Factoring Trinomials

Learning Objective(s)

·         Factor trinomials with a leading coefficient of 1.

·         Factor trinomials with a common factor.

·         Factor trinomials with a leading coefficient other than 1.

Introduction

A polynomial with three terms is called a trinomial. Trinomials often (but not always!) have the form x2 + bx + c. At first glance, it may seem difficult to factor trinomials, but you can take advantage of some interesting mathematical patterns to factor even the most difficult-looking trinomials.

So, how do you get from 6x2 + 2x – 20 to (2x + 4)(3x −5)? Let’s take a look.

Factoring Trinomials: x2 + bx + c

Trinomials in the form x2 + bx + c can often be factored as the product of two binomials. Remember that a binomial is simply a two-term polynomial. Let’s start by reviewing what happens when two binomials, such as (x + 2) and (x + 5), are multiplied.

 Example Problem Multiply (x + 2)(x + 5). (x + 2)(x + 5) Use the FOIL method to multiply binomials. x2 + 5x + 2x +10 Then combine like terms 2x and 5x. Answer x2 + 7x +10

Factoring is the reverse of multiplying. So let’s go in reverse and factor the trinomial x2 + 7x + 10. The individual terms x2, 7x, and 10 share no common factors. So look at rewriting x2 + 7x + 10 as x2 + 5x + 2x + 10.

And, you can group pairs of factors:              (x2 + 5x) + (2x + 10)

Factor each pair:                                             x(x + 5) + 2(x + 5)

Then factor out the common factor x + 5:      (x + 5)(x + 2)

Here is the same problem done in the form of an example:

 Example Problem Factor x2 + 7x +10. x2 + 5x + 2x +10 Rewrite the middle term 7x as 5x + 2x. x(x + 5) + 2(x + 5) Group the pairs and factor out the common factor x from the first pair and 2 from the second pair. (x + 5)(x + 2) Factor out the common factor (x + 5). Answer (x + 5)(x + 2)

How do you know how to rewrite the middle term? Unfortunately, you can’t rewrite it just any way. If you rewrite 7x as 6x + x, this method won’t work. Fortunately, there's a rule for that.

 Factoring Trinomials in the form x2 + bx + c   To factor a trinomial in the form x2 + bx + c, find two integers, r and s, whose product is c and whose sum is b.   Rewrite the trinomial as x2 + rx + sx + c and then use grouping and the distributive property to factor the polynomial. The resulting factors will be (x + r) and (x + s).

For example, to factor x2 + 7x +10, you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).

Look at factor pairs of 10: 1 and 10, 2 and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite 7x as 2x + 5x, and continue factoring as in the example above. Note that you can also rewrite 7x as 5x + 2x. Both will work.

Let’s factor the trinomial x2 + 5x + 6. In this polynomial, the b part of the middle term is 5 and the c term is 6. A chart will help us organize possibilities. On the left, list all possible factors of the c term, 6; on the right you'll find the sums.

 Factors whose product is 6 Sum of the factors 1 • 6 = 6 1 + 6 = 7 2 • 3 = 6 2 + 3 = 5

There are only two possible factor combinations, 1 and 6, and 2 and 3. You can see that 2 + 3 = 5. So 2x + 3x = 5x, giving us the correct middle term.

 Example Problem Factor x2 + 5x + 6. x2 + 2x + 3x + 6 Use values from the chart above. Replace 5x with 2x + 3x. (x2 + 2x) + (3x + 6) Group the pairs of terms. x(x + 2) + (3x + 6) Factor x out of the first pair of terms. x(x + 2) + 3(x + 2) Factor 3 out of the second pair of terms. (x + 2)(x + 3) Factor out (x + 2). Answer (x + 2)(x + 3)

Note that if you wrote x2 + 5x + 6 as x2 + 3x + 2x + 6 and grouped the pairs as (x2 + 3x) + (2x + 6); then factored, x(x + 3) + 2(x + 3), and factored out x + 3, the answer would be (x + 3)(x + 2). Since multiplication is commutative, the order of the factors does not matter. So this answer is correct as well; they are equivalent answers.

Finally, let’s take a look at the trinomial x2 + x – 12. In this trinomial, the c term is −12. So look at all of the combinations of factors whose product is −12. Then see which of these combinations will give you the correct middle term, where b is 1.

 Factors whose product is −12 Sum of the factors 1 • −12 = −12 1 + −12 = −11 2 • −6 = −12 2 + −6 = −4 3 • −4 = −12 3 + −4 = −1 4 • −3 = −12 4 + −3 = 1 6 • −2 = −12 6 + −2 = 4 12 • −1 = −12 12 + −1 = 11

There is only one combination where the product is −12 and the sum is 1, and that is when r = 4, and s = −3. Let’s use these to factor our original trinomial.

 Example Problem Factor x2 + x – 12 x2 + 4x + −3x – 12 Rewrite the trinomial using the values from the chart above. Use values r = 4 and s = −3. (x2 + 4x) + (−3x – 12) Group pairs of terms. x(x + 4) + (−3x – 12) Factor x out of the first group. x(x + 4) – 3(x + 4) Factor −3 out of the second group. (x + 4)(x – 3) Factor out (x + 4). Answer (x + 4)(x – 3)

In the above example, you could also rewrite x2 + x – 12 as x2 3x + 4x – 12 first. Then factor x(x – 3) + 4(x – 3), and factor out (x – 3) getting (x – 3)(x + 4). Since multiplication is commutative, this is the same answer.

Factoring Tips

Factoring trinomials is a matter of practice and patience. Sometimes, the appropriate number combinations will just pop out and seem so obvious! Other times, despite trying many possibilities, the correct combinations are hard to find. And, there are times when the trinomial cannot be factored.

While there is no foolproof way to find the right combination on the first guess, there are some tips that can ease the way.

 Tips for Finding Values that Work   When factoring a trinomial in the form x2 + bx + c, consider the following tips.   Look at the c term first. o        If the c term is a positive number, then the factors of c will both be positive or both be negative. In other words, r and s will have the same sign. o        If the c term is a negative number, then one factor of c will be positive, and one factor of c will be negative. Either r or s will be negative, but not both.   Look at the b term second. o        If the c term is positive and the b term is positive, then both r and s are positive. o        If the c term is positive and the b term is negative, then both r and s are negative. o        If the c term is negative and the b term is positive, then the factor that is positive will have the greater absolute value. That is, if |r| > |s|, then r is positive and s is negative. o        If the c term is negative and the b term is negative, then the factor that is negative will have the greater absolute value. That is, if |r| > |s|, then r is negative and s is positive.

After you have factored a number of trinomials in the form x2 + bx + c, you may notice that the numbers you identify for r and s end up being included in the factored form of the trinomial. Have a look at the following chart, which reviews the three problems you have seen so far.

 Trinomial x2 + 7x + 10 x2 + 5x + 6 x2 + x - 12 r and s values r = + 5, s = + 2 r = + 2, s = + 3 r = + 4, s = –3 Factored form (x + 5)(x + 2) (x + 2)(x + 3) (x + 4)(x – 3)

Notice that in each of these examples, the r and s values are repeated in the factored form of the trinomial.

So what does this mean? It means that in trinomials of the form x2 + bx + c (where the coefficient in front of x2 is 1), if you can identify the correct r and s values, you can effectively skip the grouping steps and go right to the factored form. You may want to stick with the grouping method until you are comfortable factoring, but this is a neat shortcut to know about!

 Jess is trying to use the grouping method to factor the trinomial v2 – 10v + 21. How should she rewrite the central b term, −10v?   A) +7v + 3v   B) −7v – 3v   C) −7v + 3v   D) +7v – 3v   Show/Hide Answer

Identifying Common Factors

Not all trinomials look like x2 + 5x + 6, where the coefficient in front of the x2 term is 1. In these cases, your first step should be to look for common factors for the three terms.

 Trinomial Factor out Common Factor Factored 2x2 + 10x + 12 2(x2 + 5x + 6) 2(x + 2)(x + 3) −5a2 − 15a − 10 −5(a2 + 3a + 2) −5(a + 2)(a + 1) c3 – 8c2 + 15c c(c2 – 8c + 15) c(c – 5)(c – 3) y4 – 9y3 – 10y2 y2(y2 – 9y – 10) y2(y – 10)(y + 1)

Notice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.

 Example Problem Factor 3x3 – 3x2 – 90x. 3(x3 – x2 – 30x) Since 3 is a common factor for the three terms, factor out the 3. 3x(x2 – x – 30) x is also a common factor, so factor out x. 3x(x2 – 6x + 5x – 30) Now you can factor the trinomial x2 – x – 30. To find r and s, identify two numbers whose product is −30 and whose sum is −1.   The pair of factors is −6 and 5. So replace –x with −6x + 5x. 3x[(x2 – 6x) + (5x – 30)] Use grouping to consider the terms in pairs. 3x[(x(x – 6) + 5(x – 6)] Factor x out of the first group and factor 5 out of the second group. 3x(x – 6)(x + 5) Then factor out x – 6. Answer 3x(x – 6)(x + 5)

Factoring Trinomials: ax2 + bx + c

The general form of trinomials with a leading coefficient of a is ax2 + bx + c. Sometimes the factor of a can be factored as you saw above; this happens when a can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an x2 term, instead of an ax2 term.

However, if the coefficients of all three terms of a trinomial don’t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.

 Factoring Trinomials in the form ax2 + bx + c   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.

This is almost the same as factoring trinomials in the form x2 + bx + c, as in this form a = 1. Now you are looking for two factors whose product is a c, and whose sum is b.

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, you need to find two factors, r and s, whose sum is b (11) and whose product is ac (or 6 • 4 = 24). You 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, you can be certain that the two factors you're looking for are also positive numbers.)

 Factors whose product is 24 Sum of the factors 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 where the product is 24 and the sum is 11, and that is when r = 3, and s = 8. Let’s use these values to factor the original trinomial.

 Example Problem Factor 6z2 + 11z + 4. 6z2 + 3z + 8z + 4 Rewrite the middle term, 11z, as 3z + 8z (from the chart above.) (6z2 + 3z) + (8z + 4) Group pairs. Use grouping to consider the terms in pairs. 3z(2z + 1) + 4(2z + 1) Factor 3z out of the first group and 4 out of the second group. (2z + 1)(3z + 4) Factor out (2z + 1). Answer (2z + 1)(3z + 4)

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! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.

 Factor 3x2 + x – 2.   A) (3x + 2)(x – 1)   B) (3x – 2)(x + 1)   C) (3x + 1)(x – 2)   D) (3x – 1)(x + 2)   Show/Hide Answer

Negative Terms

In some situations, 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)

To factor the trinomial, you need to figure out how to rewrite 11h. The product of rs = 4 • 3 = 12, and the sum of

rs = 11.

 r • s = −12 r + s = −11 −12 • 1 = −12 −12 + 1 = −11 −6 • 2 = −12 −6 + 2 = −4 −4 • 3 = −12 −4 + 3 = −1

Rewrite the middle term 11h as 12h + 1h.

−1[(4h2 – 12h) + (1h – 3)]

Group terms.

−1[4h(h – 3) + 1(h – 3)]

Factor out 4h from the first pair. The second group cannot be factored further, but you can write it as +1(h – 3) since +1(h – 3) = (h – 3). This helps with factoring in the next step.

−1[(h – 3)(4h + 1)]

Factor out a common factor of (h – 3). Notice you are left with (h – 3)(4h + 1); the +1 comes from the term +1(h – 3) in the previous step.

−1(h – 3)(4h + 1)

Note that the answer above can also be written as (−h + 3)(4h + 1) or (h – 3)( −4h – 1) if you multiply −1 times one of the other factors.

Summary

Trinomials in the form x2 + bx + c can be factored by finding two integers, r and s, whose sum is b and whose product is c. Rewrite the trinomial as x2 + rx + sx + c and then use grouping and the distributive property to factor the polynomial.

When a trinomial is in the form of ax2 + bx + c, where a is a coefficient other than 1, look first for common factors for all three terms. Factor out the common factor first, then factor the remaining simpler trinomial. If the remaining trinomial is still of the form ax2 + bx + c, 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.

When ax2 is negative, you can factor −1 out of the whole trinomial before continuing.