Video: Pack 3 β€’ Paper 1 β€’ Question 16

Pack 3 β€’ Paper 1 β€’ Question 16

02:49

Video Transcript

Factorise 80π‘₯ squared minus five fully.

To help us decide whether an expression factorises into one or two brackets, we first see if there are any common factors throughout the expression. If there are, that means we can factorise into one bracket.

In this case, 80π‘₯ squared and negative five have a common factor of five. This means we can factorise this expression into one bracket. And the highest common factor of each term, five, goes on the outside of this bracket.

To find the terms inside the bracket, we divide both of the terms in our original expression by five. 80 divided by five is 16, so 80π‘₯ squared divided by five is 16π‘₯ squared. Negative five divided by five is negative one. The factorised form of 80π‘₯ squared minus five is, therefore, five multiplied by 16π‘₯ squared minus one.

We aren’t finished yet though. Notice how the expression inside the brackets has two square numbers with a minus sign between them. This is a special kind of expression that we can call the difference of two squares.

To factorise an expression that’s the difference of two squares, we can apply this formula. π‘Ž squared minus 𝑏 squared is equal to π‘Ž plus 𝑏 multiplied by π‘Ž minus 𝑏. Note that π‘Ž and 𝑏 can be any algebraic expression, not just a list of square numbers that we’re familiar with.

Let’s take 16π‘₯ squared minus one. The square root of 16 is four, and the square root of π‘₯ squared is simply π‘₯. That means that the term in the front of each bracket is four π‘₯. The square root of one is one, so the brackets become four π‘₯ plus one multiplied by four π‘₯ minus one.

We can use the FOIL method to expand this back out to check. Four π‘₯ multiplied by four π‘₯ is 16π‘₯ squared. Four π‘₯ multiplied by negative one is negative four π‘₯. One multiplied by four π‘₯ is four π‘₯. And one multiplied by negative one is negative one. Notice how the terms negative four π‘₯ and four π‘₯ cancel each other out. Negative four π‘₯ plus four π‘₯ is zero. That leaves us with 16π‘₯ squared minus one, which is the expression we needed.

Now that we know we factorised the expression 16π‘₯ squared minus one correctly, we can pop it back into our earlier expression of five lots of 16π‘₯ squared minus one. That leaves us with five multiplied by four π‘₯ plus one multiplied by four π‘₯ minus one.

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