Question Video: Factorisation by Grouping | Nagwa Question Video: Factorisation by Grouping | Nagwa

Question Video: Factorisation by Grouping Mathematics

Factorize fully 4π‘₯π‘Ž + π‘₯𝑏 + 4π‘¦π‘Ž + 𝑦𝑏.

05:46

Video Transcript

Factorize fully the expression four π‘₯π‘Ž plus π‘₯𝑏 plus four π‘¦π‘Ž plus 𝑦𝑏.

This expression has four terms. We’re going to answer this question using the method of factorizing by grouping. I’ll begin by separating the expression into two pairs of terms and then looking at them individually. Let’s start with the first half of the expression.

Both terms here have a common factor of π‘₯, and so they can be factorized by π‘₯. We can therefore write the first two terms as π‘₯ multiplied by a bracket. Inside the bracket, we need four π‘Ž for the first term and then plus 𝑏 for the second. You could check by expanding the bracket. This does indeed give four π‘₯π‘Ž plus π‘₯𝑏. So we factorized the first pair of terms. Now let’s consider the second pair.

Both terms are positive and they have a common factor of 𝑦. So we can take 𝑦 out as a common factor, writing the second half of the expression as plus 𝑦 multiplied by a bracket. Inside the bracket, we need to have four π‘Ž for the first term so that it gives four π‘¦π‘Ž when multiplied by 𝑦, and we need to have plus 𝑏 for the second so that it gives plus 𝑦𝑏 when multiplied by 𝑦. So now we factorized the two halves of the expression.

And what you’ll notice is that they have a common bracket of four π‘Ž plus 𝑏. Now this isn’t a coincidence. This is what I hoped would happen when I decided to factorize by grouping. As the two halves of the expression have a common bracket of four π‘Ž plus 𝑏, we can bring this bracket out as a common factor. The whole expression is therefore four π‘Ž plus 𝑏 multiplied by a second bracket. The terms in the second bracket are what four π‘Ž plus 𝑏 has been multiplied by in the two halves of the expression.

So that’s π‘₯ for the first half and plus 𝑦 for the second. This gives the fully factorized form of the original four term expression. It’s equal to four π‘Ž plus 𝑏 multiplied by π‘₯ plus 𝑦. Now you may be wondering well what would happen if the terms in the original expression have been written in a different order. So I’ve chosen to swap the middle two terms around, giving four π‘₯π‘Ž plus four π‘¦π‘Ž plus π‘₯𝑏 plus 𝑦𝑏.

I’ll now attempt to use factorizing by grouping on this expression. So I’ve divided it up into two pairs of terms as before. Looking at the first pair of terms, four π‘₯π‘Ž plus four π‘¦π‘Ž, we can see that we have a common factor of four π‘Ž. Their remaining factors, which need to be included inside the bracket, are π‘₯ and plus 𝑦. Now let’s consider the second half of this expression. Both terms are positive, and they have a common factor of 𝑏.

So we can write the second half as plus 𝑏 multiplied by a bracket. The leftover factors to be included in the bracket are π‘₯ and plus 𝑦. Now let’s look at this expression. We can see that the two halves do again have a common bracket. But this time, the common bracket is π‘₯ plus 𝑦. We can bring this bracket out as a common factor, leading to π‘₯ plus 𝑦 multiplied by a second bracket, which we now need to fill in.

The terms that go in the second bracket are what π‘₯ plus 𝑦 has been multiplied by each time. So that’s four π‘Ž for the first half of the expression and plus 𝑏 for the second. Now compare this with the answer that we already found. You’ll notice that we have the same two brackets each time, but they’re just in a different order.

This means that we have the same answer, because of course multiplication is commutative. It doesn’t matter which order you multiply two numbers or two expressions together in. Therefore, we could write the brackets either way round. What we’ve seen then is that if the terms of the original expression are a different order, we still get the same result for the factorized form.

Let’s consider one final possible reordering of the terms. So I’ve brought the final term 𝑦𝑏 and moved it to become the second term of the expression. So we now have four π‘₯π‘Ž plus 𝑦𝑏 plus π‘₯𝑏 plus four π‘¦π‘Ž. Suppose we attempt to factorize by grouping as we’ve just done. Looking at the first two terms, four π‘₯π‘Ž and plus 𝑦𝑏, we see that they have no common factors. The same is true of the second pair of terms. They also have no common factors.

Does this mean that the expression doesn’t factorize? Well, no. As we’ve already seen twice, it can be factorized into two brackets. It just means that this particular ordering of the terms doesn’t work for factorizing by grouping. If you’re answering a question like this and factorizing by grouping doesn’t appear to be working, try reordering the terms. The answer to this problem was four π‘Ž plus 𝑏 multiplied by π‘₯ plus 𝑦.

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