Question Video: Expanding an Algebraic Expression Using Algebraic Identities Mathematics

Video Transcript

Expand four 𝑥 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 𝑥 squared plus three over two 𝑦 squared by four 𝑥 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 𝑥 squared multiplied by four 𝑥 squared plus three over two 𝑦 squared plus three over two 𝑦 squared multiplied by four 𝑥 squared plus three over two 𝑦 squared.

We now have two single parentheses to distribute. Four 𝑥 squared multiplied by four 𝑥 squared gives 16𝑥 to the fourth power because four multiplied by four is 16. And 𝑥 squared multiplied by 𝑥 squared is 𝑥 to the fourth power. Remember, we add the exponents. We then have four 𝑥 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 𝑥 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 𝑥 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 𝑥 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. 16𝑥 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 𝑥 squared 𝑦 squared. By exactly the same reasoning, our third term is also six 𝑥 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 𝑥 squared 𝑦 squared plus six 𝑥 squared 𝑦 squared is 12𝑥 squared 𝑦 squared. So, we have our simplified expansion; four 𝑥 squared plus three over two 𝑦 squared all squared is equal to 16𝑥 to the fourth power plus 12𝑥 squared 𝑦 squared plus nine over four 𝑦 to the fourth power.