Lesson Video: Factoring by Grouping Mathematics

In this video, we will learn how to factor expressions by grouping.

13:56

Video Transcript

In this video, we will learn how to factor expressions by grouping.

Let’s begin by revisiting the idea of factoring an expression by identifying its highest common factor. Looking at the expression three π‘₯ plus nine 𝑦 minus six, we see that each term is divisible by three. This means that three is a common factor of the expression, and we can take this outside of our parentheses. Three π‘₯ plus nine 𝑦 minus six can be rewritten as three multiplied by π‘₯ plus three 𝑦 minus two. Since the terms inside the parentheses have no common factors apart from one, we can conclude that three is the highest common factor and the expression is now in its most simplified form.

Let’s now consider a second example where the entire expression does not have a common factor apart from one. We can therefore begin by grouping pairs of terms. The terms two and two π‘₯ have a common factor of two. So we can rewrite this as two multiplied by one plus π‘₯. The terms 𝑦 and π‘₯𝑦 have a common factor of 𝑦. Therefore, this can be rewritten as 𝑦 multiplied by one plus π‘₯. We can therefore rewrite the original expression as shown.

At this stage, we notice that our two terms have a common factor of one plus π‘₯. Factoring this out, we can rewrite the expression as one plus π‘₯ multiplied by two plus 𝑦. The expression two plus two π‘₯ plus 𝑦 plus π‘₯𝑦 is therefore equal to one plus π‘₯ multiplied by two plus 𝑦. We could check this answer by distributing our parentheses using the FOIL method. We will now generalize the method used in this example to help us factor by grouping.

If we have an expression with an even number of terms that did not all share a common factor, then in certain situations we can apply the following approach to fully factor the expression. Firstly, we group the terms into pairs that each share a common factor. Secondly, we factor out the common factor from each pair. Finally, we completely factor the expression by identifying a common factor in the factored terms.

Let’s now consider another example where we can apply this method.

Factorize fully 𝑧𝑏 minus 𝑧π‘₯ plus 𝑏 minus π‘₯.

The expression in this question does not have a common factor greater than one, so we cannot directly factor all of it. However, we can factor it by grouping pairs of terms with common factors. For example, the first and third terms have a common factor of 𝑏, and the second and fourth terms have a common factor of negative π‘₯. Rewriting the expression as shown, we begin by taking out these common factors. 𝑧𝑏 plus 𝑏 can be rewritten as 𝑏 multiplied by 𝑧 plus one, and negative 𝑧π‘₯ minus π‘₯ can be rewritten as negative π‘₯ multiplied by 𝑧 plus one.

We now have an expression where the term in the parentheses, 𝑧 plus one, is common. Factoring this out gives us 𝑧 plus one multiplied by 𝑏 minus π‘₯. And as this expression cannot be factored further, it is in its fully simplified form. The expression 𝑧𝑏 minus 𝑧π‘₯ plus 𝑏 minus π‘₯ factorized fully is 𝑧 plus one multiplied by 𝑏 minus π‘₯.

It is worth noting that we could have grouped different pairs of terms in the original expression. For example, the first two terms in the initial expression have a common factor of 𝑧. So we can rewrite these as 𝑧 multiplied by 𝑏 minus π‘₯. The third and fourth terms have a common factor of one. So we can rewrite 𝑏 minus π‘₯ as one multiplied by 𝑏 minus π‘₯. At this stage, since 𝑏 minus π‘₯ is common, we can factor this out, giving us 𝑏 minus π‘₯ multiplied by 𝑧 plus one. And since multiplication is commutative, this is the same expression as we found using the first method.

We will now consider a more complicated example of this type.

Factorize fully nine π‘₯π‘š minus four 𝐿𝑧 plus four 𝐿π‘₯ minus nine π‘šπ‘§.

In the expression given, there is no common factor among all four terms apart from one. And we will therefore factorize the expression by grouping pairs of terms. If we consider the first and last terms, we note that they have a common factor of nine π‘š. So we can factor this out. Nine π‘₯π‘š minus nine π‘šπ‘§ can be rewritten as nine π‘š multiplied by π‘₯ minus 𝑧. We can use the same method when grouping the second and third terms. These have a common factor of four 𝐿. Negative four 𝐿𝑧 plus four 𝐿π‘₯ can be rewritten as four 𝐿 multiplied by negative 𝑧 plus π‘₯. Altering the order of the terms in the parentheses, we have four 𝐿 multiplied by π‘₯ minus 𝑧.

We now have an equivalent form to the original expression. At this stage, our two terms have a common factor of π‘₯ minus 𝑧. Factoring this out, we have π‘₯ minus 𝑧 multiplied by nine π‘š plus four 𝐿. This is the fully factored form of nine π‘₯π‘š minus four 𝐿𝑧 plus four 𝐿π‘₯ minus nine π‘šπ‘§.

Whilst it is not required in this question, we could check our answer by distributing the parentheses using the FOIL method. Multiplying the first terms gives us nine π‘₯π‘š. Multiplying the outer terms gives us four 𝐿π‘₯. The inner terms have a product of negative nine π‘šπ‘§. And the last terms multiply to give us negative four 𝐿𝑧. This is the same expression as in the question.

So far, in this video, we have considered expressions where no variable has been raised to a power. Sometimes, we may also be able to factor quadratic or cubic expressions by grouping terms that we can factor. Let’s now consider some specific situations where this is the case.

Factorize fully π‘₯ cubed minus two π‘₯ squared plus five π‘₯ minus 10.

The first thing we notice is that the expression is cubic since the largest exponent of a variable is three. Normally, there would be no easy way to factor this expression fully. But we notice there is a common ratio between the coefficients of the first two terms and last two terms. In the first two terms, this ratio is one to negative two. And in the last two terms, the ratio is five to negative 10, which simplifies to one to negative two. This suggests that we might be able to factor the expression by grouping the first two and last two terms together.

The highest common factor of π‘₯ cubed and negative two π‘₯ squared is π‘₯ squared. And as such, factoring π‘₯ cubed minus two π‘₯ squared gives us π‘₯ squared multiplied by π‘₯ minus two. The highest common factor of five π‘₯ and negative 10 is five. And we can rewrite five π‘₯ minus 10 as five multiplied by π‘₯ minus two. This means that we can rewrite the original expression as shown.

We now have two terms with a common factor of π‘₯ minus two. Factoring this out gives us π‘₯ minus two multiplied by π‘₯ squared plus five. As multiplication is commutative, we can rewrite this as π‘₯ squared plus five multiplied by π‘₯ minus two, which is the fully factorized form of π‘₯ cubed minus two π‘₯ squared plus five π‘₯ minus 10. It is worth noting that we can check this answer by distributing our parentheses using the FOIL method, and doing so would give us the expression in the question.

Before looking at one final example, let’s recall two special types of cubic expressions. Firstly, we recall the factored form of the sum of two cubes π‘Ž cubed plus 𝑏 cubed. This is equal to π‘Ž plus 𝑏 multiplied by π‘Ž squared minus π‘Žπ‘ plus 𝑏 squared. We also recall the factored form of the difference of two cubes. π‘Ž cubed minus 𝑏 cubed is equal to π‘Ž minus 𝑏 multiplied by π‘Ž squared plus π‘Žπ‘ plus 𝑏 squared. In our final example, part of our expression can be factored using one of these formulae.

Factorize fully π‘Ž cubed plus 𝑏 cubed plus π‘Ž plus 𝑏.

In order to factorize this expression, we begin by noticing that the first two terms are the sum of two cubes. We recall that π‘Ž cubed plus 𝑏 cubed is equal to π‘Ž plus 𝑏 multiplied by π‘Ž squared minus π‘Žπ‘ plus 𝑏 squared. This means that we can rewrite the expression π‘Ž cubed plus 𝑏 cubed plus π‘Ž plus 𝑏 as shown. Noting that π‘Ž plus 𝑏 can be rewritten as one multiplied by π‘Ž plus 𝑏, it is clear that our two terms have a common factor of π‘Ž plus 𝑏. Factoring this out gives us π‘Ž plus 𝑏 multiplied by π‘Ž squared minus π‘Žπ‘ plus 𝑏 squared plus one.

And as the expression in the second parentheses cannot be factored further, we have the fully factorized form of π‘Ž cubed plus 𝑏 cubed plus π‘Ž plus 𝑏. This is equal to π‘Ž plus 𝑏 multiplied by π‘Ž squared minus π‘Žπ‘ plus 𝑏 squared plus one. And we could check this answer by redistributing the parentheses.

We will now finish this video by summarizing the key points. We saw in this video that we can factor expressions that lack a common factor by grouping them into pairs that share common factors. We also saw that we can factor cubic expressions that have a shared ratio between the coefficients of terms using factoring by grouping. Finally, we saw that if we can apply a factoring formula to part of an expression β€” such as the sum or difference of cubes, the difference of squares, or perfect square trinomials β€” then we can group that expression and factor it separately.

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