Video: Expanding and Simplifying Algebraic Expressions Involving Factorization of the Difference of Two Squares

Expand and simplify π‘ŽΒ³ βˆ’ 9𝑏³ + π‘Žπ‘ (π‘Ž βˆ’ 9𝑏), then factorise the result.

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Video Transcript

Expand and simplify [π‘Ž] cubed minus nine 𝑏 cubed plus π‘Žπ‘ multiplied by π‘Ž minus nine 𝑏, then factorise the result.

So if we take a look at this question, it actually has two parts. First, we’ll need to expand and simplify and in the second part, we need to actually factorise. So we’re gonna begin by expanding and simplifying. So when we first expand the parentheses, we’re gonna get π‘Ž squared 𝑏. So then our second term is gonna be negative nine π‘Žπ‘ squared. That’s cause if you have π‘Žπ‘ multiplied by negative nine 𝑏, you get negative nine π‘Žπ‘ squared.

So now, well we can’t actually simplify the expression any further cause we got no like terms. But what we can do is actually rearrange it so that we have common factors. So now what I’ve done is that I’ve actually rearranged it so that actually we have two terms next to each other β€” that’s small groups β€” and actually have common factors cause what we’re gonna do now is factor it using a method of grouping.

And to do that, I’m actually gonna factor each two terms separately. So first of all, we got π‘Ž cubed plus π‘Ž squared 𝑏. So therefore, we have π‘Ž squared outside the parentheses and inside the parentheses we have π‘Ž plus 𝑏 and that’s because π‘Ž squared multiplied by π‘Ž gives us π‘Ž cubed and π‘Ž squared multiplied by 𝑏 gives us π‘Ž squared 𝑏.

And then, for our next two terms or our next group, we have negative nine 𝑏 cubed minus nine π‘Žπ‘ squared. So therefore, a common factor of both of these is negative nine 𝑏 squared, which means that inside the parentheses we’re gonna have 𝑏 plus π‘Ž and that’s because negative nine 𝑏 squared multiplied by 𝑏 is negative nine 𝑏 cubed and negative nine 𝑏 squared multiplied by π‘Ž is negative nine π‘Žπ‘ squared.

So now, what we’re gonna do is a quick check to see if we’ve done the first stage correctly. And we have because actually we have got the same factor in both of our parentheses because π‘Ž plus 𝑏 is the same as 𝑏 plus π‘Ž. And actually if you haven’t got the same factor in each parenthesis and then you need to check to make sure you have done the rest of it correctly because if you haven’t got that, then you won’t be able to use this method.

So therefore, what we have now is π‘Ž squared minus nine 𝑏 squared. And that’s because we took the factors that are outside of our parentheses and made that into one of our factors. And then, this is multiplied by π‘Ž plus 𝑏 which is what was in both parentheses. So now, we actually have the difference of two squares and that’s because π‘Ž squared is a squared term. And we’ve got nine 𝑏 squared which is also a squared term because nine is a squared number and 𝑏 is obviously squared. And then, we have a negative between them.

And when we have the difference of two squares, we can actually factor this in a special way. And when we do that, we get π‘Ž plus three 𝑏 multiplied by π‘Ž minus three 𝑏 then multiplied by π‘Ž plus 𝑏.

And we actually found that because the roots of π‘Ž squared is equal to π‘Ž. So π‘Ž multiplied by π‘Ž is π‘Ž squared. And the root of nine 𝑏 squared is three 𝑏 because three 𝑏 multiplied by three 𝑏 gives us nine 𝑏 squared. And we have negative and positive. And that’s because when we actually expand the parentheses, what we’d actually be left with is two terms that are actually gonna cancel each other out because what we’d get is a positive three π‘Žπ‘ and then minus three π‘Žπ‘. So they cancel out.

So therefore, we can say that if we expand and simplify π‘Ž cubed minus nine 𝑏 cubed plus π‘Žπ‘ multiplied by π‘Ž minus nine 𝑏, then the fully factorised result is π‘Ž plus three 𝑏 multiplied by π‘Ž minus three 𝑏 multiplied by π‘Ž plus 𝑏.

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