Video: Multiplying Binomials

In this video, we will learn how to multiply two binomials using different methods such as FOIL (first, inner, outer, last), and the area method.

14:15

Video Transcript

Multiplying binomials. In this video, we will learn how to multiply two binomials using the FOIL method and the area method. Let’s start by reminding ourselves of what a binomial is. A binomial is an expression that contains the sum or difference of two terms. So, for example, three π‘₯ squared plus five π‘₯ is a binomial. And so is 𝑦 minus seven.

So if we have two binomials and we want to multiply them, what would that look like. For example, if we have the binomial π‘₯ plus two and we multiply it with binomial π‘₯ plus three, we can write this with a multiplication symbol in the middle. Notice that we have parentheses around each binomial so that we know that we multiply all of each one by all of the other one. We most commonly see this type of problem without the multiplication symbol between the two binomials. We may also see that multiplication of two binomials refer to as expanding, or expanding the product, or using the distributive property.

Now, let’s talk about how to actually multiply our two binomials. There are a number of different methods for this. But here, we’re going to use two different ones. The first method is the FOIL method. Here, the acronym FOIL represents First, Outer, Inner, and Last and refers to the position of the terms in the binomial. The second method is the area or grid method. And this method of multiplication is commonly seen for multiplying integers.

So let’s take a look at an example of multiplying two binomials and see how we would apply both of these methods.

Expand the product two π‘₯ plus one, 3π‘₯ minus two.

In this question, the word product indicates that we’ll be multiplying our two binomials. We’re going to use two different methods to demonstrate the expansion of these binomials. The first is the FOIL method. And the second is the area or grid method. So starting with the FOIL method, this means that we take our first, then the outer, then the inner, and then the last terms and multiply. Looking at our binomials, the first terms are the two π‘₯ and the three π‘₯. So we multiply those, writing two π‘₯ times three π‘₯.

The outer terms will be our two π‘₯ and our negative two. So we add two π‘₯ times negative two. The inner terms will be our plus one and our three π‘₯. Therefore, we add our one times our three π‘₯. And finally, the last two terms will be the plus one and the negative two. So we add plus one times negative two. We may not always need to write this line of workings. But it can be helpful to illustrate our method. And now, we work out the products. Two π‘₯ times three π‘₯ is six π‘₯ squared. Two π‘₯ times negative two is negative or minus four π‘₯ plus one times three π‘₯ is equivalent to plus three π‘₯. And finally, plus one times negative two is negative two.

The next thing to do at this stage of the question is to simplify by collecting any like terms. Here, we have two terms in π‘₯. So we can write this as six π‘₯ squared minus π‘₯ minus two, because our negative four π‘₯ plus three π‘₯ is equivalent to a negative π‘₯. Let’s see how we would get the same answer using the grid method. To set up our grid, we put one binomial horizontally and one binomial vertically. The first binomial will give us two π‘₯ and one horizontally. And the second binomial of three π‘₯ minus two can be split into the terms three π‘₯ and negative two.

We then calculate the product of each row and column, starting with three π‘₯ times two π‘₯. This will give us six π‘₯ squared. Next, three π‘₯ times one is three π‘₯. In our second row, negative two times two π‘₯ is negative four π‘₯. And finally, negative two times one is negative two. We then simplify by adding all four terms and seeing if we can collect any like terms. We start with six π‘₯ squared. We then have a plus three π‘₯ and a negative four π‘₯, which simplifies to negative π‘₯ and then our final term of negative two. So using either of our methods would give us the answer of six π‘₯ squared minus π‘₯ minus two.

In the next example, we’ll see a problem where the binomials include a number of different variable terms. However, we can use the same process and the same methods to get an answer for the expansion of these two binomials.

Use the distributive property to fully expand two π‘₯ plus 𝑦, π‘₯𝑦 minus two 𝑧.

In this question, the phrase β€œto use the distributive property” means that we can multiply the sum by multiplying each addend separately and then adding the products. We could use a number of methods to expand the brackets. Here, we’re going to look at two of those. The first method is the FOIL method, which is an acronym. We can recall that it stands for the First, the Outer, the Inner, and the Last terms of our binomials.

So with our two binomials set, we start by multiplying the first two terms, which is two π‘₯ times π‘₯𝑦. Next, our outer terms will be two π‘₯ times negative two 𝑧. It’s always worth taking extra care when we have the variable 𝑧, because we don’t want to confuse it with the digit two. Next, our inner terms will be 𝑦 multiplied by π‘₯𝑦. And finally, we add on the product of our last terms, which is 𝑦 times negative two 𝑧.

We now need to simplify our products. The coefficient of our first term is two. π‘₯ multiplied by π‘₯ will be π‘₯ squared. And we have our 𝑦. So the first term simplifies to two π‘₯ squared 𝑦. The coefficient of our second term is found by multiplying plus two by negative two, giving us negative four. And we then have π‘₯ times 𝑧. Our third term simplified will be π‘₯𝑦 squared since we have 𝑦 times 𝑦. And our final term will be negative two 𝑦𝑧. At this stage, we always check if we can simplify our answer by collecting any like terms. And as there are none here, this would be our final answer.

As an alternative method, we could use the area or grid method to expand our binomials. To set up our grid, we split up our binomials into their component terms with one on the row and one on the column. And it doesn’t matter which way round we put them. To fill in each cell in our grid, we multiply the row value by the column value.

So taking our first cell, we have π‘₯𝑦 times two π‘₯, which would be two π‘₯ squared 𝑦. For our next cell, we multiply π‘₯𝑦 by 𝑦 to give us π‘₯𝑦 squared. On our next row then, we have negative two 𝑧 times two π‘₯, which would give us negative four π‘₯𝑧. And our final grid value would be negative two 𝑦𝑧. To find our answer from this grid, we add our four products, giving us two π‘₯ squared 𝑦 minus four π‘₯𝑧 plus π‘₯𝑦 squared minus two 𝑦𝑧, which is the same as we achieved using the FOIL method.

Let’s now look at a question where we subtract the expansion of two binomials from another term.

Expand and simplify seven minus three minus 𝑦, 𝑦 plus two.

In this question, we can see we have two binomials which are multiplied together. We’re going to expand these binomials first and then subtract the answer from seven. We’re going to use the grid method to multiply these binomials. And notice the negative in front of them isn’t included.

Setting up our grid, we have three minus 𝑦 as our first binomial. Our second binomial can be split into the terms 𝑦 and two. And we can write that with or without the plus sign. It doesn’t matter which way round do we put our binomials. Filling in our grid, we start with 𝑦 times three, which is three 𝑦. Next, we have 𝑦 times negative 𝑦, which will give us negative 𝑦 squared. On the next row, we have two times three, which is six. And the final term is calculated by two times negative 𝑦, which is negative two 𝑦.

To take our answer from the grid then, we add together the four products we’ve just calculated. Starting with our largest exponent of 𝑦, we have negative 𝑦 squared. Next, we notice that we have two terms in 𝑦, which we can collect together. So three 𝑦 plus negative two 𝑦 will give us plus 𝑦. And then, we add on our final term from the grid which is plus six. So to answer the question then of seven minus three minus 𝑦, 𝑦 plus two, we replace what we’ve worked out in our expansion, giving us seven minus minus 𝑦 squared plus 𝑦 plus six.

It’s important to include all of these in parentheses since we want to subtract all of these terms. It can be helpful to write the next line where we distribute the negative across all the terms. So we have seven minus minus 𝑦 squared, which is plus 𝑦 squared. A negative times plus 𝑦 will give us negative 𝑦. And finally, a negative times plus six will give us negative six.

To simplify then, we check if there are any like terms. We can see that we have a seven and a negative six, which is equivalent to one, giving us a final answer of 𝑦 squared minus 𝑦 plus one. We could write these terms in any order. But by convention, we usually write them in decreasing exponent values. Here, we have our 𝑦 squared first, then our 𝑦 term, and then our constant.

In our next example, we’ll look at binomials that have higher exponent values. And we’re going to use the rules of exponents to help us solve the problem.

Find 𝐴𝐡 given 𝐴 equals five π‘₯ cubed minus three π‘₯ and 𝐡 equals negative six π‘₯ squared plus three π‘₯.

To find 𝐴𝐡 in this question, that means we need to multiply 𝐴 and 𝐡. So 𝐴𝐡 can be found by multiplying all of five π‘₯ cubed minus three π‘₯ by all of negative six π‘₯ squared plus three π‘₯. Since 𝐴 and 𝐡 are both binomials, we can use a method for multiplying or expanding two binomials. So setting up a multiplication or an area grid, we can split up our binomials into their separate terms, remembering to include the negatives wherever they appear.

To find the value in each cell, we take the product of the row and the column value. The coefficient of the term in the first cell can be found by multiplying negative six and five, which is negative 30. For our π‘₯ term, we multiply π‘₯ squared by π‘₯ cubed. And here, we may need to pause and remind ourselves of one of the rules of exponents. And that is that π‘₯ to the power of π‘Ž times π‘₯ to the power of 𝑏 is equal to π‘₯ to the power of π‘Ž plus 𝑏. So our calculation π‘₯ squared times π‘₯ cubed is equal to π‘₯ to the power of five. Therefore, the first term in our grid is negative 30π‘₯ to the power of five.

In the next cell of our grid, we have the coefficient, which will be equal to negative six times negative three, which will be 18 since two negative values multiplied will always give a positive value. For the π‘₯-value then, we have π‘₯ squared times π‘₯, which is equivalent to π‘₯ squared times π‘₯ to the power of one, giving us π‘₯ cubed. And so the whole term will be 18π‘₯ cubed. In the next row, we have three π‘₯ times five π‘₯ cubed, which will give us 15π‘₯ to the fourth power. The final term in our grid is calculated by three π‘₯ times negative three π‘₯. And that’s equal to negative nine π‘₯ squared.

To find the answer using the grid then, we add our four products together, giving us negative 30π‘₯ to the fifth power plus 15π‘₯ to the fourth power plus 18π‘₯ cubed minus nine π‘₯ squared. As there are no like terms, we can’t simplify this any further. So this is our final answer for 𝐴𝐡.

Before we summarize what we’ve learned in this video, it may be useful to note that multiplying two binomials is part of the method we use for multiplying three or more binomials. It’s outside the scope of this video. But as a rough guide, if we were multiplying three binomials, for example, π‘₯ plus three times π‘₯ plus seven times two π‘₯ minus five, we would expand two of the binomials, for example, to give us π‘₯ squared plus 10π‘₯ plus 21 and then multiply that by two π‘₯ minus five. Here, we could use the grid method to find our final answer for this.

So in summary then, in this video, we learned how to multiply two binomials using two methods, the FOIL method and the area or grid method. We need to be careful when multiplying negative values. And finally, we may need to use an important exponent rule when multiplying binomials. That is that π‘₯ to the power of π‘Ž times π‘₯ to the power of 𝑏 is equal to π‘₯ to the power of π‘Ž plus 𝑏.

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