Write nine multiplied by two 𝑥 minus one minus three multiplied by two 𝑥 minus eight plus one in the form 𝑎 multiplied by 𝑏𝑥 plus 𝑐 for integers 𝑎, 𝑏, and 𝑐.
It might at first seem a little unclear as to what we need to do here. But at its heart, this is an expand and simplify question, at least initially. We know this because we’re given an expression with several terms involving brackets and being asked to write it as a single bracket with integers. Remember those are just whole numbers, 𝑎, 𝑏, and 𝑐.
We’ll begin then by expanding each bracket in the expression given. Remember when we expand, it means multiply. Here, we want to multiply each term inside the bracket by the number on the outside.
For the first bracket, we begin multiplying nine by two 𝑥. Nine multiplied by two is 18. So nine multiplied by two 𝑥, nine lots of two 𝑥, is 18𝑥.
Next, we’re going to multiply nine by negative one. Remember a positive multiplied by a negative is a negative. And since nine multiplied by one is nine, this means nine multiplied by negative one is negative nine. And we’ve expanded the first bracket; it can be written as 18𝑥 minus nine.
Next, we’ll expand the second bracket. We need to be extra careful here though since this bracket is being multiplied by negative three. So to begin, we’ll multiply negative three by two 𝑥. Negative three multiplied by two is negative six. So negative three multiplied by two 𝑥, that’s negative three lots of two 𝑥, is negative six 𝑥.
Next, we’ll multiply negative three by negative eight. A negative multiplied by a negative is a positive. So negative three multiplied by negative eight is 24. And our second bracket expands to negative six 𝑥 plus 24.
Let’s rewrite the expression from our question using these expanded brackets. Nine multiplied by two 𝑥 minus one was 18𝑥 minus nine. And negative three multiplied by two 𝑥 minus eight was negative six 𝑥 plus 24. And of course, we mustn’t forget this one.
We’re going to simplify this expression, that’s the simplified part of expand and simplify, by collecting like terms, collecting together the things that are the same. Here, we have 𝑥s and integers, whole numbers.
Let’s begin by adding the 𝑥s together. We have 18𝑥 and we’re going to add negative six 𝑥. Remember adding a negative is just the same as subtracting. So we’re doing 18𝑥 minus six 𝑥 which is 12𝑥.
Next, we’ll add the integers. To do this, we’ll imagine we start at negative nine on the number line. We’re going to add 24, which means moving up the number line 24 places.
If we do this, we get to 15. Then, we need to add one more. And that takes us to 16. Our expression, therefore, simplifies to 12𝑥 plus 16.
Let’s now compare this to the form we’re being asked to write our answer in. Notice how there’s a bracket with an integer on the outside. To get our expression to look like this, we’re going to need to factorise it.
When we factorise, we put it back into brackets. And to do this, we look for the highest common factor of the terms involved. We’re going to need to find the highest common factor, the HCF, of 12𝑥 and 16.
Well, the largest number that can be divided evenly into 12 and 16 with no remainder is four. And in fact, the highest common factor of 12𝑥 and 16 is also four. This means that four is the number that goes on the outside of our bracket.
To work out the terms that go on the inside of the bracket, we’re going to divide both 12𝑥 and 16 by four. 12 divided by four is three. So 12𝑥 divided by four is three 𝑥. And 16 divided by four is four. Our final answer is four multiplied by three 𝑥 plus four.
We can perform a little check by multiplying this back out and seeing whether we get the expression we had before. Four multiplied by three 𝑥 is 12𝑥 and four multiplied by four is 16. This is the expression we’ve just factorised. So we’ve performed the factorisation correctly.
And we’re done. Our answer is four multiplied by three 𝑥 plus four.