Addition and Subtraction of Polynomials



Once you understand what like terms are, then you can begin adding and subtracting the terms of polynomials. This will allow you to simplify somewhat difficult-to-understand polynomials into much more compact forms, a process known as simplification. Some of this you will already understand intuitively from dealing with the process of solving simple Algebra 1 equations.

Consider the polynomial 3a + 4a. Intuitively, we understand that this is 7a. We can prove it, however, using the distributive property:

3a + 4a
a(3+4)
a(7)
7a

What you will notice is that 3a and 4a are like terms. Note that if we had 3a + 4b instead, we would not be able to simplify the polynomial at all. This idea gives us a simple rule to remember when we simplify the terms of polynomials: You can only add and subtract like terms.

For a more difficult example, consider what would happen if we want to simplify the following polynomial:

2x²y³ + 3xy³ – 4x² – 5xy³

First, we need to find out if there are any like terms. In this example, 3xy³ and -5xy³ are the only like terms, since 2x²y³ and -4x² don’t have any other terms with the same variable component. We should rearrange the order of the terms to put the like terms beside of each other, like the following:

2x²y³ + 3xy³ – 4x² – 5xy³
3xy³ – 5xy³ + 2x²y³ – 4x²

With the like terms beside of each other, we can apply the distributive property to add or subtract them:

3xy³ – 5xy³ + 2x²y³ – 4x²
xy³(3-5) + 2x²y³ – 4x²
xy³(-2) + 2x²y³ – 4x²
-2xy³ + 2x²y³ – 4x²

And we’re finished. Intuitively, you can skip the distribution steps and just go straight to identifying the like terms and doing the addition or subtraction if you like. You will usually want to rewrite the polynomial first, however, continuing to group the like terms together to make it more difficult for you to make simple mistakes. For example:

2ab – 3ab³ – 4a³b² + 5ab – 6a³b²
2ab + 5ab – 4a³b² – 6a³b² – 3ab³
7ab – 10a³b² – 3ab³

This is much easier to understand once you’ve done a few practice problems. Like many topics and operations in Algebra 1, this is much less difficult to do than it first seems.