Video: AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 3 β€’ Question 14

AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 3 β€’ Question 14

05:44

Video Transcript

The variable π‘˜ is a positive integer. Prove that 12π‘˜ cubed plus 24π‘˜ squared plus 12π‘˜ over four π‘˜ squared plus four π‘˜ is always a multiple of three.

This word variable just means that π‘˜ can take different values. However, it must always be a positive integer; that’s, a positive whole number. Now we’re asked to prove that this expression involving π‘˜ is always a multiple of three. When we’re asked to prove something like this, we need to use algebra. It isn’t enough to show that this expression just works for certain values of π‘˜, as this wouldn’t be a proof. We need to prove that the same is true for all possible values of π‘˜. So in this case, that’s all positive integers.

As we’ve been asked to prove that this expression is always a multiple of three, we want to write it in the form three multiplied by something. So we’ll look at how we can simplify the expression we’ve been given to try and bring it into this form. First, we’ll look at factorizing both the numerator and denominator of this fraction. Let’s look at the numbers in the numerator first. We have 12, 24, and 12. And as the highest common factor of these three numbers is 12, we can bring this out as a common factor first of all.

Next, we’ll look at the powers of π‘˜. We have π‘˜ cubed, π‘˜ squared, and then π‘˜, which is π‘˜ to the power of one. We bring out the lowest power of π‘˜. So in this case that’s just π‘˜. We therefore have 12π‘˜ on the outside of the bracket, and there are no further common factors in the numerator. Inside the bracket, we need to write down the terms that we have to multiply 12π‘˜ by to give the original expression. To give 12π‘˜ cubed, we have to multiply 12π‘˜ by π‘˜ squared because π‘˜ multiplied by π‘˜ squared is π‘˜ cubed, so 12π‘˜ multiplied by π‘˜ squared gives 12π‘˜ cubed. To get 24π‘˜ squared, we have to multiply 12π‘˜ by two π‘˜ as 12 multiplied by two gives 24 and π‘˜ multiplied by π‘˜ gives π‘˜ squared.

Finally, to get 12π‘˜, we have to multiply 12π‘˜ by one, so we need to include plus one as the third term in the bracket. The numerator therefore factorizes as 12π‘˜ multiplied by π‘˜ squared plus two π‘˜ plus one. In the denominator, if we look at the numbers first of all, we can take out a common factor of four. And then looking at π‘˜ squared and π‘˜, we can again take out the lower power of π‘˜, so that’s just π‘˜, giving four π‘˜ as a factor on the outside of the bracket. Inside the bracket, we need to have π‘˜ plus one so that when we multiply four π‘˜ by π‘˜ we get four π‘˜ squared and when we multiply four π‘˜ by plus one we get plus four π‘˜.

We’ve now fully factorized both the numerator and denominator of this expression. Now already you may be able to spot that there’s some cancellation that we can do here. There’s a factor of π‘˜ in both the numerator and denominator, so this can be cancelled. We also have a 12 in the numerator and a four in the denominator. So dividing these both by four, we now have a three in the numerator and a one in the denominator. So writing out the simplified expression, we have three multiplied by π‘˜ squared plus two π‘˜ plus one all over π‘˜ plus one.

Now this is looking positive because we have this factor of three here, but we don’t know that the rest of the expression, so that’s π‘˜ squared plus two π‘˜ plus one over π‘˜ plus one, will give an integer for every value of π‘˜, which would mean that if we’re multiplying three by something that isn’t an integer, this would not give a multiple of three. We need to see how we can simplify this expression further. Let’s see if we can factorize that quadratic expression in the numerator. As the coefficient of π‘˜ squared is just one, this means that we just have π‘˜ at the front of each bracket because π‘˜ multiplied by π‘˜ gives π‘˜ squared.

To complete the brackets, we need to look for two numbers which add to the coefficient of π‘˜, so that’s positive two, and multiplied to the constant term, that’s positive one. There aren’t a lot of possibilities here because the only factors of one are one and one, but one and one add to two. So we can complete each bracket with a plus one, and we see that π‘˜ squared plus two π‘˜ plus one factorizes as π‘˜ plus one multiplied by π‘˜ plus one. We can also write this as π‘˜ plus one squared. Making this substitution for π‘˜ squared plus two π‘˜ plus one, we now have three multiplied by π‘˜ plus one π‘˜ plus one all over π‘˜ plus one.

One of the brackets in the numerator can cancel with the bracket in the denominator, and we’re just left with three multiplied by π‘˜ plus one. We aren’t quite finished though we’ve written the expression as three multiplied by something, but we need to show that that something is a positive integer in order to give a multiple of three. Well if π‘˜ is a positive integer, then adding one will just give the next positive integer. So if π‘˜ is a positive integer, then so is π‘˜ plus one. Multiplying any positive integer by three will always give a multiple of three. So three multiplied by π‘˜ plus one is therefore a multiple of three.

So we’ve proven that the original expression, 12π‘˜ cubed plus 24π‘˜ squared plus 12π‘˜ over four π‘˜ squared plus four π‘˜, is always a multiple of three when π‘˜ is a positive integer.

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