r + 7 d )

( x + y ) ( x 2 − x y + y 2 ) ( x + y ) ( x 2 − x y + y 2 )

Real-World Applications

A developer wants to purchase a plot of land to build a house. The area of the plot can be described by the following expression: ( 4 x + 1 ) ( 8 x − 3 ) ( 4 x + 1 ) ( 8 x − 3 ) where x is measured in meters. Multiply the binomials to find the area of the plot in standard form.

A prospective buyer wants to know how much grain a specific silo can hold. The area of the floor of the silo is ( 2 x + 9 ) 2 . ( 2 x + 9 ) 2 . The height of the silo is 10 x + 10 , 10 x + 10 , where x is measured in feet. Expand the square and multiply by the height to find the expression that shows how much grain the silo can hold.

For the following exercises, perform the given operations.

( 4 t − 7 ) 2 ( 2 t + 1 ) − ( 4 t 2 + 2 t + 11 ) ( 4 t − 7 ) 2 ( 2 t + 1 ) − ( 4 t 2 + 2 t + 11 )

( 3 b + 6 ) ( 3 b − 6 ) ( 9 b 2 − 36 ) ( 3 b + 6 ) ( 3 b − 6 ) ( 9 b 2 − 36 )

( a 2 + 4 a c + 4 c 2 ) ( a 2 − 4 c 2 ) ( a 2 + 4 a c + 4 c 2 ) ( a 2 − 4 c 2 )

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Unit 4: Module 4: Polynomial and quadratic expressions, equations, and functions

About this unit.

"In earlier modules, students analyze the process of solving equations and developing fluency in writing, interpreting, and translating between various forms of linear equations (Module 1) and linear and exponential functions (Module 3). These experiences combined with modeling with data (Module 2), set the stage for Module 4. Here students continue to interpret expressions, create equations, rewrite equations and functions in different but equivalent forms, and graph and interpret functions, but this time using polynomial functions, and more specifically quadratic functions, as well as square root and cube root functions." Eureka Math/EngageNY (c) 2015 GreatMinds.org

Topic A: Lessons 1-2: Factoring monomials

• Intro to factors & divisibility (Opens a modal)
• Which monomial factorization is correct? (Opens a modal)
• Factoring monomials (Opens a modal)
• Worked example: finding the missing monomial factor (Opens a modal)
• Worked example: finding missing monomial side in area model (Opens a modal)
• Factors & divisibility Get 3 of 4 questions to level up!
• Factor monomials Get 3 of 4 questions to level up!

Topic A: Lessons 1-2: Common factor

• Greatest common factor of monomials (Opens a modal)
• Factoring with the distributive property (Opens a modal)
• Factoring polynomials by taking a common factor (Opens a modal)
• Taking common factor from binomial (Opens a modal)
• Taking common factor from trinomial (Opens a modal)
• Taking common factor: area model (Opens a modal)
• Factoring polynomials: common binomial factor (Opens a modal)
• Greatest common factor of monomials Get 3 of 4 questions to level up!
• Factor polynomials: common factor Get 3 of 4 questions to level up!

Topic A: Lessons 1-2: Factoring binomials intro

• Difference of squares intro (Opens a modal)
• Factoring quadratics: Difference of squares (Opens a modal)
• Perfect square factorization intro (Opens a modal)
• Factoring quadratics: Perfect squares (Opens a modal)
• Factoring quadratics as (x+a)(x+b) (Opens a modal)
• Factoring quadratics: leading coefficient = 1 (Opens a modal)
• Factoring quadratics as (x+a)(x+b) (example 2) (Opens a modal)
• More examples of factoring quadratics as (x+a)(x+b) (Opens a modal)
• Difference of squares intro Get 3 of 4 questions to level up!
• Perfect squares intro Get 3 of 4 questions to level up!
• Warmup: factoring quadratics intro Get 3 of 4 questions to level up!
• Factoring quadratics intro Get 3 of 4 questions to level up!

Topic A: Lessons 3-4: Special forms

• Factoring difference of squares: leading coefficient ≠ 1 (Opens a modal)
• Factoring difference of squares: analyzing factorization (Opens a modal)
• Factoring difference of squares: missing values (Opens a modal)
• Factoring difference of squares: shared factors (Opens a modal)
• Factoring perfect squares (Opens a modal)
• Identifying perfect square form (Opens a modal)
• Factoring higher-degree polynomials: Common factor (Opens a modal)
• Factoring perfect squares: negative common factor (Opens a modal)
• Factoring perfect squares: missing values (Opens a modal)
• Factoring perfect squares: shared factors (Opens a modal)
• Difference of squares Get 3 of 4 questions to level up!
• Perfect squares Get 3 of 4 questions to level up!

Topic A: Lessons 3-4: Factoring by grouping

• Intro to grouping (Opens a modal)
• Factoring by grouping (Opens a modal)
• Factoring quadratics by grouping (Opens a modal)
• Factoring quadratics: leading coefficient ≠ 1 (Opens a modal)
• Strategy in factoring quadratics (part 1 of 2) (Opens a modal)
• Strategy in factoring quadratics (part 2 of 2) (Opens a modal)
• Factoring quadratics in any form (Opens a modal)
• Factor quadratics by grouping Get 3 of 4 questions to level up!

Topic A: Lesson 5: The zero product property

• Zero product property (Opens a modal)
• Solving quadratics by factoring (Opens a modal)
• Solving quadratics by factoring: leading coefficient ≠ 1 (Opens a modal)
• Solving quadratics using structure (Opens a modal)
• Zero product property Get 3 of 4 questions to level up!
• Quadratics by factoring (intro) Get 3 of 4 questions to level up!
• Quadratics by factoring Get 3 of 4 questions to level up!
• Solve equations using structure Get 3 of 4 questions to level up!

Topic A: Lessons 6-7: Solving basic one-variable quadratic equations

• Solving quadratics by taking square roots (Opens a modal)
• Solving quadratics by taking square roots examples (Opens a modal)
• Solving quadratics by taking square roots: with steps (Opens a modal)
• Quadratics by taking square roots: strategy (Opens a modal)
• Quadratic equations word problem: triangle dimensions (Opens a modal)
• Quadratic equations word problem: box dimensions (Opens a modal)
• Quadratics by taking square roots (intro) Get 3 of 4 questions to level up!
• Quadratics by taking square roots Get 3 of 4 questions to level up!
• Quadratics by taking square roots: strategy Get 3 of 4 questions to level up!
• Quadratics by taking square roots: with steps Get 3 of 4 questions to level up!

Topic A: Lessons 8-10: Parabolas intro

• Parabolas intro (Opens a modal)
• Graphing quadratics in factored form (Opens a modal)
• Quadratic word problems (factored form) (Opens a modal)
• Parabolas intro Get 3 of 4 questions to level up!
• Warmup: graphing quadratics in factored form Get 3 of 3 questions to level up!
• Graph quadratics in factored form Get 3 of 4 questions to level up!
• Quadratic word problems (factored form) Get 3 of 4 questions to level up!

Topic B: Lessons 11-13: Completing the square

• Completing the square (Opens a modal)
• Solving quadratics by completing the square (Opens a modal)
• Worked example: Completing the square (intro) (Opens a modal)
• Worked example: Rewriting expressions by completing the square (Opens a modal)
• Worked example: Rewriting & solving equations by completing the square (Opens a modal)
• Worked example: completing the square (leading coefficient ≠ 1) (Opens a modal)
• Solving quadratics by completing the square: no solution (Opens a modal)
• Completing the square (intro) Get 3 of 4 questions to level up!
• Completing the square (intermediate) Get 3 of 4 questions to level up!
• Completing the square Get 3 of 4 questions to level up!

Topic B: Lessons 14-15: The quadratic formula

• The quadratic formula (Opens a modal)
• Using the quadratic formula (Opens a modal)
• Worked example: quadratic formula (Opens a modal)
• Worked example: quadratic formula (example 2) (Opens a modal)
• Worked example: quadratic formula (negative coefficients) (Opens a modal)
• Using the quadratic formula: number of solutions (Opens a modal)
• Proof of the quadratic formula (Opens a modal)
• Quadratic formula Get 3 of 4 questions to level up!
• Number of solutions of quadratic equations Get 3 of 4 questions to level up!

Topic B: Lesson 16: Graphing quadratic equations from the vertex form

• Vertex form introduction (Opens a modal)
• Vertex & axis of symmetry of a parabola (Opens a modal)
• Graphing quadratics: vertex form (Opens a modal)
• Quadratic word problems (vertex form) (Opens a modal)
• Warmup: graphing quadratics in vertex form Get 3 of 4 questions to level up!
• Graph quadratics in vertex form Get 3 of 4 questions to level up!
• Quadratic word problems (vertex form) Get 3 of 4 questions to level up!

Topic B: Lesson 17: Graphing quadratic functions from the standard form

• Finding the vertex of a parabola in standard form (Opens a modal)
• Graphing quadratics: standard form (Opens a modal)
• Quadratic word problem: ball (Opens a modal)
• Forms & features of quadratic functions (Opens a modal)
• Worked examples: Forms & features of quadratic functions (Opens a modal)
• Finding features of quadratic functions (Opens a modal)
• Graph quadratics in standard form Get 3 of 4 questions to level up!
• Quadratic word problems (standard form) Get 3 of 4 questions to level up!
• Features of quadratic functions: strategy Get 3 of 4 questions to level up!
• Warmup: Features of quadratic functions Get 3 of 3 questions to level up!
• Features of quadratic functions Get 3 of 4 questions to level up!
• Graph parabolas in all forms Get 3 of 4 questions to level up!

Topic C: Lessons 18-19: Translating graphs of functions

• Intro to parabola transformations (Opens a modal)
• Shifting parabolas (Opens a modal)
• Shifting functions examples (Opens a modal)
• Graphing shifted functions (Opens a modal)
• Shift parabolas Get 3 of 4 questions to level up!
• Shift functions Get 3 of 4 questions to level up!

Topic C: Lessons 20-22: Scaling and transforming graphs

• Scaling & reflecting parabolas (Opens a modal)
• Identifying function transformations (Opens a modal)
• Identifying horizontal squash from graph (Opens a modal)
• Reflecting & compressing functions (Opens a modal)
• Transforming the square-root function (Opens a modal)
• Graphs of square-root functions (Opens a modal)
• Square-root functions & their graphs (Opens a modal)
• Radical functions & their graphs (Opens a modal)
• Scale & reflect parabolas Get 3 of 4 questions to level up!
• Identify function transformations Get 3 of 4 questions to level up!
• Graphs of square and cube root functions Get 3 of 4 questions to level up!

Chapter 3: Polynomial Functions

3.4.1: the division of polynomials, learning outcomes.

• Divide a polynomial by a monomial
• Use long division to divide polynomials

Dividing a Polynomial by a Monomial

The distributive property of multiplication over addition also works for division, since division is just multiplication by the reciprocal.

The distributive property

[latex]a(b+c)=ab+ac[/latex]

[latex]\dfrac{a+b}{c}=\dfrac{1}{c}(a+b)=\dfrac{a}{c}+\dfrac{b}{c}[/latex]

where [latex]a,\;b,[/latex] and [latex]c[/latex] are algebraic terms.

To divide a polynomial by a monomial, we can write the division as a fraction, and then decompose the fraction into the sum of fractions, and then simplify each fraction. For example, to divide the polynomial [latex]3x^3-6x^2+18x-7[/latex] by [latex]2x^2[/latex]:

[latex]\begin{aligned}&\;\;\;\;\,\dfrac{3x^3-6x^2+18x-7}{2x^2} \\ \\&= \dfrac{3x^3}{2x^2}-\dfrac{6x^2}{2x^2}+\dfrac{18x}{2x^2}-\dfrac{7}{2x^2} \\ \\&= \dfrac{3}{2}x-3+\dfrac{9}{x}-\dfrac{7}{2x^2}\end{aligned}[/latex]

Divide [latex]12x^4-18x^3+36x^2-24x[/latex] by [latex]-6x^2[/latex].

Divide each term in the first polynomial by the monomial divisor:

[latex]\begin{aligned}&\;\;\;\;\,\dfrac{12x^4-18x^3+36x^2-24x}{-6x^2}\\ \\&=\dfrac{12x^4}{-6x^2}-\dfrac{18x^3}{-6x^2}+\dfrac{36x^2}{-6x^2}-\dfrac{24x}{-6x^2}\\ \\&=-2x^2+3x-6+\dfrac{4}{x}\end{aligned}[/latex]

Divide [latex]60x^5-30x^4+25x^3-35x+10[/latex] by [latex]10x^3[/latex].

[latex]6x^2-3x+\dfrac{5}{2}-\dfrac{7}{2x^2}+\dfrac{1}{x^3}[/latex]

Dividing Polynomials using Long Division

Long division revision.

When we divide two numbers, we often end up with a remainder. For example, [latex]17\div 3 = 5\;\text{R}(3)[/latex].

This can also be thought of as:

[latex]\dfrac{17}{3}=5+\dfrac{2}{3}[/latex]

Equivalently, multiplying both sides by 3:

[latex]17=3\cdot 5 + 2[/latex]

The process used to accomplish division of polynomials is similar to long division of whole numbers, so it is essential that we are confident with long division of whole numbers. If you haven’t performed long division for a while, take a moment to refresh before reading this section.

The Long Division Algorithm for Arithmetic

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

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

[latex]\begin{aligned}\left(\text{Divisor }\cdot \text{Quotient}\right)\text{ + Remainder}&=\text{Dividend}\\ \left(3\cdot 59\right)+1&= 178\end{aligned}[/latex]

We call this the Division Algorithm :

[latex]\dfrac{\text{Dividend}}{\text{Divisor}}=\text{Quotient}+\dfrac{\text{Remainder}}{\text{Divisor}}[/latex]

Equivalently:

[latex]\text{Dividend}=\text{Quotient}\cdot\text{Divisor}+\text{Remainder}[/latex]

The dividend, divisor, and remainder are whole numbers, and the remainder < divisor.

Use long division to divide: [latex]\dfrac{242}{15}[/latex]

[latex]\begin{aligned}16\\15) \overline{242}\\15\;\;\\ \overline{92}\\ 90\\ \overline{2}\end{aligned}[/latex]

So, [latex]\dfrac{242}{15}=16 +\dfrac{2}{15}[/latex]

Equivalently, [latex]242=15\cdot 16 + 2[/latex]

Use long division to divide: [latex]\dfrac{463}{12}[/latex]

[latex]\dfrac{463}{12}=38+\dfrac{7}{12}[/latex]

The Division Algorithm

The division algorithm for arithmetic can be adapted to polynomials.

division algorithm for polynomials

Let [latex]p(x)[/latex] and [latex]d(x)[/latex] be polynomials where the degree of [latex]p(x)[/latex] ≥ the degree of [latex]d(x)[/latex], then

[latex]\dfrac{p(x)}{d(x)}=q(x)+\dfrac{r(x)}{d(x)}[/latex]

where the degree of [latex]r(x)[/latex] < the degree of [latex]d(x)[/latex].

Equivalently, multiplying both sides by [latex]d(x)[/latex]:

[latex]p(x)=d(x)q(x) + r(x)[/latex]

The division algorithm allows us to divide two polynomials. To set up this division there are two basic rules that must be adhered to:

• The polynomials of the divisor and dividend need to be in descending order , which means the terms of a polynomial are written in the order of descending powers of variables.
• There should be no gaps in the descending order of powers. Any gaps need to be filled with zeros.

For example, if the dividend is given as [latex]4+5x-3x^3[/latex], the first step tells us to write the polynomial in descending order:

[latex]-3x^3+5x+4[/latex]

Since there is no [latex]x^2[/latex] term in this polynomial (i.e. there is a gap), we must include [latex]0x^2[/latex] as a term to fill the gap:

[latex]-3x^3\color{blue}{+0x^2}+5x+4[/latex]

Write the polynomial in descending order, filling any gaps with zeros.

1. [latex]7x^3-8+2x^2[/latex]

2. [latex]6x-4x^2-3x^3+x^5[/latex]

1. First write [latex]7x^3-8+2x^2[/latex] in descending order: [latex]7x^3+2x^2-8[/latex]

Now look for missing terms: there is no [latex]x[/latex] term. Write [latex]0x[/latex] where the [latex]x[/latex] term should be:

[latex]7x^3+2x^2\color{blue}{+0x}-8[/latex]

2. First write [latex]6x-4x^2-3x^3+x^5[/latex] in descending order: [latex]x^5-3x^3-4x^2+6x[/latex]

Now look for missing terms: there is no [latex]x^4[/latex] term and no constant term. Write [latex]0x^4[/latex] where the [latex]x^4[/latex] term should be, and write 0 for the constant term at the end of the polynomial:

[latex]x^5\color{blue}{+0x^4}-3x^3-4x^2+6x\color{blue}{+0}[/latex]

1. [latex]7x^4-8x+2x^2[/latex]

2. [latex]-2x-4x^4-3x^3+x^5[/latex]

• [latex]7x^4\color{blue}{+0x^3}+2x^2-8x\color{blue}{+0}[/latex]
• [latex]x^5-4x^4-3x^3\color{blue}{+0x^2}-2x\color{blue}{+0}[/latex]

The process for dividing polynomials is basically the same as it was for whole numbers.

We can identify the dividend, divisor, quotient, and remainder by writing this answer in its equivalent form:

[latex]\begin{array}{lllllll}&x^2+x+1&=&(x-1)&(x+2)&+&3\\&\text{Dividend}&&\text{Divisor}&\text{Quotient}&&\text{Remainder}\end{array}[/latex]

Divide [latex]2{x}^{3}-3{x}^{2}+4x+5[/latex] by [latex]x+2[/latex] using the long division algorithm.

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

or equivalently,

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

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

Notice that the remainder in this example is negative. Even when the remainder is negative, we add it as a negative remainder.

using long division to divide polynomials

• Set up the division problem based on the rules of descending powers and no gaps.
• Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
• Put the answer as the first term in the quotient.
• Multiply the answer by the divisor and write it below the like terms of the dividend.
• Subtract the product in step 4 from all the terms above it.
• Repeat steps 2–5 until reaching the last term of the dividend.
• If the remainder is non-zero, express as a fraction using the divisor as the denominator.

Divide [latex]5{x}^{2}+3x - 2[/latex] by [latex]x+1[/latex].

[latex]\dfrac{5{x}^{2}+3x - 2}{x+1}=5x - 2[/latex]

Analysis of the Solution

This division problem had a remainder of 0. This tells us that the dividend is divided exactly by the divisor and that, therefore,  the divisor and the quotient are factors of the dividend.   [latex]5{x}^{2}+3x - 2=\left(x+1\right)\left(5x - 2\right)[/latex]

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

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

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

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

[latex]\;\;\;\left(3x - 2\right)\left(2{x}^{2}+5x - 7\right)+1\\=3x\left(2{x}^{2}+5x - 7\right)-2\left(2{x}^{2}+5x - 7\right)+1\\=6x^3+15x^2-21x-4x^2-10x+14+1\\=6{x}^{3}+11{x}^{2}-31x+15[/latex]

Notice, as we write our result,

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

Divide [latex]16{x}^{3}-12{x}^{2}+20x - 3[/latex] by [latex]4x+5[/latex].

[latex]\dfrac{16x^3-12x^2+20x-3}{4x+5}=4x^2-8x+15+\left (\dfrac{-78}{4x+5}\right )[/latex]

• Two Rules for Setting Up a Long Division. Authored by : Leo Chang and Hazel McKenna. Provided by : Utah Valley University. License : CC BY: Attribution
• Revision and Adaptation. Authored by : Hazel McKenna. Provided by : Lumen Learning. License : CC BY: Attribution
• Divide a Polynomial by a Monomial. Authored by : Leo Chang and Hazel McKenna. Provided by : Utah Valley University. License : CC BY: Attribution
• Division algorithm for polynomials; Examples and Try Its: hjm202, hjm871; hjm761; hjm096. Authored by : Hazel McKenna. Provided by : Utah Valley University. License : CC BY: Attribution
• Question ID 29482. Authored by : McClure,Caren. License : Other . License Terms : IMathAS Community License CC-BY + GPL
• College Algebra. Authored by : Abramson, Jay et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
• Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program.. Provided by : Monterey Institute of Technology and Education. Located at : http://nrocnetwork.org/dm-opentext.%20 . License : CC BY: Attribution

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Factoring to Solve Problems

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TEKS Standards and Student Expectations

A(10)  Number and algebraic methods. The student applies the mathematical process standards and algebraic methods to rewrite in equivalent forms and perform operations on polynomial expressions. The student is expected to:

A(10)(E) factor, if possible, trinomials with real factors in the form a x 2 + b x + c including perfect square trinomials of degree two

A(10)(F)  decide if a binomial can be written as the difference of two squares and if possible, use the structure of a difference of two squares to rewrite the binomial

A(10)(D) rewrite polynomial expressions of degree one and degree two in equivalent forms using the distributive property

A(8)  Quadratic functions and equations. The student applies the mathematical process standards to solve, with and without technology, quadratic equations and evaluate the reasonableness of their solutions. The student formulates statistical relationships and evaluates their reasonableness based on real-world data. The student is expected to:

A(8)(A) solve quadratic equations having real solutions by factoring, taking square roots, completing the square, and applying the quadratic formula

Resource Objective(s)

The student will use a variety of methods to solve problems by factoring, including models, guess and check, grouping, and special cases.

Essential Questions

What is the process for factoring a trinomial with a leading coefficient of 1?

What is the process for factoring a trinomial when the leading coefficient is not 1?

How can you determine if the rules for difference of two squares, difference of two cubes, or perfect squares  can be used to factor?

When does the grouping method for factoring apply?

• GCF (Greatest Common Factor)
• Leading Coefficient
• Quadratic Equation

Introduction: Factoring by Modeling

One way to find the factors of a value is to use models to create a rectangle. The length and width represent the factors.

Example 1 : Factors of 12 can be shown three different ways. 

12 has a total of 6 different factors: 1, 2, 3, 4, 6, and 12.

Example 2:  Algebraic expressions can also be factored using algebra tiles to create rectangles. In the example below, 2 x + 4 is used to create a rectangle. 

As you can see, the length and width of the rectangle represent the factors of 2 x + 4. The length is ( x + 2) and the width is 2. 

The factors of 2 x + 4 are 2 and ( x + 2).

Example 3:  Trinomials can also be factored using algebra tiles. The trinomial x 2 + x - 6 can be modeled by the following rectangle.

The side lengths of the rectangle show the factors of x 2  + x - 6  are ( x + 3) and ( x - 2) .

When you are ready, complete the two practice examples below to check your understanding.

Factoring Rule 1: Greatest Common Factor (GCF)

Always check to see if you can factor something out!

The first rule to factoring is to find the greatest common factor (GCF) of each term in the polynomial .

• If there is any factor in common in the polynomial, divide each term by that factor.
• Then, rewrite the polynomial using the distributive property with the common factor on the outside of the parenthesis.

2 x 3 -   2 x 2 - 12 x

All terms in the example have a coefficient that is divisible by 2. Each term also has an x . Therefore, the GCF is 2 x .

When 2 x  is factored out and the distributive property is applied the result is 

2 x ( x 2 -   x   -   6 ) .

Check your understanding with the two practice examples  below.

Factoring Rule 2: Special Products

If you are given a binomial to factor, the first special product to check for is the difference of two squares.

Difference of Two Squares:

  a 2 - b 2 =   ( a   -   b ) ( a   +   b ) .

If both terms are separated by a subtraction symbol and they are both perfect squares, the square roots can be used to write the factors.

Example 1: Factor  4 x 2 - 9 y 2 :

• The terms are separated by a subtraction symbol.
• Both terms are perfect squares.

    ( 2 x ) 2   a n d   ( 3 y ) 2

According to the rule for the difference of two squares, the factors will be  ( 2 x - 3 y ) ( 2 x + 3 y ) .

If you are given a trinomial to factor, the first special product to check for is a perfect square trinomial.

Perfect Square Trinomials:

a 2 + 2 a b + b 2   =   ( a + b ) 2   o r   a 2 - 2 a b + b 2   =   ( a - b ) 2

If the first term is a square, the last term is a square, and the middle term is two times the square root of the first and the last term, then it is a perfect square. There are two forms: one with a positive middle term and one with a negative middle term.

Example 2: Factor  x 2 + 6 x + 9

Notice the first and last terms are both perfect squares. When you multiply the square roots of x  and 3 times 2, you get 6 x   which is the middle term. According to the rule, t he factors are 

( x + 3 ) ( x + 3 )   o r   ( x + 3 ) 2

Example 3: Factor  4 x 2 - 20 x + 25

Notice the first and last terms are both perfect squares. When you multiply the square roots of 2 x and 5 times -2, you get -20 x which is the middle term. According to the rule, the factors are 

( 2 x - 5 ) ( 2 x - 5 )   o r   ( 2 x - 5 ) 2

Check your understanding by completing the practice example below. Determine which polynomials can be factored with a special product rule. Match each polynomial with the type of special product (first blank) and its factors (second blank).

Factoring Rule 3: Trinomials with a Leading Coefficient of 1

After checking for a GCF and Special Products, look for polynomials with three terms that have a leading coefficient of 1. If the leading coefficient is 1, put x 's as the first term in each set of parentheses. Then you factor the last term and find the two factors that add or subtract to make the middle term.

Use the practice example to complete the activity below.

Factoring Rule 4: Other Trinomials

If the leading coefficient is not 1, you factor both the first and last terms and find the two products that add or subtract to make the middle term. Check your answer by using the smiley face check.

Factoring Rule 5: Four or More Terms

If there are four or more terms, you need to group them together, and then factor each part.

You can do this two ways—either by using the box method or parentheses.

Box Method: 

• Put all four terms in a two by two box.
• Look for the GCF in each column and row.
• Include signs for each of the factors.
• The result will be two factors: one on the length and the other on the width.

Parentheses Method: 

• Put parentheses around the first two terms and another set around the last two terms.
• Find the GCF of each set of parentheses. 
• Once the GCF is factored out, the terms inside each set of parentheses should match. 
• The factors will be the terms inside the parentheses (from step 3) and the GCF outside each set of parentheses (in step 2). 

Check your understanding by factoring the following polynomial (using either method) and answer the questions below. 

Solving Problems by Factoring

Some problems may ask for the solution of a quadratic equation . To solve these problems, set the factors equal to 0 and solve for x . 

     2 x 3   -    2 x 2   - 12 x   =   2 x   ( x 2   -    x   -    6 )   =   2 x   ( x   -    3 ) ( x   +   2 )

To solve this problem, you would need to set all factors to 0 and solve for x . There will be three solutions.

Fill in the blanks below to find the three solutions to the polynomial. 

Check polynomials in your graphing calculator:

Your solutions will be the x -intercepts on the graph.

• Input your equation using the equation editor [ Y  =] key on the graphing calculator.
• Press [GRAPH].
• Press [2nd] [TRACE] to access the CALC menu.
• Press [2] to select the zero command.
• Use the arrow keys to select a left bound and a right bound.

Vocabulary Activity

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