Video: GCSE Mathematics Foundation Tier Pack 4 β€’ Paper 3 β€’ Question 16

GCSE Mathematics Foundation Tier Pack 4 β€’ Paper 3 β€’ Question 16

06:09

Video Transcript

Part a) Fully factorise 20𝑝 minus four.

To factorise an expression means to write it as a product of terms or expressions which multiply together to give the original expression. The word β€œfully” is usually a clue that we can take out more than one common factor or that the expression can also be partially factorised. We need to make sure that we fully factorise this expression in order to get all of the marks.

First, let’s look at the number part of the two terms. We need to find the highest common factor of 20 and four. In fact, this is equal to four as four is a factor of four and is also a factor of 20, as four multiplied by five gives 20. So we can take four out as a common factor.

Now, next, we would usually look at the letters. But in this expression, only one of the two terms actually has any letters, which means there are no further common factors that we can take out. We can, therefore, open our bracket and we just need to work out what goes inside it.

Now, the inside of this bracket is what we need to multiply four by to give the original expression. To get 20𝑝, we have to multiply four by five 𝑝. As four multiplied by five is 20, so four multiplied by five 𝑝 is 20𝑝. There’s then a negative sign between the two terms and then we need to include a one, as four multiplied by one gives four.

If we look at the bracket, there are no common factors between five 𝑝 and one other than one. So we can’t factorise this expression any further. We can of course check that our factorisation is correct by expanding the bracket out again. And it gives 20𝑝 minus four.

Now, you may not have spotted that the highest common factor of 20 and four was in fact four. But instead, you may have realised that 20 and four are both even numbers. You may then have taken out a factor of two, which would have left you with two on the outside and then 10𝑝 minus two inside the bracket.

However, this is only partially factorised as if we look inside the bracket, 10 and two are both even numbers. So we can take out a further common factor of two. On the outside of the bracket, we’d now have four as this is two multiplied by two and on the inside five 𝑝 minus one. So we have the same answer as we found initially.

What this means is that even if you don’t take out the highest common factor of both terms initially, you can still get the correct answer, as long as you check whether your bracket can be factorised further at every stage.

Part b) Factorise π‘ž squared plus four π‘ž minus 12.

Now, this is a quadratic expression as the highest power of π‘ž is π‘ž squared. So it’s going to factorise into two brackets. As we only have one lot of π‘ž squared, this means that we’re just going to have one π‘ž or π‘ž at the start of each of our brackets as π‘ž multiplied by π‘ž gives π‘ž squared.

We then need to work out what the numbers are that go in each of the brackets. Now, when we’re factorising a quadratic like this, there is a rule that we need to remember. And it is that the two numbers we’re looking for need to sum to the coefficient or number in front of π‘ž β€” so in this case, that’s positive four β€” and they need to multiply to the constant term, which is the number that doesn’t have any π‘žs or π‘ž squareds. So in this case, they need to multiply to negative 12.

Now, we’ll actually ignore that negative part of negative 12 first and we’ll begin by just listing out the factors of 12. So we have 12 and one, six and two, or four and three. In fact, as the numbers need to multiply to negative 12, this means that they need to have different signs. So one will be positive and one will be negative.

We then need to see how we can make a sum of four with one of these factor pairs by making one positive and the other negative. You’ll notice that six minus two is equal to four. So if you make the two negative, but keep the six positive, then this pair of numbers will sum to four, but also multiply to negative 12.

So the two numbers we’re looking for in our bracket are positive six and negative two and our factorised form of the quadratic is π‘ž plus six π‘ž minus two.

Now, we want to check our factorisation is correct. So we can expand the brackets back out again using the acronym FOIL. Here, F stands for first. So we multiply the first term in each bracket together, which was we’ve already seen π‘ž multiplied by π‘ž is equal to π‘ž squared. O stands for outers. So we multiply the π‘ž in the first bracket by the negative two in the second, which gives negative two π‘ž. I stands for inners. So we multiply the six in the first bracket by the π‘ž in the second, which gives six π‘ž. And finally, L stands for last. So we multiply the six in the first bracket by the negative two in the second, giving negative 12.

Our expansion is, therefore, π‘ž squared minus two π‘ž plus six π‘ž minus 12. And in the centre of this expansion, negative two π‘ž plus six π‘ž simplifies to positive four π‘ž. So we have π‘ž squared plus four π‘ž minus 12, which is what we were hoping to get.

The factorised form of the quadratic π‘ž squared plus four π‘ž minus 12 then is π‘ž plus six π‘ž minus two. You could of course write these two brackets the other way around with π‘ž minus two first and π‘ž plus six second, as it doesn’t matter which order we multiply in.

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