Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
Algebra 1

6.1.4 Adding and Subtracting Polynomial Functions

Algebra 16.1.4 Adding and Subtracting Polynomial Functions
Search for key terms or text.

Activity

Add the polynomial functions. Evaluate the sum for each given value.

z ( x ) = 8 x 2 + 4 x − 6 z ( x ) = 8 x 2 + 4 x − 6

w ( x ) = − 5 x 2 + 2 x + 3 w ( x ) = − 5 x 2 + 2 x + 3

1. ( z + w ) ( x ) = ( z + w ) ( x ) = ___________

2. ( z + w ) ( 6 ) = ( z + w ) ( 6 ) = ___________

3. ( z + w ) ( − 9 ) = ( z + w ) ( − 9 ) = ___________

4. ( z + w ) ( 0 ) ( z + w ) ( 0 ) ___________

Subtract the polynomial functions. Evaluate the sum for each given value.

m ( x ) = − 6 x 2 + 2 x − 8 m ( x ) = − 6 x 2 + 2 x − 8

n ( x ) = − x 2 − 7 x − 6 n ( x ) = − x 2 − 7 x − 6

5. ( m − n ) ( x ) = ( m − n ) ( x ) = ___________

6. ( m − n ) ( − 2 ) = ( m − n ) ( − 2 ) = ___________

7. ( m − n ) ( 0 ) = ( m − n ) ( 0 ) = ___________

8. ( m − n ) ( 4 ) = ( m − n ) ( 4 ) = ___________

9. What is a good way to check that you found the correct sum or difference when adding or subtracting polynomial functions?

Video: Combining and Evaluating Polynomial Functions

Watch the following video to learn more about how to add and subtract polynomial functions and evaluate them at a given value:

Access multimedia content

Self Check

For functions f ( x ) = 5 x 2 − 4 x − 1 and g ( x ) = x 2 + 3 x + 8 , find:

         ( f + g ) ( x ) = _______ 

         ( f + g ) ( 5 ) = _______

  1. ( f + g ) ( x ) = 6 x 2 − x + 7

    ( f + g ) ( 5 ) = 32

  2. ( f + g ) ( x ) = 4 x 2 − 7 x − 9

    ( f + g ) ( 5 ) = 56

  3. ( f + g ) ( x ) = 6 x 2 − x + 7

    ( f + g ) ( 5 ) = 152

  4. ( f + g ) ( x ) = 4 x 2 − x + 7

    ( f + g ) ( 5 ) = 102

Additional Resources

Adding and Subtracting Polynomial Functions

Just as polynomials can be added and subtracted, polynomial functions can also be added and subtracted.

For functions f ( x ) f ( x ) and g ( x ) g ( x ) ,

  • ( f + g ) ( x ) = f ( x ) + g ( x ) ( f + g ) ( x ) = f ( x ) + g ( x )
  • ( f − g ) ( x ) = f ( x ) − g ( x ) ( f − g ) ( x ) = f ( x ) − g ( x )

Let’s look at some examples.

For the functions f ( x ) = 3 x 2 − 5 x + 7 f ( x ) = 3 x 2 − 5 x + 7 and g ( x ) = x 2 − 4 x − 3 g ( x ) = x 2 − 4 x − 3 , find:

  1. ( f + g ) ( x ) ( f + g ) ( x )
  2. ( f + g ) ( 3 ) ( f + g ) ( 3 )
  3. ( f − g ) ( x ) ( f − g ) ( x )
  4. ( f − g ) ( − 2 ) ( f − g ) ( − 2 )

The following is a step-by-step breakdown.

a. ( f + g ) ( x ) = f ( x ) + g ( x ) f + g ) ( x ) = f ( x ) + g ( x )

Step 1 - Substitute

f ( x ) = 3 x 2 − 5 x + 7 f ( x ) = 3 x 2 − 5 x + 7 and g ( x ) = x 2 − 4 x − 3 g ( x ) = x 2 − 4 x − 3 .

( f + g ) ( x ) = ( 3 x 2 − 5 x + 7 ) + ( x 2 − 4 x − 3 ) ( f + g ) ( x ) = ( 3 x 2 − 5 x + 7 ) + ( x 2 − 4 x − 3 )

Step 2 - Rewrite without the parentheses.

( f + g ) ( x ) = 3 x 2 − 5 x + 7 − x 2 − 4 x − 3 ( f + g ) ( x ) = 3 x 2 − 5 x + 7 − x 2 − 4 x − 3

Step 3 - Put like terms together.

( f + g ) ( x ) = 3 x 2 + x 2 − 5 x − 4 x + 7 − 3 ( f + g ) ( x ) = 3 x 2 + x 2 − 5 x − 4 x + 7 − 3

Step 4 - Combine like terms.

( f + g ) ( x ) = 4 x 2 − 9 x + 4 ( f + g ) ( x ) = 4 x 2 − 9 x + 4

In part (a), we found ( f + g ) ( x ) ( f + g ) ( x ) , and now are asked to find ( f + g ) ( 3 ) ( f + g ) ( 3 ) .

b. ( f + g ) ( x ) = 4 x 2 − 9 x + 4 ( f + g ) ( x ) = 4 x 2 − 9 x + 4

Step 1 - To find ( f + g ) ( 3 ) ( f + g ) ( 3 ) , substitute x = 3 x = 3 .

( f + g ) ( 3 ) = 4 ( 3 ) 2 − 9 · 3 + 4 ( f + g ) ( 3 ) = 4 ( 3 ) 2 − 9 · 3 + 4

Step 2 - Simplify.

( f + g ) ( 3 ) = 4 · 9 − 9 · 3 + 4 ( f + g ) ( 3 ) = 4 · 9 − 9 · 3 + 4

( f + g ) ( 3 ) = 36 − 27 + 4 ( f + g ) ( 3 ) = 36 − 27 + 4

( f + g ) ( 3 ) = 13 ( f + g ) ( 3 ) = 13

Notice that we could have found ( f + g ) ( 3 ) ( f + g ) ( 3 ) by first finding the values of f ( 3 ) f ( 3 ) and g ( 3 ) g ( 3 ) separately and then adding the results.

Step 1 - Find f ( 3 ) f ( 3 ) .

f ( x ) = 3 x 2 − 5 x + 7 f ( x ) = 3 x 2 − 5 x + 7

f ( 3 ) = 3 ( 3 ) 2 − 5 ( 3 ) + 7 f ( 3 ) = 3 ( 3 ) 2 − 5 ( 3 ) + 7

f ( 3 ) = 19 f ( 3 ) = 19

Step 2 - Find g ( 3 ) g ( 3 ) .

g ( x ) = x 2 − 4 x − 3 g ( x ) = x 2 − 4 x − 3

g ( 3 ) = − 6 g ( 3 ) = − 6

Step 3 - Find ( f + g ) ( 3 ) ( f + g ) ( 3 ) .

( f + g ) ( x ) = f ( x ) + g ( x ) ( f + g ) ( x ) = f ( x ) + g ( x )

( f + g ) ( 3 ) = f ( 3 ) + g ( 3 ) ( f + g ) ( 3 ) = f ( 3 ) + g ( 3 )

Step 4 - Substitute f ( 3 ) = 1 9 f ( 3 ) = 1 9 and g ( 3 ) = − 6 g ( 3 ) = − 6 .

( f + g ) ( 3 ) = 19 + ( − 6 ) ( f + g ) ( 3 ) = 19 + ( − 6 )

( f + g ) ( 3 ) = 13 ( f + g ) ( 3 ) = 13

c. ( f − g ) ( x ) = f ( x ) − g ( x ) ( f − g ) ( x ) = f ( x ) − g ( x )

Step 1 - Substitute

f ( x ) = 3 x 2 − 5 x + 7 f ( x ) = 3 x 2 − 5 x + 7 and g ( x ) = x 2 − 4 x − 3 g ( x ) = x 2 − 4 x − 3 .

( f − g ) ( x ) = ( 3 x 2 − 5 x + 7 ) − ( x 2 − 4 x − 3 ) ( f − g ) ( x ) = ( 3 x 2 − 5 x + 7 ) − ( x 2 − 4 x − 3 )

Step 2 - Rewrite without the parentheses.

( f − g ) ( x ) = 3 x 2 − 5 x + 7 − x 2 + 4 x + 3 ( f − g ) ( x ) = 3 x 2 − 5 x + 7 − x 2 + 4 x + 3

Step 3 - Put like terms together.

( f − g ) ( x ) = 3 x 2 − x 2 − 5 x + 4 x + 7 + 3 ( f − g ) ( x ) = 3 x 2 − x 2 − 5 x + 4 x + 7 + 3

Step 4 - Combine like terms.

( f − g ) ( x ) = 2 x 2 − x + 10 ( f − g ) ( x ) = 2 x 2 − x + 10

d. ( f − g ) ( x ) = 2 x 2 − x + 10 ( f − g ) ( x ) = 2 x 2 − x + 10

Step 1 - To find ( f − g ) ( − 2 ) ( f − g ) ( − 2 ) , substitute x = − 2 x = − 2 .

( f − g ) ( − 2 ) = 2 ( − 2 ) 2 − ( − 2 ) + 10 ( f − g ) ( − 2 ) = 2 ( − 2 ) 2 − ( − 2 ) + 10

Step 2 - Simplify.

( f − g ) ( − 2 ) = 2 · 5 − ( − 2 ) + 10 ( f − g ) ( − 2 ) = 2 · 5 − ( − 2 ) + 10

( f − g ) ( − 2 ) = 20 ( f − g ) ( − 2 ) = 20

Try It: Adding and Subtracting Polynomial Functions

For functions f ( x ) = 2 x 2 − 4 x + 3 f ( x ) = 2 x 2 − 4 x + 3 and g ( x ) = x 2 − 2 x − 6 g ( x ) = x 2 − 2 x − 6 , find:

  1. ( f + g ) ( x ) ( f + g ) ( x )
  2. ( f + g ) ( 3 ) ( f + g ) ( 3 )
  3. ( f − g ) ( x ) ( f − g ) ( x )
  4. ( f − g ) ( − 2 ) ( f − g ) ( − 2 )

PreviousNext
Citation/Attribution

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:

    Access for free at https://openstax.org/books/algebra-1/pages/about-this-course

  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:

    Access for free at https://openstax.org/books/algebra-1/pages/about-this-course

Citation information

© May 21, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.