GED Mathematical Reasoning: Factoring Expressions II – Grouping

Example 1

Factor by Grouping: a^3 + 3a^2 - 5a - 15

 

First notice that the expression shown in the first example contains 4 terms which means factoring by grouping is an appropriate strategy to use. To factor by grouping, think of the first two terms as being one group and the second two terms as being another group.

Next, we’ll factor out the GCF from each group separately.

Focusing on the first group of two terms: a^3 + 3a^2

a^3 factors into: a^3 = a \cdot a \cdot a

3a^2 factors into:  3a^2 = 3 \cdot a \cdot a

Both factorizations have two a’s in common, therefore the GCF of the first group is: a^2. When we factor a^2 from the first two terms, we’re left with a + 3 in parenthesis.

Now let’s look at the second group of two terms: -5a - 15.  Since we have a leading negative, we already know that the GCF of these second two terms will be negative.

Since 5a factors into: 5a = 5 \cdot a

And 15 factors into:  15 = 3 \cdot5

The GCF of the second group is: -5

When we factor -5 from the second two terms, we’re left with a + 3 in parenthesis.

As a side note: Be mindful of the positive and negative signs here – when we factor out -5 and divide each term of the original group by -5, the negatives become positive. At this point, we can rewrite our original expression using the factored groups.

a^3 + 3a^2 -5a - 15 = a^2(a + 3) -5(a + 3)

Notice that we use the sign of the -5 as a subtraction sign to separate the two factored groupings. And we haven’t fully factored the expression because it is not yet written as a PRODUCT.

Do you see that the binomials in the two sets of parenthesis match? This is critical to the factor by grouping process.

After step 2, if the parentheses do not match go back and check for an error. Or, you may re-arrange the terms in the original expression and redo step 2.

To complete the factor by grouping process, we’ll factor out the common binomial by writing the common binomial in front of a set of parenthesis and dividing the expression by the binomial to determine what goes inside the parenthesis.

a^3 + 3a^2 -5a - 15 = a^2(a + 3) - 5(a + 3)

a^2(a + 3) - 5(a + 3) = (a+3)\left( \frac{a^2(a + 3)}{(a + 3)} -\frac{5(a + 3)}{(a + 3)} \right)

a^2(a + 3) - 5(a + 3) = (a + 3)(a^2 - 5)

Essentially, the common binomial goes in one set of parenthesis and the remaining terms go in a second set of parenthesis. We know we have reached factored form because the expression is written as a product of two binomials. And just so you know, the order of the parenthesis doesn’t matter – this answer can be switched around to read: (a^2 - 5)(a + 3)

The final step is to check our result using multiplication. Multiplying out our final answer should result in the original expression. Remember: to multiply a binomial times a binomial, we can use the distributive property twice, which is also known as the FOIL method.

algeb77a \cdot a^2 = a^3
a \cdot -5 = -5a
3 \cdot a = 3a^2
3 \cdot -5 = -15

Although the terms in our check are in a different order, the terms are the same as in the original expression so our factored answer is correct.

 

Example 2

Factor by Grouping: 3xy + 9x + y + 3

 

Since the expression shown in this second example contains 4 terms, it is appropriate to use the factor by grouping approach. For our first step, we’ll group the first two terms together and the second two terms together.

Following step 2 of the process, let’s factor the GCF out of each group.

Focusing on the first group of two terms: 3xy + 9x

3xy factors into: 3xy = 3 \cdot x \cdot y

9x factors into: 9x = 3 \cdot 3 \cdot x

Therefore the GCF of the first group is: 3x. When we factor 3x from the first two terms, we’re left with y + 3 in parenthesis.

Now let’s look at the second group of two terms: y + 3. Neither of these terms factors and they have nothing in common except for the number one, since one is a factor of everything. So we’ll say that the GCF of this second group is 1. When we factor from the second two terms, we’re left with y +3 in parenthesis.

At this point, we can rewrite our original expression using the factored groups.

3xy + 9x + y + 3 = 3x(y + 3) + 1(y + 3)

Notice that we use the sign of the positive 1 as an addition sign to separate the two factored groupings.

At this point, we haven’t fully factored the expression because it isn’t yet written as a PRODUCT. So let’s verify that the two sets of parenthesis match, which they do, and move on to Step 3.

For Step 3 we’ll factor out the common binomial by writing it in front of a set of parenthesis and divide the expression by the binomial to determine what goes inside the parenthesis.

3x(y + 3) + 1(y + 3) = (y + 3)\left( \frac{3x(y + 3)}{(y + 3)} + \frac{1(y + 3)}{(y + 3)} \right)

(y + 3)\left( \frac{3x(y + 3)}{(y + 3)} + \frac{1(y + 3)}{(y + 3)} \right) = (y + 3)(3x + 1)

We know we have reached factored form because the expression is written as a product. And remember, the order of the parenthesis doesn’t matter – this answer can be switched around to read: (3x + 1)(y + 3)

The fourth and final step is to check our result using multiplication.

algeb78y \cdot 3x = 3xy
y \cdot 1 = y
3 \cdot 3x = 9x
3 \cdot 1 = 3

In closing, factoring out the GCF and factor by grouping are two strategies for factoring polynomials and in your further studies you will learn much more about this topic. But it’s important to know that not every polynomial can be factored.

Just like the numbers 3, 5, 7, and 11 cannot be factored and are called “prime”, some polynomials cannot be factored and are called “prime.”

For a quick example, consider the polynomial:   7 a + 5 b - 2 c + 11d

None of the terms have any factors in common, so there is no GCF to factor out. And even though this expression contains four terms, we can’t factor it by grouping for the same reason – no two terms have a factor in common. So this polynomial cannot be factored and we say it is “prime.”

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