Video: Squaring Binomials

In this video, we will learn how to expand the square of the difference or sum of two monomials.

17:39

Video Transcript

In this video, we will learn how to expand the square of the difference or sum of two monomials. A monomial is just a single algebraic term such as π‘₯, two, or 𝑦. The difference or sum of two monomials gives an expression such as π‘₯ plus two or π‘Ž minus 𝑏, which is referred to as a binomial. Usually, when we see words with bi- at the start, there is some link with the number two, for example, bicycles which have two wheels. So, a binomial is simply an algebraic expression with two terms.

Squaring a binomial means to multiply a binomial by itself. So, for example, we’re looking for the result of π‘₯ plus two multiplied by π‘₯ plus two, which we can write as π‘₯ plus two all squared. This is an important algebraic skill because it allows us to manipulate more complex algebraic expressions and will often be required in problems in other areas of mathematics, such as problems involving area of two-dimensional shapes. In this video, we’ll look at a variety of different methods for squaring binomials, so you can pick the one that makes the most sense to you.

So, let’s begin by looking at our first example.

Expand π‘š plus four squared.

So, we have a binomial, π‘š plus four. And the first really important thing to notice is that we are squaring this entire binomial. This means that we’re taking the expression π‘š plus four and multiplying it by itself. So, another way of writing this would be as π‘š plus four multiplied by π‘š plus four. The word expand is another way of saying distribute. So, we want to expand the brackets or distribute the parentheses. We need to multiply both terms in the first binomial by both terms in the second.

Let’s look at a physical interpretation of this first using area. Suppose we have a length of π‘š plus four units. We can break this down into a length of π‘š units and a length of four units. When we multiply π‘š plus four by itself, this is equivalent to finding an expression for the area of a square with side lengths of π‘š plus four units. We can divide this area up into four smaller regions and find each of their individual areas. They’re each either rectangles or squares, so we find their areas by multiplying their dimensions together.

The first region has an area of π‘š multiplied by π‘š which is π‘š squared. The second region has an area of π‘š multiplied by four, which is four π‘š. The third region has an area of four multiplied by π‘š, which is another lot of four π‘š. And the final region has an area of four multiplied by four, which is 16. The total area can be found by summing the four individual areas, giving π‘š squared plus four π‘š plus four π‘š plus 16. Now, we must remember to simplify this expression by grouping like terms. And the only like terms are those in the center of our expansion, plus four π‘š plus four π‘š, which makes positive eight π‘š. Our expression, therefore, simplifies to π‘š squared plus eight π‘š plus 16. And this is the expanded form of π‘š plus four all squared.

Now, notice that, at this stage here, we have four terms in our expansion, and the middle two are exactly the same. Once we’ve grouped the like terms, we have three terms in our expansion. This will always be the case when squaring a binomial. So, if you find you have a different number of terms or you don’t have two terms which are exactly the same, then something is gone wrong. So, you need to check your work. So, that’s one approach to consider this expansion as the area of a square with side lengths of π‘š plus four units. This is also sometimes referred to as a grid method because it follows the same principles as the grid method for multiplication.

A second method we could consider would be to use the distributive property of multiplication. If we’re multiplying π‘š plus four by π‘š plus four, then we can write this as π‘š multiplied by π‘š plus four plus four multiplied by π‘š plus four. So, we’ve distributed the terms in our first binomial over the second. We then have to multiply out or expand or distribute each of these single sets of parentheses. π‘š multiplied by π‘š gives π‘š squared. π‘š multiplied by four gives four π‘š. We then have four multiplied by π‘š, which gives another lot of four π‘š, and finally four multiplied by four, which gives 16.

Our expression is now identical to the one that we had at this stage here in our first method. And the only remaining step is to group the like terms in the center of our expansion. As before, we find that π‘š plus four all squared is equal to π‘š squared plus eight π‘š plus 16. So, that’s two possible methods, but there is a third. And this one is perhaps the most popular. It’s called the FOIL method. And the word FOIL is an acronym, meaning that each of its letters stands for something. It’s just a way of ensuring that we multiply all of the correct pairs of terms together, and we don’t miss any out.

The letter F in the word FOIL stands for first. So, this means we multiply the first term in the first binomial by the first term in the second. That’s π‘š multiplied by π‘š, which is, of course, π‘š squared. The letter O stands for outers or outside. So, we multiply the terms that are on the outside of our expansion. That’s the π‘š in the first binomial and the four in the second, π‘š multiplied by four, which is four π‘š. You’ve probably guessed that the I stands for inners or inside, so we multiply together the terms in the inside of the expansion. That’s the four in the first binomial and the π‘š in the second, four multiplied by π‘š, which gives four π‘š. And finally, L, which stands for lasts. We multiply together the last term in each binomial. That’s the four in the first binomial and the four in the second, four multiplied by four, which is 16.

As before, we should always have four terms in our expansion at this point. And we have two like terms in the center of the expansion, which can be combined. Using all three methods then, we’ve arrived at the same result. When we expand π‘š plus four all squared, which means to multiply the binomial π‘š plus four by itself, we get the answer π‘š squared plus eight π‘š plus 16. It is usual, although not essential, to write the terms in our expansion in this order, that is, decreasing powers of π‘š. So, we have our π‘š squared term first, then our π‘š term, and then finally the constant term.

Before we move on to looking at other examples, let’s address a really common mistake. We saw in our first example that the correct expansion of π‘š plus four all squared is π‘š squared plus eight π‘š plus 16. Unfortunately, the mistake that gets made so very often is to think that when we square a binomial such as π‘š plus four, we simply square each term, giving π‘š squared plus four squared. This would, of course, simplify to π‘š squared plus 16. But as we can see, this is not the same as the answer we’ve already found. We’re missing that term of positive eight π‘š.

This is equivalent to forgetting about two regions of the area in the first method we used. We have the π‘š squared and the 16. But we don’t have the two areas that correspond to multiplying π‘š by four each time. But this is such a common mistake, even made by students of advanced mathematics when they’re not thinking properly. So, make sure that you don’t make it. Another common mistake worth mentioning would be to think that π‘š multiplied by π‘š gives two π‘š. Remember, though, that we are multiplying the terms not adding, so the correct answer is π‘š squared.

Let’s now go on and consider some more examples.

Expand negative π‘₯ plus two 𝑦 all squared.

In this question then, we have a binomial expression, negative π‘₯ plus two 𝑦. And we are squaring it. That means we’re multiplying this binomial by itself. So, we’re looking for the result of multiplying negative π‘₯ plus two 𝑦 by negative π‘₯ plus two 𝑦. There are numerous different methods that we can use. In this question, I’m going to choose to use the FOIL method. Now, we just need to be a little bit careful because one of the terms in our binomial is negative. And we don’t want to let this trip us up. We need to be really careful with the signs when we’re multiplying each pair of terms together.

So, F, remember, stands for firsts. We multiply the first term in each binomial together. That’s negative π‘₯ multiplied by negative π‘₯, which gives π‘₯ squared. Remember, a negative multiplied by a negative gives a positive. Then, the letter O stands for outers or outsides. We multiply the terms on the outside of our expansion. That’s the negative π‘₯ in the first binomial by the positive two 𝑦 in the second, giving negative two π‘₯𝑦. I stands for inners or inside. So, we multiply the terms in the center of our expansion. That’s the two 𝑦 in the first binomial and the negative π‘₯ in the second, giving another lot of negative two π‘₯𝑦. Finally, the letter L stands for lasts, so we multiply the last term in each binomial together. That’s positive two 𝑦 multiplied by positive two 𝑦, which is four 𝑦 squared.

So, after completing all four of our multiplications, we now have four terms in our expansion, π‘₯ squared minus two π‘₯𝑦 minus two π‘₯𝑦 plus four 𝑦 squared. Remember, there should always be two identical terms in the center of our expansion. And indeed, there are. We have negative two π‘₯𝑦 minus another lot of two π‘₯𝑦. We can therefore simplify our expansion by grouping like terms, and we have our final answer to the problem. The simplified expansion of negative π‘₯ plus two 𝑦 all squared is π‘₯ squared minus four π‘₯𝑦 plus four 𝑦 squared.

Great, we’re gaining some confidence in applying these methods now. Let’s go on and look at another example which is slightly more complex, but only because the terms in the binomial we’re squaring are not quite so straightforward.

Expand four π‘₯ squared plus three over two 𝑦 squared all squared.

Now, at first glance, this question looks relatively complicated. But, in fact, what we have is a binomial. That’s the sum of two algebraic terms. And then we are squaring it. That means we’re multiplying this binomial by itself. So, we’re looking for an algebraic expression for the result of multiplying four π‘₯ squared plus three over two 𝑦 squared by four π‘₯ squared plus three over two 𝑦 squared. Although this may look complicated, the standard processes we follow are exactly the same. Let’s use the distributive method. So, we take the two terms in our first binomial and we distribute them over the second, giving four π‘₯ squared multiplied by four π‘₯ squared plus three over two 𝑦 squared plus three over two 𝑦 squared multiplied by four π‘₯ squared plus three over two 𝑦 squared.

We now have two single parentheses to distribute. Four π‘₯ squared multiplied by four π‘₯ squared gives 16π‘₯ to the fourth power because four multiplied by four is 16. And π‘₯ squared multiplied by π‘₯ squared is π‘₯ to the fourth power. Remember, we add the exponents. We then have four π‘₯ squared multiplied by three over two 𝑦 squared. And we can simplify the coefficient here in the moment. For now, we have four multiplied by three over two π‘₯ squared 𝑦 squared.

That’s the first set of parentheses expanded. Now, let’s consider the second. We have positive three over two 𝑦 squared multiplied by four π‘₯ squared. And again, we’ll simplify the coefficient here in a moment. For now, we’ll just write it as three over two multiplied by four π‘₯ squared 𝑦 squared. Finally, we have three over two 𝑦 squared multiplied by three over two 𝑦 squared, which we’ll write as three over two multiplied by three over two 𝑦 to the fourth power. Again, remember, we add the exponents.

Notice that we have four terms in our expansion at this stage and the two central terms are identical to one another, although the four and the three over two have been written in the opposite order. Now, we can simplify. 16π‘₯ to the fourth power requires no simplification. In our second term, we have four multiplied by three over two. So, we can cancel a factor of two in the numerator and denominator to give two multiplied by three over one, which is simply six.

Our second term is therefore six π‘₯ squared 𝑦 squared. By exactly the same reasoning, our third term is also six π‘₯ squared 𝑦 squared. And to simplify the coefficient in our final term, we multiply the numerators together, giving nine, and multiply the denominators, giving four. So, our final term is nine over four 𝑦 to the fourth power.

All that remains is to simplify our expansion by grouping the like terms in the center. Six π‘₯ squared 𝑦 squared plus six π‘₯ squared 𝑦 squared is 12π‘₯ squared 𝑦 squared. So, we have our simplified expansion; four π‘₯ squared plus three over two 𝑦 squared all squared is equal to 16π‘₯ to the fourth power plus 12π‘₯ squared 𝑦 squared plus nine over four 𝑦 to the fourth power.

So, whilst that example may have looked a little more complicated, the processes we use are exactly the same. In our final example, we’ll see how to apply these methods to a question involving problem solving.

If π‘₯ plus 𝑦 all squared is equal to 100 and π‘₯𝑦 equals 20, what is the value of π‘₯ squared plus 𝑦 squared?

So, we’ve been given two pieces of information about these numbers π‘₯ and 𝑦 and asked to use them to determine the value of π‘₯ squared plus 𝑦 squared. Now, your first thought may be that π‘₯ plus 𝑦 all squared is just equal to π‘₯ squared plus 𝑦 squared. In which case, the value we’re looking for is the value given in the question; it’s 100. But if this is the case, why have we also been given the value of π‘₯𝑦?

In fact, if we were to answer the question this way, we’d have made one of the most common mistakes in mathematics because we’ve incorrectly expanded the binomial. Remember that π‘₯ plus 𝑦 all squared means π‘₯ plus 𝑦 multiplied by π‘₯ plus 𝑦. So, we are, in fact, multiplying a binomial by itself, not just squaring each of the individual terms.

Let’s see what happens if we correctly expand π‘₯ plus 𝑦 all squared. Using the FOIL method which, remember, stands for firsts, outers, inners, lasts, this gives π‘₯ squared plus π‘₯𝑦 plus π‘₯𝑦 plus 𝑦 squared, which simplifies to π‘₯ squared plus two π‘₯𝑦 plus 𝑦 squared. What we now have is an equation connecting π‘₯ plus 𝑦 all squared, whose value we know, π‘₯𝑦, whose value we know, and π‘₯ squared plus 𝑦 squared, whose value we wish to calculate.

Substituting 100 for π‘₯ plus 𝑦 all squared and 20 for π‘₯𝑦, we have 100 equals π‘₯ squared plus 𝑦 squared plus two multiplied by 20. That simplifies to 100 equals π‘₯ squared plus 𝑦 squared plus 40. And subtracting 40 from each side, we find that π‘₯ squared plus 𝑦 squared is equal to 60. So, we’ve solved the problem. By correctly expanding the binomial π‘₯ plus 𝑦 all squared and then substituting the values given in the question, we found that π‘₯ squared plus 𝑦 squared is equal to 60.

Let’s now review the key points we’ve seen in this video. One really useful method for squaring binomials is the FOIL method, where each letter stands for a different pair of terms we need to multiply: firsts, outers, inners, lasts. The grid or area method can be helpful for visualizing why we have to multiply each term in the first binomial by each term in the second. We picture this as finding an expression for the area of the square whose side lengths are each equal to the binomial we’re squaring. We will always have four terms in our initial expansion. The two central terms will be identical. And we then simplify by collecting like terms to give an algebraic expression with three terms.

The general form when expanding a binomial of the form π‘₯ plus 𝑦 all squared is π‘₯ squared plus two π‘₯𝑦 plus 𝑦 squared. And finally, an incredibly common misconception which we must do everything in our power to avoid: when we square a binomial, we don’t simply square the individual terms. π‘₯ plus 𝑦 all squared is not equal to π‘₯ squared plus 𝑦 squared. Instead, we need to use one of the formal methods for squaring a binomial such as the FOIL method, the grid or area method, or the distributive property.

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