Question Video: Expanding and Factorising Algebraic Expressions Involving Perfect-Square Trinomials | Nagwa Question Video: Expanding and Factorising Algebraic Expressions Involving Perfect-Square Trinomials | Nagwa

Question Video: Expanding and Factorising Algebraic Expressions Involving Perfect-Square Trinomials Mathematics • Second Year of Preparatory School

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Expand π‘Ž(π‘Ž βˆ’ 16𝑏) + 64𝑏² βˆ’ 81, and then factorise the result completely.

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Video Transcript

Expand π‘Ž multiplied by π‘Ž minus 16𝑏 plus 64𝑏 squared minus 81, and then factorise the result completely.

This question actually takes the form of two parts. The first part is where we’re gonna expand and in the second part where we’re gonna factor. So what we’re gonna do deal with is first part, which is expand. So what we’re gonna do is expand the parentheses and then actually simplify where we can before we move on to the second part. So the first term that we’re gonna have is π‘Ž squared, and that’s because we got π‘Ž multiplied by π‘Ž, which gives us π‘Ž squared. And then we’re gonna minus 16 π‘Žπ‘, and we get this because π‘Ž multiplied by negative 16𝑏 is negative 16π‘Žπ‘. And then we have our plus 64𝑏 squared and minus 81.

So now we take a look and can see well actually there aren’t any like terms. So we actually can’t simplify this any further, and we’ve done the first part cause we’ve actually expanded the expression. So now what we want to do is actually to move on and actually look at how we’d factor it. Well if we’re gonna look to actually factor the expression that we’ve got, we’re actually going to start to use something called a perfect square trinomial, because actually three of our terms can help us to do that. Because if we actually look at the part which is π‘Ž squared minus 16π‘Žπ‘ plus 64𝑏 squared, then we’ve got a relationship that we know when we’re looking at perfect square trinomials, because if we’ve got something in the form π‘Ž squared plus two multiplied by π‘Ž multiplied by 𝑏 plus 𝑏 squared, then when you factor this, you’re gonna get π‘Ž plus 𝑏 multiplied by π‘Ž plus 𝑏.

Or if you have π‘Ž squared minus two multiplied by π‘Ž multiplied by 𝑏 plus 𝑏 squared, this is gonna be equal to π‘Ž minus 𝑏 multiplied by π‘Ž minus 𝑏. But how does this relate to the terms that I just mentioned in our expression? So first of all, we look at our relationship, and it says well something squared, so capital 𝐴 squared. Well in our case, what that’s gonna be is actually just π‘Ž squared. So that first term is nice and simple because we’ve actually already got it in the form that is asked for.

But when we look at the final term, it’s going to be slightly different cause actually we’ve got 64𝑏 squared. And what we say when we look at our relationship is that it’s again something squared, this time we called it capital 𝐡, but it means something else squared. Well 64𝑏 squared can actually be written as something squared because that something can be eight 𝑏 because eight multiplied by eight is 64 and 𝑏 multiplied by 𝑏 is 𝑏 squared. So we can say that it’s eight 𝑏 and then all squared.

So now if we actually look at the middle term, we can see that we have negative and then we’ve got two multiplied by π‘Ž multiplied by eight 𝑏 that we can actually see that we’re gonna get our negative 16π‘Žπ‘. Okay, great! So what we can actually see is we’ve actually now got in the form of our second relationship. So therefore, in that case, it means that we now know how to factor. Because following the rule above, we’d actually have π‘Ž minus 𝑏 multiplied by π‘Ž minus 𝑏. So in our case, that’s our π‘Ž minus eight 𝑏 multiplied by π‘Ž minus eight 𝑏.

Okay, great! So we’ve actually factored the first three terms. So we’re now at the stage where we have π‘Ž minus 𝑏 multiplied by π‘Ž minus eight 𝑏 minus 81. But is this fully factored? Can we do anything else? Well if we’re actually dealing with squares, let’s have a look at the final part, which is negative 81. Well if you think about negative 81, this is the same as negative and then nine squared. So if we actually think about that, then that would be negative and then in parentheses nine squared, so negative 81.

So therefore, we think okay, great! Well let’s add this into our parentheses as part of the factoring. So I’ve done that, so we’ve got π‘Ž minus eight 𝑏, and I’ve got a nine in the first parentheses, and I’ve got π‘Ž minus eight 𝑏 and again nine in the second parentheses. However, we don’t want the nine to actually affect any of the terms. So how could we deal with this? Well the way that we actually deal with this is by having positive nine and negative nine, and that’s because positive nine multiplied by negative nine is going to give us our negative 81 that we need to actually fulfill the original expression.

However, because we have positive nine and negative nine, any other multiplication that takes place are actually going to cancel themselves out. And I’ll show this by checking our answer using expansion. So first of all, we can have π‘Ž multiplied by π‘Ž, which gives us π‘Ž squared. And then we’re going to have π‘Ž multiplied by negative eight 𝑏, which gives us negative eight π‘Žπ‘. And then we’ll have π‘Ž multiplied by negative nine, which gives us negative nine π‘Ž.

And then we move on to the second term in the first parentheses. So we have negative eight 𝑏 multiplied by π‘Ž, which gives us negative eight π‘Žπ‘. And then we have negative eight 𝑏 multiplied by negative eight 𝑏, which gives is positive 64𝑏 squared. And then we have negative eight 𝑏 multiplied by negative nine, which gives us positive 72𝑏. So then we move on to the final term in the first parentheses. So we have positive nine multiplied by π‘Ž, which gives us positive nine π‘Ž. Then we have positive nine multiplied by negative eight 𝑏, which gives us negative 72𝑏 because we have a negative multiplied by a positive. And then finally, we have positive nine multiplied by negative nine, which gives us negative 81.

Okay, so we’ve now fully expanded in our check. Now let’s try and actually simplify and collect like terms. Well there’s only one π‘Ž squared, so we’ll have π‘Ž squared. And then we’ve got negative eight π‘Žπ‘ and then minus eight π‘Žπ‘, which will give us negative 16π‘Žπ‘. Then we have negative nine π‘Ž plus nine π‘Ž, which is zero, so they cancel out. Then there’s positive 64𝑏 squared. Then positive 72𝑏 minus 72𝑏, which again will give us zero, so that cancels itself out. And then we’re left with negative 81.

So therefore, we get π‘Ž squared minus 16π‘Žπ‘ plus 64𝑏 squared minus 81. And a quick check back, and this is where we started. So therefore, we can say that if you expand π‘Ž multiplied by π‘Ž minus 16𝑏 plus 64𝑏 squared minus 81 and then factor the result completely, you get π‘Ž minus eight 𝑏 plus nine multiplied by π‘Ž minus eight 𝑏 minus nine.

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