A binomial is a polynomial with two terms being summed. Below are some examples of what constitutes a binomial:
- 4x2 - 1
- -⅓x5 + 5x3
- 2(x + 1) = 2x + 2
- (x + 1)(x - 1) = x2 - 1
The last example is is worth noting because binomials of the form
x2 - y2
can be factored as (x + y)(x - y). Recognizing binomials of this form can save you time when working on algebra problems because this form is fairly common.
Performing operations such as addition/subtraction, multiplication, and exponentiation of binomials requires us to combine like terms and use the distributive property, as described below.
Adding and subtracting binomials
Adding and subtracting binomials requires us to combine like terms. This means that we can only add terms that have the same exponent. For example, we can't add x2 and x, because they have different exponents, but we can combine x + 3x = 4x or 7x2 - 3x2 = 4x2.
Add x + 1 and 2x2 - 3x:
(x + 1) + (2x2 - 3x)
2x2 + x - 3x + 1
2x2 - 2x + 1
Multiplying binomials involves the use of the distributive property, as well as combining like terms to simplify the resulting expression. The distributive property states:
a(b + c) = (a × b) + (a × c)
When using the distributive property on an expression of the form above, the term outside the parenthesis is multiplied by each of the terms inside the parenthesis. When multiplying binomials, which by definition have two terms each, we use the same property, except that we must remember to multiply all the terms such that:
(a + b)(c + d) = (a × c) + (a × d) + (b × c) + (b × d)
This is just using the distributive property. One method for remembering how to multiply terms is the FOIL method. FOIL stands for "First Outer Inner Last," and is a mnemonic for remembering which terms need to be multiplied. The order does not particularly matter as long as all the appropriate terms are multiplied and added.
Another way to look at multiplying binomials is breaking one of the factors up:
|(a + b)(c + d)||= a(c + d) + b(c + d)|
|= (a × c) + (a × d) + (b × c) + (b × d)|
This is the same result (and really the same process) as the FOIL method, just presented differently. Use whatever method you are most comfortable with. As long all of the terms are multiplied appropriately, the answer will be the same.
One other thing we need to use when multiplying binomials, or variables in general, is exponent rules. In the case of multiplying polynomials, the rule you'll most commonly need to use concerns how to handle exponents when multiplying like bases. When multiplying like bases, add the exponents.
Multiply (x + 1)(x2 - 4):
|(x + 1)(x2 - 4)||= (x1)(x2) + (x)(-4) + (1)(x2) + (1)(-4)|
|= x3 -4x + x2 - 4|
|= x3 + x2 - 4x - 4|
A variable with no exponent written implies an exponent of 1. Typically this 1 is not written, but for this example we explicitly wrote the exponent of 1 on the "x" just to show that we are adding an exponent of 1 and 2 to get x3 when multiplying x1x2.
Expanding binomials raised to an exponent
Handling exponents on binomials can be done by just multiplying the terms using the distributive property, with algorithms such as the binomial theorem, or using Pascal's triangle. Refer to the mentioned pages for more information on using the binomial theorem or Pascal's triangle. This page will just show an example of using of the distributive property, which works for smaller exponents, but quickly gets tedious.
Expand (x + 1)3:
(x + 1)3 = (x + 1)(x + 1)(x + 1)
We can just break this multiplication problem into two parts, just multiplying (x + 1)(x + 1) first, then multiplying the result by the third (x + 1). Using FOIL or whatever method you prefer:
|(x + 1)(x + 1)||= x2 + x + x + 1|
|= x2 + 2x + 1|
To multiply (x2 + 2x + 1)(x + 1), we just extend the distributive property and ensure we multiply each term by (x + 1).
|x2(x + 1) + 2x(x + 1) + 1(x + 1)||= x3 + x2 + 2x2 + 2x + x + 1|
|= x3 + 3x2 + 3x + 1|
This quickly becomes cumbersome with a higher exponent, but there is a pattern that can be seen in Pascal's triangle that allows us to use a formulaic approach to expanding binomials using the binomial theorem. Once we learn how to use the binomial theorem, expanding binomials raised to the nth power, even if n is very large, is relatively simple when compared to having to use the distributive property repeatedly.