# Video: GCSE Mathematics Foundation Tier Pack 2 β’ Paper 3 β’ Question 10

GCSE Mathematics Foundation Tier Pack 2 β’ Paper 3 β’ Question 10

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### Video Transcript

Part a) Factorise 44π plus 22. Part b) Fully factorise one over seven π cubed π minus one over seven π squared π squared.

So first with part a, what we need to do is actually factorise 44π plus 22. So the first thing we do is we actually look at the numbers in each of the terms in our expression. So we got 44 and 22. So what we want to do is factorise to actually find the highest common factor of these two values. And we can see that the highest common factor is going to be 22. And thatβs because 22 multiplied by one is equal to 22 and 22 multiplied by two is equal to 44.

Okay, great, so we found the factor thatβs gonna be outside our brackets. Thatβs all weβre gonna have outside our bracket with this expression because actually we donβt have any shared letters. So there isnβt a π in both of them. There is only π in the first term.

Okay, so now, what goes inside the bracket? Well, the first term inside the bracket is gonna be two π. And thatβs because 22 multiplied by two π will give us our 44π that weβre looking for in the original expression. And then, the next term will be one or positive one. And thatβs because as we said, 22 multiplied by one gives us the 22 that we had in the original expression. So therefore, weβve actually factorised 44π plus 22. And the result is 22 multiplied by two π plus one.

For part b, what we need to do is fully factorise one over seven π cubed π minus one over seven π squared π squared. And what Iβm gonna do is actually doing stages just so you can see how we actually pull out each of our factors.

Well, our first shared factor in each of our terms is one over seven or one-seventh. So we can actually take this outside of our bracket. So weβve got one-seventh. And then inside, weβll have π cubed π because one seventh multiplied by π cubed π gives us one-seventh π cubed π then minus π squared π squared cause again if we multiply one-seventh by π squared π squared, we get a seventh π squared π squared.

Okay, great, what would be the next factor we could take out? Now, the next factor weβre gonna take out is π squared because this is the highest power of π thatβs actually in both of our terms. So Iβve shown here why. So if we have π cubed, thatβs equal to π squared multiplied by π. If we had π squared, well, thatβs just π squared. So we can see that the highest shared power of π is actually π squared.

So therefore, if I take this outside the bracket alongside the one over seven or one-seventh, we get one over seven π squared. And then inside the bracket, we have ππ. And thatβs because π squared multiplied by π gives us the π cubed we want. And then obviously, weβve got the π and then minus π squared because actually we donβt need the π term there because π squared is already there because weβve π squared outside the bracket.

Okay, great, can we go any further? Are there any other factors that we can take out? Well, the final factor we can actually take out is π. And thatβs because thereβs actually an π in each of our terms. And π is the highest power of π in both of them because weβve got π squared in one of them, which is the same as π multiplied by π and then just π in the other. So as we see, π will be the highest factor of these.

So then, when we take this outside the bracket, what we actually have is one over seven π squared π multiplied by π minus π. So we have π minus π inside the bracket. So therefore, we can say that fully factorised one over seven π cubed π minus one over seven π squared π squared is equal to one over seven π squared π multiplied by π minus π.

Okay, we solved the problem. But what I want to do is actually show a check so just to check that weβve got the right final answer. Well, the first thing we do to check is see can we actually factorise it anymore? Well, no, we canβt because itβs fully factorised.

So then, the next check we can do is by actually expanding. So in the first term, what weβre gonna get is one over seven π cubed π. And thatβs because if you multiply one over seven π squared π by π, weβre gonna get one over seven. And then we got π cubed and thatβs because π squared multiplied by π is π cubed because if you multiplied them at the same base, we add the powers. So two add one gives us three and then weβve got our π.

And then, our second term is gonna be one over seven π squared π squared. And thatβs because if we got one over seven π squared π and multiply it by π, well weβre gonna get one over seven then π squared. And then, π multiplied by π gives us π squared. And this is the expression we started with. So we can say, βYes, weβre happyβ. Weβve fully factorised and weβve checked our answer.