Video: AQA GCSE Mathematics Higher Tier Pack 3 β€’ Paper 1 β€’ Question 16

Expand and simplify (2π‘₯ βˆ’ 3𝑦)Β³

03:37

Video Transcript

Expand and simplify two π‘₯ minus three 𝑦 all to the power of three.

Since we have two π‘₯ minus three 𝑦 to the power of three, that means we’re multiplying it to itself three times. So it’s equal to two π‘₯ minus three 𝑦 times two π‘₯ minus three 𝑦 times two π‘₯ minus three 𝑦. In order to expand, we first need to multiply the first two set of brackets together. And we can use the FOIL method.

So to multiply two π‘₯ minus three 𝑦 times two π‘₯ minus three 𝑦, we multiply the first terms together. That’s what F stands for. And two π‘₯ times two π‘₯ is four π‘₯ squared. Then we multiply the outside terms together: two π‘₯ times negative three 𝑦, negative six π‘₯𝑦. Then we multiply the inside terms together: negative three 𝑦 times two π‘₯, which is negative six π‘₯𝑦.

Now notice we have negative three 𝑦 times two π‘₯. Negative three times two is negative six. And then 𝑦 times π‘₯ we could write it as 𝑦π‘₯. But it’s good to keep the variables in alphabetical order. That way, it is easier to combine like terms, to recognize which terms are alike. And now we multiply the last terms together: negative three 𝑦 times negative three 𝑦, which is positive nine 𝑦 squared.

Now notice the middle terms are alike. Negative six π‘₯𝑦 minus six π‘₯𝑦 is negative 12π‘₯𝑦. So now we’ve replaced two π‘₯ minus three 𝑦 times two π‘₯ minus three 𝑦 with four π‘₯ squared minus 12π‘₯𝑦 plus nine 𝑦 squared. Now let’s multiply these polynomials together.

So let’s begin expanding. We need to take each of the blue terms and multiply them to each of the pink terms. So we have four π‘₯ squared times two π‘₯. That’s eight π‘₯ cubed. And now we take four π‘₯ squared times negative three 𝑦, which is negative 12π‘₯ squared 𝑦. Now we take negative 12π‘₯𝑦 times two π‘₯, which is negative 24π‘₯ squared 𝑦, and then negative 12π‘₯𝑦 times negative three 𝑦, which is positive 36π‘₯𝑦 squared. Now we take nine 𝑦 squared times two π‘₯, which is positive 18π‘₯𝑦 squared, and lastly nine 𝑦 squared times negative three 𝑦. And we have negative 27𝑦 to the power of three.

So now it’s time to combine like terms. Notice we have variables π‘₯ and 𝑦. π‘₯ comes first alphabetically. So we will have its highest power first and go down from there for the π‘₯. So the highest power that we have for π‘₯ is three, π‘₯ to the power of three. And there’s only one term like that, eight π‘₯ to the power of three.

Now we can combine negative 12π‘₯ squared 𝑦 and negative 24π‘₯ squared 𝑦. It’s π‘₯ to the power of two. So we have negative 36π‘₯ squared 𝑦. Now we have 36π‘₯𝑦 squared plus 18π‘₯𝑦 squared. That gives us 54π‘₯𝑦 squared, π‘₯ to the power of one. And then lastly, we have π‘₯ to the power of zero. There’s no π‘₯. So we have minus 27𝑦 to the power of three, making our final answer eight π‘₯ to the power of three minus 36π‘₯ squared 𝑦 plus 54π‘₯𝑦 squared minus 27𝑦 to the power of three.

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