Multiplying Binomials
Multiplying binomials (a polynomial with two terms), could be thought of as th key to what is perhaps the single most important thing to know. With a real understanding of the patterns that come from two binomials being multiplied comes access to many other, very important skills.
The most important skill to be learned from the Polynomials Unit is factoring. Factoring is like the opposite of distributing. So, just like it would be very difficult to learn division without first knowing how to multiply, factoring is really difficult if you do not understand what happens when you distribute. It is failure to understand this skill that creates a gateway, of sorts, into higher mathematics.
Carefully read, and annotate, the information in the first tab. Then watch the videos and try the practice problems.
When opportunity knocks, it is too late to prepare.
NOTE: This is a skill you must possess to the extent that you can do it backwards! Factoring is taking a polynomial and un-multiplying it, in essence. In order to do that, you must really understand multiplication. It is not at all unlike not knowing how to divide without first knowing the multiplication rules.
There are short-cuts for multiplying binomials that we will learn soon, especially the short-cut called F.O.I.L. However, that short-cut only works for multiplying binomials together. We will also need to know how to multiply any two (or more polynomials) together, so learning what is really happening is of utmost importance.
Not only will you have to un-do multiplication (factoring), you’ll also need to know how to multiply any polynomials together.
Let’s start with something familiar, and build on that.
3(4x)
This is notation used for multiplying. It says to multiply 3 and 4x, which yields 12x. This is the product of two monomials. Let’s see a monomial multiplying with a binomial.
3(x2 – 2x)
Distributing is just multiplication. Here we must multiply the 3 and both terms of the following binomial. Do not think of the binomial as two separate things, the x2 – 2x are not two values, but a single value. It cannot be written as a single term because we do not know the value of x. Regardless, there are three of them, so we can distribute the three, giving us the following.
3(x2 – 2x) = 3×x2 – 3×2x
We write this as: 3x2 – 6x, of course.
Now let’s take a look at two binomials.
(x + 1)(x2 – 2x)
This is the product of two numbers, each number with two parts (binomials). In order to carry out the calculation we need to distribute both parts of the first number to the second. That means the x from the first binomial times (x2 – 2x) and +1 times (x2 – 2x).
If we distribute each of those we get the following.
As is usually the case, after distribution you must collect (combine) like terms. Here we have a -2x2, and a +1x2, which make –x2.
(x + 1)(x2 – 2x) = x3 – x2 – 2x
The main issue students run into here is paying attention to details. Watch out for mistakes involving:
- Sign Errors
- Exponents
- Combining Like Terms
The most common conceptual mistake that students make is when a binomial is squared. Let’s take a look at this carefully.
The only other issue comes into play when dealing with exponents and variables. At this point you should be proficient with manipulating both, but let’s take a quick review anyway, just to be sure.
If we set this one up, we get:
Distributing each term, we get:
Now, if we combine like terms we get:
Just remember to be careful and that exponents are a notation for repeated multiplication. The term a2 means a × a. Then (a2)2 would mean a2 × a2, which is a × a × a × a.
Note: This product of two binomials has a special name, it is called Difference of Squares. The first and second term are perfect squares and it is something you’ll need to note when you begin factoring.
While there are strong patterns with a lot of the problems you’ll see involving the multiplication of binomials, don’t feel every product must fit the pattern. Let’s see an example that has an a-typical product.
Summary: Setting up the distributing is how these products are calculated without short-cut. This is also how it happens when multiplying larger polynomials.
Coming Soon