Category: Polynomials

What are Monomials? – Answered

If you want to know what monomials are, this page will help you identify and understand these algebraic terms.

Monomials have only one algebraic term. A monomial can be a number, a number with variables, or a number with variables raised to an exponential power.

What are Monomials – Quiz

Instructions: Identify the monomials in the questions in the quiz below.

[WpProQuiz 20]

Monomials – Examples

As stated above, a monomial has one term. A monomial cannot have negative exponents, nor can it have fractional exponents.

The following are examples of monomials:

3

15b

xy

43ab

5x²

62x²y²

Terms that are not monomials – Examples

The following are not monomials:

x²y + y²  →  This is a binomial.

3x² + xy + y²  →  This is a trinomial.

Binomials and trinomials are classified as polynomials.

So, two or more monomials are needed to make a polynomial.

Polynomials – Further Exercises

This post is intended as an introductory exercise.

You should also view the following posts for further practice for your algebra test:

Binomials

Adding Polynomials

Subtracting Polynomials

Multiplying Polynomials

Dividing Polynomials

Get the download

Subtracting Polynomials – Examples

You should study problems on subtracting polynomials with examples for your algebra test.

To see the examples, please go to the sections below the following quiz.

Subtracting Polynomials – Quiz

Instructions: Subtract the polynomials in the quiz below after studying the examples in the next section. You will see the answers after each question.

[WpProQuiz 31]

Polynomials with Negatives

When you see a problem about subtracting polynomials, you need to be especially careful with polynomial expressions that contain negative numbers.

You should also view our post on adding polynomials if you need practice with this skill.

Grouping Like Terms

If you see a problem on subtracting polynomials, you need to be sure to group like terms together.

Click here to learn how to group like terms.

The steps for subtracting polynomials are similar to those for adding polynomials because you have to group like terms.

Subtracting Polynomials - Example

When subtracting polynomials you also need to remember that two negative signs together make a positive.

So, when you have to subtract a negative number you actually need to add that number, as shown in step 2 below.

Now study the steps in the following example problem.

Instructions:  Subtract the following polynomials.

(2x² + 3xy + 5y²) – (4x² – 4xy – 6y²) = ?

Answer:   –2x² + 7xy + 11y²  

STEP 1:  

Remove the parentheses, paying attention to any negatives.

(2x² + 3xy + 5y²) – (4x² – 4xy – 6y²) =

2x² + 3xy + 5y² – 4x² – –4xy – –6y²

STEP 2:  

Change subtraction to addition when a negative number is subtracted.

2x² + 3xy + 5y² – 4x² – –4xy – –6y² =

2x² + 3xy + 5y² – 4x² + 4xy + 6y²

STEP 3:  

Group the like terms together using sets of parentheses.

2x² + 3xy + 5y² – 4x² + 4xy + 6y² =

2x² – 4x² + 3xy + 5y² – 4x² + 4xy + 6y² =

2x² – 4x² + 3xy + 5y² + 4xy + 6y² =

2x² – 4x² + 3xy + 4xy  + 5y² + 4xy + 6y² =

2x² – 4x² + 3xy + 4xy  + 5y² + 6y²

STEP 4:  

Perform the operations on the terms inside each set of parentheses to solve the problem.

2x² – 4x² + 3xy + 4xy  + 5y² + 6y² =

(2x² – 4x²) + (3xy + 4xy)  + (5y² + 6y²) =

–2x² + 7xy + 11y²

Polynomials - Further Practice

If you thought these problems on subtracting polynomials were helpful, you may also want to look at the following posts on polynomials:

Multiplying polynomials

Dividing polynomials

Get the downloads

Multiplying Polynomials – Examples Page

Look at these multiplying polynomials examples to prepare for your math exam.

Multiplying Polynomials – Quiz

Instructions: Multiply the polynomials in the questions in the quiz below. To see the free samples first, please scroll down.

[WpProQuiz 21]

Multiplying Polynomials – Example

A polynomial contains two or more algebraic terms, so you will need to know how to multiply the terms of the polynomial together to get the correct result. Now look at the example that follows.

Instructions:  Simplify the following expression.

3a²b(3ab + 4a²b² – 5ab)

Answer:  12a⁴b³ – 6a³b²

STEP 1:  Set up the distribution of the multiplication.

3a²b(3ab + 4a²b² – 5ab) =

(3a²b × 3ab) + (3a²b × 4a²b²) + (3a²b × –5ab)

STEP 2: Multiply one term by the other in each set of parentheses.

Multiply the numbers by each other and the variables by each other, paying attention to the exponents and negatives.

(3a²b × 3ab) + (3a²b × 4a²b²) + (3a²b × –5ab) =

9a³b² + 12a⁴b³ + –15a³b²

STEP 3: Simplify further by grouping like terms together, if possible.

9a³b² + 12a⁴b³ + –15a³b² =

9a³b² + 12a⁴b³ – 15a³b² =

9a³b² – 15a³b² + 12a⁴b³ =

–6a³b² + 12a⁴b³ =

12a⁴b³ – 6a³b²

Multiplying Polynomials – Step by Step

Multiplying polynomials is performed by using the distributive property of multiplication.

Polynomials include quadratics, binomials, trinomials, and polynomials with more than three terms.

You will usually have to deal with exponents in order to solve polynomial multiplication problems.

So, be sure that you are comfortable with Exponent Laws for these types of algebra questions.

You may see the words “simplify” or “expand” in the instructions when you have problems on multiplying polynomials.

“Simplify” and “expand” both mean that you have to perform multiplication on the polynomial.

When multiplying polynomials, you need to multiply one term by the other in each set of parentheses. To do so, multiply the numbers by each other and the variables by each other. Be sure to pay special attention to the exponents and negatives.

If you have had difficulties with this example on multiplying polynomials, please view our posts on:

Algebraic distribution

Grouping like terms

Multiplying Polynomials – Further Practice

Multiplying polynomials in quadratic equations is performed by using the FOIL Method.

You may also want to look at out posts on:

Quadratic Equations

FOIL Method

More free algebra practice

Dividing Polynomials Examples Using Long Division

This page provides dividing polynomials examples and problems, using long division.

You may wish to view the examples on dividing polynomials below the exercises before attempting to answer the questions.

Dividing Polynomials – Exercises

Instructions:  Perform the division as shown in each problem.

1)   (3x – 12) ÷ 3 = ?

2)   (16x³ + 8x² + 4x) ÷ (2x) = ?

3)   (84a²b + 16ab² + 36ab – 12ab³) ÷ (4ab) = ?

4)   (15a² + 22a + 8) ÷ (3a + 2) = ?

5)    (x² – 2x + 5) ÷ (x – 1) = ?

Dividing Polynomials – Answers

1)   The correct answer is:  x – 4

(3x – 12) ÷ 3 =

(3x ÷ 3) – (12 ÷ 3) =

(3x ÷ 3) – (4) =

x – 4

2)   The correct answer is:  8x² + 4x + 2

(16x³ + 8x² + 4x) ÷ (2x) =

(16x³ ÷ 2x) + (8x² ÷ 2x) + (4x ÷ 2x) =

(16x³ ÷ 2x) + (8x² ÷ 2x) + (2) =

(16x³ ÷ 2x) + (4x) + 2 =

8x² + 4x + 2

3)   The correct answer is:  21a – 3b² + 4b + 9

(84a²b + 16ab² + 36ab – 12ab³) ÷ (4ab) =

(84a²b ÷ 4ab) + (16ab² ÷ 4ab) + (36ab ÷ 4ab) – (12ab³ ÷ 4ab) =

(84a²b ÷ 4ab) + (16ab² ÷ 4ab) + (36ab ÷ 4ab) – (3b²) =

(84a²b ÷ 4ab) + (16ab² ÷ 4ab) + (9) – 3b² =

(84a²b ÷ 4ab) + (16ab² ÷ 4ab) + 9 – 3b² =

(84a²b ÷ 4ab) + (4b) + 9 – 3b² =

(21a) + 4b + 9 – 3b² =

21a – 3b² + 4b + 9

4)   The correct answer is:  5a + 4

   5a + 4
3a + 2)    15a² +  22a + 8
–(15a² + 10a)
     12a + 8
–(12a + 8)

0

5)   The correct answer is: \(\text x – 1 + \frac{4}{(x – 1)}\)

  x – 1
x– 1)     x²  –  2x  +  5
–(x² – x)
–x   +  5
–(–x  +  1)
4

As shown in the solution to problem 5, solutions to questions on dividing polynomials can sometimes have a remainder.

To see more free math sample questions, please click here.

To learn how to divide polynomial equations using factoring or synthetic division, please see our posts on:

Synthetic Division

Factoring

Dividing Polynomials by Monomials – Example

A monomial is a single algebraic term.

In other words, a monomial is not an expression, so it won’t have the plus or minus sign.

A monomial can be a number or a number with a variable.

Divide a polynomial by a monomial using following steps.

Example:   (6x² – 2x) ÷ 2x = ?

Answer:  3x – 1

STEP 1:  Expand the problem using sets of parentheses.

(6x² – 2x) ÷ 2x =

(6x² ÷ 2x) – (2x ÷ 2x)

STEP 2:  Find the factors for the terms in each set of parentheses.

3x × 2x = 6x²   so  (6x² ÷ 2x) = 3x

1 × 2x = 2x   so  (2x ÷ 2x) = 1

STEP 3:  Take the equation from step 1 and simplify for each set of parentheses.

(6x² ÷ 2x) – (2x ÷ 2x) =

(6x² ÷ 2x) – 1 =

3x – 1

Dividing Polynomials by Binomials – Examples

To divide a polynomial by a binomial or by another polynomial, you can use long division.

Example:   (m³ – m) ÷ (m + 1) = ?

Answer:   m² – m

STEP 1:  Set up the long division. Be sure to put in the missing terms.

For example, if you have a polynomial with m³ but not m² , like this example, then you can put in placeholders.

To use placeholders, be sure that your dividend has all of the variables with the exponents in descending order, and put in zeros where necessary.

m + 1)  m³ +  0m² – m

STEP 2:  Divide the first term from the divisor (m) into the first term of the dividend (m³ ) using long division.

According to the laws of exponents we know that:

m³ ÷ m = m²

So, use m² as the first term in the quotient.

m²
m + 1)  m³ +  0m² – m

STEP 3:  Then multiply the first term that you put in the quotient (m²) by the divisor (m + 1) and write the result down below the divisor.

  m²
m + 1)    m³ +  0m² – m
–(m³ + m²)

STEP 4:  Now subtract the result of step 3 from the dividend.

  m²
m + 1)    m³ +  0m² – m
–(m³ + m²)
–m²

STEP 5:  Bring down the next term from the dividend (–m).

  m²
m + 1)    m³ +  0m² – m
–(m³ + m²)
–m² – m

STEP 6:  Divide the first term from the divisor (m) into the first term from step 5 (–m²).

–m² ÷ m = –m

Place this result (–m) in the quotient at the top of the problem.

  m² – m
m + 1)    m³ +  0m² – m
–(m³ + m²)
–m² – m

STEP 7:  Then multiply the second number  in the quotient (–m) by the divisor  (m + 1) and write the result down below.

  m² – m
m + 1)    m³ +  0m² – m
–(m³ + m²)
–m² – m
–(–m²– m)

STEP 8:  Now subtract to get your result.

  m² – m
m + 1)    m³ +  0m² – m
–(m³ + m²)
–m² – m
–(–m² – m)
0

For an alternative to solve these types of questions on polynomial division, please see our post entitled:  Remainder Theorem

Dividing Polynomials – Quadratics

You might see problems on polynomials in the quadratic form on your algebra test.

These polynomial division problems have equations in one of the following forms:

ax² ± bx ± c = 0

x² ± bx ± c = 0

x² ± bx = n

x² = n

To learn how to solve polynomial equations that are in the quadratic form, please see our posts on:

Factoring Quadratics

Solving Quadratic Equations

Problems on dividing polynomials can also solved by setting up the polynomial expressions as a fraction.

You may also want to have a look at our post on algebraic fractions.

For information on our downloadable PDFs, please click on the image below.

Binomial Expressions – Definition and Examples

Definition: Binomial expressions are polynomials that consist of two algebraic terms. You will need to find the greatest common factor in order to to factor binomial expressions.

Example: Binomial expressions may have a common factor, such as the variable a in the expression:  ab + ac

Binomial Expressions – Greatest Common Factor Quiz

Choose the answer which has factored out the greatest common factor. You may wish to scroll down to see the further examples on this page before attempting the quiz below.

[WpProQuiz 6]

Types of Binomials – Further Examples

There are different types of binomial expressions.

Perhaps the most common types of binomial expressions are: factored quadratics, the difference of two perfect squares, the difference of two perfect cubes, the sum of two perfect cubes, and binomials that share a common factor.

Factored Quadratic

A factored quadratic consists of two binomials that are multiplied using the FOIL method.

A factored quadratic is in the following format:

(x ± n)(x ± n)

If you need help with this skill, please visit our post:

Factoring Quadratics

Difference of two perfect squares

The difference of two perfect squares consists of a binomial in the format:

a² – b²

When the difference of two perfect squares is factored, it consists of two binomial expressions in the following format:

(a + b)(a – b)

If you need to review this skill, please see our post:

Factoring the difference of two perfect squares

Difference of two perfect cubes

The difference of two perfect cubes consists of a binomial in the format:

a³ – b³

When we factor the difference of two perfect cubes, the result is a binomial expression and a polynomial expression in this format:

(a – b)(a² + ab + b²)

Sum of two perfect cubes

The sum of two perfect cubes consists of a binomial in the format:

a³ + b³

When we factor the sum of two perfect cubes, we have a binomial expression and a polynomial expression in this format:

(a + b)(a² – ab + b²)

If you need to review how to factor the sum or difference of two perfect cubes, please see our post:

Factoring the sum or difference of two perfect cubes

Binomials that Share a Common Factor

Binomials that share a common factor are in the form:

ab + ac

To factor this type of binomial, we expand the expression and factor as show below.

ab + ac =

(a × b) + (a × c) =

a[(a × b) + (a × c)] =

a(b + c)

Polynomials – Further Exercises

If you found the previous problems difficult, you may want to try the following exercises:

Algebraic Terms

Algebraic Distribution

Finding the Greatest Common Factor

Adding Polynomials – Examples

To see the examples and solutions on polynomials, please scroll down. Then try your skills on the addition of polynomials by completing the exercises below.  The answers and explanations are provided after the last exercise.

Adding Polynomials – Exercises

1)   (6x² + 9xy + 7y²) + (8x² + 2xy + 5y²) = ?

2)   (2x² + 3xy + 5y²) + (2x² – 6xy – 7y²) = ?

3)   (3x³ – 5xy – 2y + 3y²) + (2x³ + 4x² + 5x + 3y – 8xy – 9y²) = ?

4)   (3a² + 4ab + 2b²) + (7a² – ab – 6b²) + (8a² – 5ab – 4b²) = ?

5)   (4x² + 3xy  + 6y + 7y²) + (3x² + 7xy + 5x – 9y²) + (6x² – 8y + 2xy) = ?

Adding Polynomials – Answers

Answer 1

1)   The answer is:  14x² + 11xy + 12y²

Step 1: Remove the parentheses.

(6x² + 9xy + 7y²) + (8x² + 2xy + 5y²) =

6x² + 9xy + 7y² + 8x² + 2xy + 5y² 

Step 2: Then put like terms together.

6x² + 8x² + 9xy + 7y² + 8x² + 2xy + 5y² =

6x² + 8x² + 9xy + 7y² + 2xy + 5y² =

6x² + 8x² + 9xy + 2xy + 7y² + 2xy + 5y² =

6x² + 8x² + 9xy + 2xy + 7y² + 5y² 

       Step 3: Then perform the operations to solve.

(6x² + 8x²) + (9xy + 2xy) + (7y² + 5y²) =

14x² + 11xy + 12y²

Answer 2

2)   The answer is:  4x² – 3xy – 2y²

(2x² + 3xy + 5y²) + (2x² – 6xy – 7y²) =

2x² + 3xy + 5y² + 2x² – 6xy – 7y² 

Step 2: Then put like terms together.

2x² + 2x² + 3xy + 5y² + 2x² – 6xy – 7y² =

2x² + 2x² + 3xy + 5y² – 6xy – 7y² =

2x² + 2x² + 3xy – 6xy + 5y² – 6xy – 7y² =

2x² + 2x² + 3xy – 6xy + 5y² – 7y² 

Step 3: Then perform the operations to solve.

(2x² + 2x²) + (3xy – 6xy) + (5y² – 7y²) =

4x² – 3xy – 2y²

Answer 3

3)   The answer is:  5x³ + 4x² + 5x – 13xy – 6y²+ y

Step 1: Remove the parentheses.

(3x³ – 5xy – 2y + 3y²) + (2x³ + 4x² + 5x + 3y – 8xy – 9y²) =

3x³ – 5xy – 2y + 3y² + 2x³ + 4x² + 5x + 3y – 8xy – 9y² 

Step 2: Then put like terms together.

3x³ + 2x³ – 5xy – 2y + 3y² + 2x³ + 4x² + 5x + 3y – 8xy – 9y² =

3x³ + 2x³ – 5xy – 2y + 3y² + 4x² + 5x + 3y – 8xy – 9y² =

3x³ + 2x³  + 4x² – 5xy – 2y + 3y² + 4x² + 5x + 3y – 8xy – 9y² =

3x³ + 2x³  + 4x² – 5xy – 2y + 3y² + 5x + 3y – 8xy – 9y² =

3x³ + 2x³  + 4x² + 5x – 5xy – 2y + 3y² + 5x + 3y – 8xy – 9y² =

3x³ + 2x³  + 4x² + 5x – 5xy – 2y + 3y² + 3y – 8xy – 9y² =

3x³ + 2x³  + 4x² + 5x – 5xy – 8xy – 2y + 3y² + 3y – 8xy – 9y² =

3x³ + 2x³  + 4x² + 5x – 5xy – 8xy – 2y + 3y² + 3y – 9y² =

3x³ + 2x³  + 4x² + 5x – 5xy – 8xy – 2y + 3y² + 3y – 2y – 9y² =

3x³ + 2x³  + 4x² + 5x – 5xy – 8xy + 3y² + 3y – 2y – 9y² =

3x³ + 2x³  + 4x² + 5x – 5xy – 8xy + 3y² – 9y² + 3y – 2y – 9y² =

3x³ + 2x³  + 4x² + 5x – 5xy – 8xy + 3y² – 9y² + 3y – 2y 

Step 3: Then perform the operations to solve.

(3x³ + 2x³ ) + 4x² + 5x + (–5xy – 8xy) + (3y² – 9y²) + (3y – 2y) =

5x³ + 4x² + 5x – 13xy – 6y²+ y

Answer 4

4)   The answer is:  18a² – 2ab – 8b²

Step 1: Remove the parentheses.

(3a² + 4ab + 2b²) + (7a² – ab – 6b²) + (8a² – 5ab – 4b²) =

3a² + 4ab + 2b² + 7a² – ab – 6b² + 8a² – 5ab – 4b² 

Step 2: Then put like terms together.

3a² + 7a² + 4ab + 2b² + 7a²– ab – 6b² + 8a² – 5ab – 4b² =

3a² + 7a² + 4ab + 2b² – ab – 6b² + 8a² – 5ab – 4b² =

3a² + 7a² + 8a²  + 4ab + 2b² – ab – 6b² + 8a² – 5ab – 4b² =

3a² + 7a² + 8a²  + 4ab + 2b² – ab – 6b² – 5ab – 4b² =

3a² + 7a² + 8a²  + 4ab – ab + 2b² – ab – 6b² – 5ab – 4b² =

3a² + 7a² + 8a²  + 4ab – ab + 2b² – 6b² – 5ab – 4b² =

3a² + 7a² + 8a²  + 4ab – ab – 5ab + 2b² – 6b² – 5ab – 4b² =

3a² + 7a² + 8a²  + 4ab – ab – 5ab + 2b² – 6b² – 4b² 

Step 3: Then perform the operations to solve.

(3a² + 7a² + 8a²) + (4ab – ab – 5ab) + (2b² – 6b² – 4b²) =

18a² – 2ab – 8b²

Answer 5

5)   The answer is:  13x² + 5x + 12xy – 2y² – 2y

Step 1: Remove the parentheses.

(4x² + 3xy  + 6y + 7y²) + (3x² + 7xy + 5x – 9y²) + (6x² – 8y + 2xy) =

4x² + 3xy  + 6y + 7y² + 3x² + 7xy + 5x – 9y² + 6x² – 8y + 2xy 

Step 2: Then put like terms together.

4x² + 3x² + 3xy  + 6y + 7y² + 3x² + 7xy + 5x – 9y² + 6x² – 8y + 2xy =

4x² + 3x² + 3xy  + 6y + 7y² + 7xy + 5x – 9y² + 6x² – 8y + 2xy =

4x² + 3x² + 6x² + 3xy + 6y + 7y² + 7xy + 5x – 9y² + 6x² – 8y + 2xy =

4x² + 3x² + 6x² + 3xy + 6y + 7y² + 7xy + 5x – 9y² – 8y + 2xy =

4x² + 3x² + 6x² + 5x + 3xy  + 6y + 7y² + 7xy + 5x – 9y² – 8y + 2xy =

4x² + 3x² + 6x² + 5x + 3xy + 6y + 7y² + 7xy – 9y² – 8y + 2xy =

4x² + 3x² + 6x² + 5x + 3xy + 7xy + 6y + 7y² + 7xy – 9y² – 8y + 2xy =

4x² + 3x² + 6x² + 5x + 3xy + 7xy + 6y + 7y² – 9y² – 8y + 2xy =

4x² + 3x² + 6x² + 5x + 3xy + 7xy + 2xy + 6y + 7y² – 9y² – 8y + 2xy =

4x² + 3x² + 6x² + 5x + 3xy + 7xy + 2xy + 6y + 7y² – 9y² – 8y =

4x² + 3x² + 6x² + 5x + 3xy + 7xy + 2xy + 6y + 7y² – 9y² – 8y + 6y =

4x² + 3x² + 6x² + 5x + 3xy + 7xy + 2xy + 7y² – 9y² – 8y + 6y 

Step 3: Then perform the operations to solve.

(4x² + 3x² + 6x²) + 5x + (3xy  + 7xy + 2xy) + (7y² – 9y²) + (–8y + 6y) =

13x² + 5x + 12xy – 2y² – 2y

Adding Polynomials – Examples of Addition of Polynomials

For problems on the addition of polynomials, you need to group like terms together.

Click here to see our post on grouping like terms.

Now look at the adding polynomials example that follows.

Instructions:  Study the following adding polynomials example.

(x² + xy + y²) + (3x² – 4xy + 2y²) = ?

Answer:  4x² – 3xy + 3y² 

STEP 1:  Remove the parentheses, paying attention to any negatives.

(x² + xy + y²) + (3x² – 4xy + 2y²) =

x² + xy + y² + 3x² – 4xy + 2y²

STEP 2:  Group the like terms together using sets of parentheses.

x² + xy + y² + 3x² – 4xy + 2y² =

x² + 3x² + xy + y² + 3x² – 4xy + 2y² =

x² + 3x² + xy + y² – 4xy + 2y² =

x² + 3x² + xy – 4xy + y² – 4xy + 2y² =

x² + 3x² + xy – 4xy + y² + 2y² =

(x² + 3x²) + (xy – 4xy) + (y² + 2y²)

STEP 3:  Perform the addition or subtraction of the terms inside each set of parentheses to solve.

(x² + 3x²) + (xy – 4xy) + (y² + 2y²) =

(4x²) + (–3xy) + (3y²) =

4x² – 3xy + 3y²

Addition of Polynomials on Your Exam

You will see several problems on adding polynomials on your exam.

If you liked this post on examples of adding polynomials, you should also see our post on subtracting polynomials to review this skill.

Also note that problems on the addition of polynomials may involve adding expressions that contain a negative number.

Addition of Polynomials – Further Practice

If you have found these questions and examples on adding polynomials useful, you may also want to view our other posts on polynomials:

Multiplying polynomials

Dividing polynomials

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