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.
