Factoring Trinomials by Grouping 2

 

Learning Objective(s)

·         Factor polynomials of the form ax2 + bx + c by grouping.

 

Introduction

 

Knowing how to factor trinomials in the form x2 + bx + c is useful, but what if there happens to be a coefficient, a, before the leading x2 term?

 

The general form of trinomials with a leading coefficient of a is ax2 + bx + c. To factor them, we will draw upon what we know about trinomials that contain a single x2 term and extend that knowledge to account for the coefficient a. We will also consider some tricks and tips for factoring these trinomials.

 

Thinking about ax2

 

Trinomials can include a coefficient before the x2 term. If we call this coefficient a, the general form of these sorts of trinomials is ax2 + bx + c.  This coefficient adds a bit of complexity to factoring, but we can still find the factors by using the grouping technique and then using the Distributive Property.

 

Here's the strategy for factoring trinomials in the form ax2 + bx + c, with a 1:

 

Factoring Trinomials

 

To factor a trinomial in the form ax2 + bx + c, find two integers, r and s, whose sum is b and whose product is ac. Rewrite the trinomial as ax2 + rx + sx + c and then use grouping and the Distributive Property to factor the polynomial.

 

 

It's not all that different from factoring trinomials without a leading coefficient.

 

Let’s see how this strategy works by factoring 6z2 + 11z + 4.

 

In this trinomial, a = 6, b = 11, and c = 4. According to the strategy, we need to find two factors, r and s, whose sum is b (11) and whose product is ac (or 6 • 4 = 24). We can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since ac is positive and b is positive, we can be certain that the two factors we're looking for are also positive numbers.)

 

r s = 24

r + s

1 24 = 24

1 + 24 = 25

2 12 = 24

2 + 12 = 14

3 8 = 24

3 + 8 = 11

4 6 = 24

4 + 6 = 10

 

There is only one combination in which rs = 24 and r + s = 11: when r = 3, and s = 8. Let’s use these values to factor our original trinomial.

 

 

Example

Problem

 

Factor 6z2 + 11z + 4

 

 

 

 

6z2 + 3z + 8z + 4

 

To make grouping and factoring easier, rewrite the trinomial ax2 + bx + c as ax2 + rx + sx + c. Use values from the chart above: r = 3 and s = 8.

 

In this trinomial, replace 11z with 3z + 8z

 

 

(6z2 + 3z) + (8z + 4)

 

 

Use grouping to consider the terms in pairs.

 

 

3z(2z + 1) + (8z + 4)

 

Use the Distributive Property to pull the common factor, 3z, out of first group.

 

 

3z(2z + 1) + 4(2z + 1)

 

Use the Distributive Property to pull the common factor, 4, out of second group.

 

 

(2z + 1)(3z + 4)

 

Use the Distributive Property to pull the common factor, (2z + 1), out of the pairs.

 

The original trinomial is now fully factored.

Answer

(2z + 1)(3z + 4)

 

 

 

There you have it: the fully factored form of 6z2 + 11z + 4 is (2z + 1)(3z + 4). We used grouping to organize our terms and the Distributive Property to pull out common factors.

 

We can also use this method to factor trinomials in the form x2 + bx + c. If we think about x2 + bx + c as ax2 + bx + c in which the coefficient of x2 (or a) is 1, then ac = 1 • c, or simply c. So the strategy that we use for factoring trinomials in the form x2 + bx + c, finding two numbers whose sum is b and whose product is c, is just a special case of the general method for factoring trinomials in the form ax2 + bx + c.

 

Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial 2z2 + 35z + 7 for instance. Can you think of two integers whose sum is b (35) and whose product is ac (2 · 7 = 14)? There are none! We call this type of trinomial, which cannot be factored using integers, a prime trinomial.

 

Greatest Common Factors

 

Sometimes the easiest factors to pull out of a trinomial are integers. For example, take the case of 6x2 – 26x – 20. Do you notice any common factors among the three terms? a = 6, b = -26, and c = -20, so a common factor is 2. Using the Distributive Property again, we can pull a factor of 2 out of each term and rewrite the trinomial as 2(3x2 – 13x – 10). Then we can factor 3x2 – 13x – 10.

 

Let’s take a look at how to factor 6x2 – 26x – 20 in two ways: pulling the common factor out first, or waiting until the end to do it.

 

Example

Problem

 

Factor 6x2 – 26x – 20

 

 

2(3x2 – 13x – 10)

 

Solution 1: Start by using the Distributive Property to pull out the common factor of 2.

 

 

 

2(3x2 – 15x + 2x – 10)

 

Rewrite the trinomial in parentheses in the form ax2 + rx + sx + c.

 

Think: what numbers have a sum of -13 and a product of -30? -15 and 2

 

2[(3x2 – 15x) + (2x – 10)]

 

Group the pairs of terms.

 

2[3x(x – 5) + 2(x – 5)]

 

Use the Distributive Property to pull the common factors out of each group.

 

2[(x – 5)(3x + 2)]

 

Use the Distributive Property to pull the common factor (x – 5) from both groups.

Answer

2(x – 5)(3x + 2)

 

 

 

 

Example

Problem

 

Factor 6x2 – 26x – 20

 

 

 

 

 

6x2 – 30x + 4x – 20

 

Solution 2: Start by rewriting the trinomial in parentheses in the form ax2 + rx + sx + c.

 

Think: what numbers have a sum of -26 and a product of -120? -30 and 4

 

Rewrite – 26x as – 30x + 4x.

 

(6x2 – 30x) + (4x – 20)

 

Group the pairs of terms.

 

 

6x(x – 5) + 4(x – 5)

 

Use the Distributive Property to pull the common factors out of each group.

 

 

(x – 5)(6x + 4)

 

 

Use the Distributive Property to pull the common factor (x – 5) from both groups.

 

 

 

(x – 5)[2(3x + 2)]

 

 

2(x – 5)(3x + 2)

 

Notice that 6x + 4 can be factored further. Use the Distributive Property to pull the common factor of 2 out of 6x + 4.

 

Use the Commutative Property to put the monomial (2) as the first factor in your answer.

 

Answer

2(x – 5)(3x + 2)

 

 

 

 

We arrived at the same solution using both methods. However, look back to where we had to consider how to expand the bx term in both solution methods. When we pulled out the factor of 2 at the beginning, we had to look for two numbers with a sum of -13 and a product of -30; but when we left the factor of two within the problem, as we did in Solution 2, we had to find two numbers with a sum of -26 and a product of -120.

 

This is the reason why many people choose to pull out the greatest common factor before factoring. Pulling out the GCF first typically makes factoring the trinomial easier, since the resulting values for a, b, and c will all be lower, and so there will be fewer combinations of integers whose sum is b and whose product is ac.

 

Let’s look at one more example: 5t3 + 15t2 – 20t. This one may seem different because the highest exponent of the variable is a 3, but we'll see this is really the same kind of problem as in the previous examples.

 

Example

Problem

 

Factor 5t3 + 15t2 – 20t

 

 

 

5t(t2 + 3t – 4)

 

Use the Distributive Property to pull the common factor out. Both the integer 5 and the variable t are common factors, so 5t is the greatest common factor.

 

 

5t(t2t + 4t – 4)

 

Think: what numbers have a sum of 3 and a product of -4? (-1 and 4.) Rewrite 3t as t + 4t.

 

5t[(t2t) + (4t – 4)]

 

Group the pairs of terms.

 

 

5t[t(t – 1) + 4(t – 1)]

 

Use the Distributive Property to pull the common factors out of each group.

 

 

5t[(t – 1)(t + 4)]

 

Use the Distributive Property to pull the common factor (t – 1) from both groups.

Answer

 

5t(t – 1)(t + 4)

 

 

 

Negative Terms

 

We will also face some problems in which a is negative, as in -4h2 + 11h + 3. It often makes sense to factor out -1 as the first step in factoring, as doing so will change the sign of ax2 from negative to positive, making the remaining trinomial easier to factor.

 

Example

Problem

 

Factor -4h2 + 11h + 3

 

 

-1(4h2 – 11h – 3)

 

Factor -1 out of the trinomial. Notice that the signs of all three terms have changed.

 

 

-1(4h2 – 12h + 1h – 3)

 

Expand the term -11h, by finding two integers whose sum is -11 and whose product is -12.

(-12 and 1)

 

-1[(4h2 – 12h) + (1h – 3)]

 

Group terms.

 

 

 

-1[4h(h – 3) + (h – 3)]

 

Use Distributive Property to pull the common factor, 4h, out of first grouped term. The second grouped term cannot be factored further. (Note that 1h is the same as h, so we don’t need to keep writing the 1.)

 

 

-1[(h – 3)(4h + 1)]

 

Use Distributive Property to pull common factor, (h – 3), out of both terms.

Answer

-1(h – 3)(4h + 1)

 

 

 

Summary

 

Sometimes trinomials will be in the form ax2 + bx + c, in which a is a coefficient other than 1. To factor these trinomials, find two integers, r and s, whose sum is b and whose product is ac. Then rewrite the trinomial as ax2 + rx + sx + c and use grouping and the Distributive Property to factor the polynomial.

 

In cases when a, b, and c are all multiples of the same integer, it often makes sense to pull that integer out as the first step in factoring. Extra variables common to the terms may also be pulled out in the same way. Similarly, when ax2 is negative, a common first step is to factor -1 out of the whole trinomial before continuing.