Video: GCSE Mathematics Foundation Tier Pack 4 β€’ Paper 1 β€’ Question 28

GCSE Mathematics Foundation Tier Pack 4 β€’ Paper 1 β€’ Question 28

03:48

Video Transcript

Factorise π‘₯ squared minus 25.

To factorise an algebraic expression means to write it as the product of two or more factors, which are multiplied to give the original expression. This expression is a quadratic expression as it has an π‘₯ squared term and no higher powers of π‘₯. It doesn’t, however, have an π‘₯ term and it’s actually a special type of quadratic expression.

The key to answering this question is to spot that 25 is actually a square number. It’s equal to five squared. So we can write π‘₯ squared minus 25 as π‘₯ squared minus five squared. Now, the reason this helps is because an expression that can be written in exactly this form is called a difference of two squares, which we sometimes see abbreviated to D O T S or DOTS. This just means that we’re subtracting one square from another.

A difference of two squares always factorises in the same way. If we have π‘Ž squared minus 𝑏 squared, then this factorises as π‘Ž plus 𝑏 multiplied by π‘Ž minus 𝑏. So the two brackets are almost identical. But one has a positive sign between the terms and one has a negative sign.

We can check this is indeed the correct factorisation of π‘Ž squared minus 𝑏 squared by expanding our brackets out again using the FOIL method. The F stands for first. So we need to multiply the first term in each bracket together. We have π‘Ž multiplied by π‘Ž which is equal to π‘Ž squared. The O stands for outers. So we’re multiplying the terms on the outside together. That’s the π‘Ž in the first bracket and the negative 𝑏 in the second, which gives negative π‘Žπ‘.

Next, I stands for inners. So we’re multiplying the terms on the inside of this expansion together. So that’s the 𝑏 from the first bracket and the π‘Ž from the second, which gives π‘Žπ‘. Finally, L stands for last. So we’re multiplying the 𝑏 in the first bracket with the negative 𝑏 in the second, which gives negative 𝑏 squared.

Now, here is the key point. The two terms in the center of our expansion are identical, apart from the fact they have different signs. We have negative π‘Žπ‘ plus π‘Žπ‘. And so these two terms cancel each other out directly. And that’s due to the fact that the two brackets were very similar, but just had different signs.

So the two terms in the center cancel each other out. And we’ve just left to the π‘Ž squared minus 𝑏 squared, which confirms that our factorisation is correct.

So now, if we apply this rule to the expression that we’re asked to factorise, where we don’t have π‘Ž squared minus 𝑏 squared, but π‘₯ squared minus five squared, well the π‘₯ is like the π‘Ž and the five is like the 𝑏. So our first bracket is π‘₯ plus five and our second bracket is π‘₯ minus five.

You could expand these brackets out if you wanted to check that the factorisation is correct. And what you’d get is π‘₯ squared minus five π‘₯ plus five π‘₯ minus 25. And as before, you’d see that those two terms in the middle would cancel each other out, just leaving us with π‘₯ squared minus 25. So our factorisation is correct.

So whenever you’re asked to factorise a quadratic expression, which is just a squared term minus a number, check whether that number is a square number because if it is, it’s a difference of two squares. And we can factorise the expression easily using the method that we have here.

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