# Video: Pack 2 β’ Paper 1 β’ Question 14

Pack 2 β’ Paper 1 β’ Question 14

04:07

### Video Transcript

Prove algebraically that four lots of π plus one all squared minus π minus two all squared is always a multiple of three.

First, letβs note that any multiple of three can be written as three π, where π is an integer. For example, three equals three times one. Six equals three times two. And 18 equals three times six. And we could do this for any multiple of three. We will use this rule for writing multiples of three in order to help us answer the question.

Letβs now look at the expression given in the question. We will start by expanding π plus one squared and π minus two squared, using the FOIL method, starting with π plus one squared, which we can also write as π plus one times π plus one.

And now we will use the FOIL method. So we start by multiplying the first term in each bracket. Thatβs π times π, giving us π squared. Then we multiply the outside terms. So thatβs π times one, giving us plus π. Then we multiply the inside terms. Thatβs one times π, giving us another plus π. And then, finally, we multiply the last term in each bracket. So thatβs one times one, giving us plus one.

Now we notice that we can combine the π plus π to give us two π. And we obtain that π plus one all squared is equal to π squared plus two π plus one. Now we will expand π minus two squared, which can be written as π minus two times π minus two. And now we will use the FOIL method.

So we start by multiplying the first term in each bracket. This is π times π, giving us π squared. Next, we multiply the outside term in each bracket. So this is π timesed by minus two, giving us minus two π. Next, we multiply the inside terms. So thatβs minus two times π, giving us another minus two π. And then, finally, we multiply the last term in each bracket. So this is minus two times minus two. And since a negative times a negative gives us a positive, this is equal to plus four.

Now we notice that we have minus two π minus two π, giving us minus four π. And so we get that π minus two all squared is equal to π squared minus four π plus four. Now we can substitute the expanded forms of π plus one squared and π minus two squared back into the expression in the question. This gives us that four lots of π plus one squared minus π minus two squared is equal to four timesed by π squared plus two π plus one minus π squared minus four π plus four.

Now we simply multiply every term in the left-hand bracket by four and every term in the right-hand bracket by negative one. This gives us four π squared plus eight π plus four minus π squared plus four π minus four.

Now we can combine the terms with the same powers of π. For the π squared terms, we have a four π squared minus π squared. And this leaves us with three π squared. For the π terms, we have eight π plus four π, giving us plus 12π. And for the units, we have plus four minus four, giving us plus zero. However, this has no effect here, so we can ignore it.

We are left with three π squared plus 12π. And we notice that each term here is a multiple of three. So we can factorise out three. So we are left with three lots of π squared plus four π. And now since π is an integer, this means that π squared plus four π will also be an integer. And since we are multiplying an integer by three, that means that this is always a multiple of three, using the rule which we stated at the start. And so we have shown that four lots of π plus one all squared minus π minus two all squared is always a multiple of three.