Video: AQA GCSE Mathematics Foundation Tier Pack 1 β€’ Paper 3 β€’ Question 27

(a) Fully factorise 8𝑝² βˆ’ 12𝑝³. (b) Solve π‘₯Β² βˆ’ 16π‘₯ + 60 = 0.

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Video Transcript

Part a) Fully factorise eight 𝑝 squared minus 12𝑝 cubed. Part b) Solve π‘₯ squared minus 16π‘₯ plus 60 is equal to zero.

In part a, we’re being asked to factorise this expression, which means we want to write it as the product of terms or expressions which are multiplied together to give eight 𝑝 squared minus 12𝑝 cubed. We’ll look for common factors of the two terms. The word β€œFully” suggests that this expression can also be partially factorised. We must make sure that we factorise by taking out the highest common factor of the two terms.

Let’s begin by looking at the numbers in the two terms first of all. We have eight and then 12. They’re both even numbers, which means they can both be divided by two. But they’re also both numbers in the four times table. And four is bigger than two. Four is the highest common factor of eight and 12. So we begin by taking a factor of four from the two terms.

Next, let’s consider the letters. We have 𝑝 squared and 𝑝 cubed. Remember 𝑝 squared means 𝑝 multiplied by 𝑝 and 𝑝 cubed means 𝑝 multiplied by 𝑝 multiplied by 𝑝. What do these two terms have in common? Well, they both have one factor of 𝑝 and then a second factor of 𝑝. In fact, 𝑝 cubed can be written as 𝑝 squared multiplied by 𝑝. So the highest common factor of 𝑝 squared and 𝑝 cubed is 𝑝 squared.

In fact, there’s a general rule that you can remember if you have different powers of the same variable, so the variable here is 𝑝, then the highest common factor will be the lower power of this variable. So in our case, we have 𝑝 squared and 𝑝 cubed and the highest common factor was 𝑝 squared, the lower power being two.

So we found the highest common factor of the terms. And now, we need to work out what goes inside the bracket. These will be the terms that we need to multiply four 𝑝 squared by to give the original expression. Well, four multiplied by two is eight. So four 𝑝 squared multiplied by two will give eight 𝑝 squared. And so, the first term in the bracket is just two.

There’s then a minus sign between the terms. And we need to consider what we multiply four 𝑝 squared by to give 12𝑝 cubed. Well, four multiplied by three gives 12 and 𝑝 squared multiplied by 𝑝 gives 𝑝 cubed. So four 𝑝 squared multiplied by three 𝑝 will give 12𝑝 cubed. And the negative sign in front of the three 𝑝 will make it negative 12𝑝 cubed overall.

We now check whether we can factorise this expression any further by looking to see whether there are any common factors in the terms inside the bracket. There aren’t. The highest common factor of two and negative three 𝑝 is just one. So this expression can’t be factorised further. The fully factorised form of eight 𝑝 squared minus 12𝑝 cubed then is four 𝑝 squared multiplied by two minus three 𝑝. Of course, we can check this answer by expanding the bracket if we wish.

Now, just suppose that we hadn’t actually spotted the highest common factor of these two terms to begin with and suppose we thought that the highest common factor was just two 𝑝 rather than four 𝑝 squared. Inside the bracket, we would then have four 𝑝 minus six 𝑝 squared because two 𝑝 multiplied by four 𝑝 gives eight 𝑝 squared and two 𝑝 multiplied by negative six 𝑝 squared gives negative 12𝑝 cubed.

This would be a factorised form of the original expression, but not a fully factorised form because there are still some common factors within the bracket. If we look at the numbers inside the bracket, four and six, we see that they are both even numbers. So they have a common factor of two. We can, therefore, take this common factor of two outside the bracket.

Then, if we look at the letters, we can see that the two terms still share a common factor of 𝑝. So we can also take this outside the bracket. Inside the bracket, we will be left with two minus three 𝑝.

Now, have a look at this term outside the bracket here. We have two 𝑝 multiplied by two which gives four 𝑝 and then multiplied by 𝑝 which gives four 𝑝 squared. So this simplifies to four 𝑝 squared multiplied by two minus three 𝑝. But notice this is exactly the same answer as we got when we factorised in one step.

What this means is that it doesn’t actually matter if you don’t factorise by the highest common factor straightaway as long as you keep checking whether there are any common factors left in the bracket.

Now, let’s have a look at part b which is asking us to solve the quadratic equation π‘₯ squared minus 16π‘₯ plus 60 is equal to zero. There are three methods which we can use to solve a quadratic equation. These are factorising, applying the quadratic formula, or completing the square. Now, if a quadratic does factorise, this is always going to be the quickest method. So it’s always worth performing a quick check whether a quadratic does factorise before we go for one of the other methods.

As the coefficient of π‘₯ squared in this equation is one, we know that if it does factorise, the first term in each bracket will just be π‘₯ because π‘₯ multiplied by π‘₯ gives π‘₯ squared. We’re then looking for two numbers which sum to the coefficient of π‘₯ β€” that’s negative 16 β€” and multiplied to the constant term, which is positive 60.

Now, you may be able to recognise these two numbers pretty quickly. But if not, let’s start by listing the factors of 60. 60 is equal to one multiplied by 60, two multiplied by 30, three multiplied by 20, four multiplied by 15, five multiplied by 12, and six multiplied by 10.

We notice that the sign of this 60 is positive which means that the signs of these two numbers must be the same as each other. They’re either both positive or both negative so that when we multiply them together, they give a positive answer. We also need to choose these two numbers so that their sum is negative 16. If we take our last pair of factors and make them both negative, then negative six multiplied by negative 10 gives 60 and negative six plus negative 10 gives negative 16. So these are the numbers that we’re looking for.

Our quadratic, therefore, factorises as π‘₯ minus six multiplied by π‘₯ minus 10. You can check this factorisation by expanding the brackets, perhaps using the FOIL method if you wish. So our quadratic equation does factorise. And now, we need to continue to solve it.

We do this by setting each of these brackets in turn equal to zero this is because if two things multiplied together to give zero, then at least one of them must be zero these two equations can be solved in a relatively straightforward way. The first equation we solve by adding six to each side giving π‘₯ is equal to six. And the second equation we solve by adding 10 to each side giving π‘₯ is equal to 10.

So our solution to this quadratic equation is π‘₯ is equal to six or 10. We can of course check our answer by substituting each value back into the equation. When we substitute π‘₯ equals 10 into the left-hand side, we get 10 squared minus 16 multiplied by 10 plus 60. This is equal to 100 minus 160 plus 60. This simplifies to 160 minus 160 which is indeed equal to zero.

This confirms that π‘₯ equals 10 is one of the solutions to this quadratic equation. And by substituting π‘₯ equals six, we can check this value in the same way.

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