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6.6: Factoring Strategy

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When you are concentrating on factoring problems of a single type, after doing a few you tend to get into a rhythm, and the remainder of the exercises, because they are similar, seem to flow. However, when you encounter a mixture of factoring problems of different types, progress is harder. The goal of this section is to set up a strategy to follow when attacking a general factoring problem.

If it hasn’t already been done, it is helpful to arrange the terms of the given polynomial in some sort of order (descending or ascending). Then you want to apply the following guidelines.

Factoring Strategy

These steps should be followed in the order that they appear.

  1. Factor out the greatest common factor (GCF).
  2. Look for a special form.
    1. If you have two perfect squares separated by a minus sign, use the difference of squares pattern to factor: a2b2=(a+b)(ab)
    2. If you have a trinomial whose first and last terms are perfect squares, you should suspect that you have a perfect square trinomial. Take the square roots of the first and last terms and factor as follows.a2+2ab+b2=(a+b)2Be sure to check that the middle term is correct.
  3. If you have a trinomial of the form ax2+bx+c, use the ac-method to factor.
  4. If you have a four-term expression, try to factor by grouping.

Once you’ve applied the above strategy to the given polynomial, it is quite possible that one of your resulting factors will factor further. Thus, we have the following rule.

Factor completely

The factoring process is not complete until none of your remaining factors can be factored further. This is the meaning of the phrase, “factor completely.”

Finally, a very good word of advice.

Check your factoring by multiplying

Once you’ve factored the given polynomial completely, it is a very good practice to check your result. If you multiply to find the product of your factors, and get the original given polynomial as a result, then you know your factorization is correct.

It’s a bit more work to check your factorization, but it’s worth the effort. It helps to eliminate errors and also helps to build a better understanding of the factoring process. Remember, factoring is “unmultiplying,” so the more you multiply, the better you get at factoring.

Let’s see what can happen when you don’t check your factorization!

Warning! The following solution is incorrect!

Factor: 2x4+8x2

Solution: Factor out the GCF

2x4+8x2=2x2(x2+4)=2x2(x+2)2

Note that this student did not bother to check his factorization. Let’s do that for him now.

Check: Multiply to check. Remember, when squaring a binomial, there is a middle term.

2x2(x+2)2=2x2(x2+4x+4)=2x4+8x3+8x2

This is not the same as the original polynomial 2x4+8x2, so the student’s factorization is incorrect. Had the student performed this check, he might have caught his error, provided of course, that he multiplies correctly during the check.

The correct factorization follows.

2x4+8x2=2x2(x2+4)

The sum of squares does not factor, so we are finished.

Check: Multiply to check.

2x2(x2+4)=2x4+8x2

This is the same as the original polynomial 2x4+8x2, so this factorization is correct.

Example 6.6.1

Factor completely: 3x6+3x2

Solution

The first rule of factoring is “Factor out the GCF.” The GCF of 3x6 and 3x2 is 3x2, so we could factor out 3x2.3x6+3x2=3x2(x4+1)This is perfectly valid, but we don’t like the fact that the second factor starts with x4. Let’s factor out 3x2 instead.3x6+3x2=3x2(x41)The second factor is the difference of two squares. Take the square roots, separating one pair with a plus sign, one pair with a minus sign. =3x2(x2+1)(x21) The sum of squares does not factor. But the last factor is the difference of two squares. Take the square roots, separating one pair with a plus sign, one pair with a minus sign.=3x2(x2+1)(x+1)(x1)

Check: Multiply to check the result.

3x2(x2+1)(x+1)(x1)=3x2(x2+1)(x21)=3x2(x41)=3x6+3x2

The factorization checks.

Exercise 6.6.1

Factor completely: 4x7+64x3

Answer

4x3(x2+4)(x+2)(x2)

Example 6.6.2

Factor completely: x3y+9xy3+6x2y2

Solution

The first rule of factoring is “Factor out the GCF.” The GCF of x3y, 9xy3, and 6x2y2 is xy, so we factor out xy.x3y+9xy3+6x2y2=xy(x2+9y2+6xy)Let’s order that second factor in descending powers of x.=xy(x2+6xy+9y2)The first and last terms of the trinomial factor are perfect squares. We suspect we have a perfect square trinomial, so we take the square roots of the first and last terms, check the middle term, and write:=xy(x+3y)2Thus, x3y+9xy3+6x2y2=xy(x+3y)2.
Check: Multiply to check the result.xy(x+3y)2=xy(x2+6xy+9y2)=x3y+6x2y2+9xy3
Except for the order, this result is the same as the given polynomial. The factorization checks.

Exercise 6.6.2

Factor completely: 3a2b4+12a4b212a3b3

Answer

3a2b2(2ab)2

Example 6.6.3

Factor completely: 2x348x+20x2

Solution

In the last example, we recognized a need to rearrange our terms after we pulled out the GCF. This time, let’s arrange our terms in descending powers of x right away.2x348x+20x2=2x3+20x248xNow, let’s factor out the GCF.=2x(x2+10x24) The last term of the trinomial factor is not a perfect square. Let’s move to the ac-method to factor. The integer pair 2,12 has a product equal to ac=24 and a sum equal to b=10. Because the coefficient of x2 is one, this is a “drop in place” situation. We drop our pair in place and write:=2x(x2)(x+12) Thus, 2x348x+20x2=2x(x2)(x+12).

Check: Multiply to check the result. We use the FOIL method shortcut and mental calculations to speed things up.2x(x2)(x+12)=2x(x2+10x24)=2x3+20x248xExcept for the order, this result is the same as the given polynomial. The factorization checks.

Exercise 6.6.3

Add exercises text here.

Answer

3x2(x4)(x5)

Example 6.6.4

Factor completely: 2a213ab24b2

Solution

There is no common factor we can factor out. We have a trinomial, but the first and last terms are not perfect squares, so let’s apply the ac-method. Ignoring the variables for a moment, we need an integer pair whose product is ac=48 and whose sum is 13. The integer pair 3,16 comes to mind (if nothing comes to mind, start listing integer pairs). Break up the middle term into a sum of like terms using the integer pair 3,16, then factor by grouping

2a213ab24b2=2a2+3ab16ab24b2=a(2a+3b)8b(2a+3b)=(a8b)(2a+3b)

Thus, 2a213ab24b2=(a8b)(2a+3b).

Check: Multiply to check the result. We use the FOIL method shortcut and mental calculations to speed things up.(a8b)(2a+3b)=2a213ab24b2 This result is the same as the given polynomial. The factorization checks.

Exercise 6.6.4

Factor completely: 8x2+14xy15y2

Answer

(2x+5y)(4x3y)

Example 6.6.5

Factor completely: 30x4+38x320x2

Solution

The first step is to factor out the GCF, which in this case is 2x2.30x4+38x320x2=2x2(15x2+19x10)The first and last terms of the trinomial factor are not perfect squares, so let’s move again to the ac-method. Comparing 15x2+19x10 with ax2+bx+c, note that ac=(15)(10)=150. We need an integer pair whose product is 150 and whose sum is 19. The integer pair 6 and 25 satisfies these requirements. Because a1, this is not a “drop in place” situation, so we need to break up the middle term as a sum of like terms using the pair 6 and 25.=2x2(15x26x+25x10)Factor by grouping. Factor 3x out of the first two terms and 5 out of the third and fourth terms.=2x2(3x(5x2)+5(5x2))Finally, factor out the common factor 5x2.=2x2(3x+5)(5x2)Thus, 30x4+38x320x2=2x2(3x+5)(5x2).

Check: Multiply to check the result. Use the FOIL method to first multiply the binomials.2x2(3x+5)(5x2)=2x2(15x2+19x10)Distribute the 2x2.=30x4+38x320x2This result is the same as the given polynomial. The factorization checks.

Exercise 6.6.5

Factor completely: 36x3+60x2+9x

Answer

3x(6x+1)(2x+3)

Example 6.6.6

Factor completely: 8x5+10x472x390x2

Solution

Each of the terms is divisible by 3x3. Factor out 3x3.15x633x5240x4+528x3=3x3[5x311x280x+176]The second factor is a four-term expression. Factor by grouping.

=3x3[x2(5x11)16(5x11)]=3x3(x216)(5x11)

The factor x216 is a difference of two squares. Take the square roots, separate one pair with a plus, one pair with a minus.=3x3(x+4)(x4)(5x11) Thus, 15x633x5240x4+528x3=3x3(x+4)(x4)(5x11).

Check: Multiply to check the result.

3x3(x+4)(x4)(5x11)=3x3(x216)(5x11)=3x3(5x311x280x+176=15x633x5240x4+528x3

This result is the same as the given polynomial. The factorization checks.

Exercise 6.6.6

Factor completely: 15x633x5240x4+528x3

Answer

2x2(x3)(x+3)(4x+5)

Using the Calculator to Assist the ac-Method

When using the ac-method to factor ax2+bx+c and ac is a very large number, then it can be difficult to find a pair whose product is ac and whose sum in b. For example, consider the trinomial:12y211y36We need an integer pair whose product is ac=432 and whose sum is b=11. We begin listing integer pair possibilities, but the process quickly becomes daunting.

1,4322,216

Note that the numbers in the second column are found by dividing ac=432 by the number in the first column. We’ll now use this fact and the TABLE feature on our calculator to pursue the desired integer pair.

  1. Enter the expression 432/X into Y1 in the Y= menu (see the first image in Figure 6.6.1).
  2. Above the WINDOW button you’ll see TBLSET. Use the 2nd key, then press the WINDOW button to access the menu shown in the second image of Figure 6.6.1. Set TblStart=1, Tbl=1, then highlight AUTO for both the independent and dependent variables.
  3. Above the GRAPH button you’ll see TABLE. Use the 2nd key, then press the GRAPH button to access the table shown in the third image in Figure 6.6.1. Use the up- and down-arrow keys to scroll through the contents of the table. Note that you can ignore most of the pairs, because they are not both integers. Pay attention only when they are both integers. In this case, remember that you are searching for a pair whose sum is b=11. Note that the pair 16,27 shown in the third image of Figure 6.6.1 is the pair we seek.
fig 6.6.1.png
Figure 6.6.1: Using the TABLE feature to assist the ac-method.

Now we can break the middle term of 12y211y36 into a sum of like terms using the ordered pair 16,27, then factor by grouping.

12y211y36=12y2+16y27y36=4y(3y+4)9(3y+4)=(4y9)(3y+4)

Check: Use the FOIL method shortcut and mental calculations to multiply.(4y9)(3y+4)=12y211y36The factorization checks.


This page titled 6.6: Factoring Strategy is shared under a CC BY-NC-ND 3.0 license and was authored, remixed, and/or curated by David Arnold via source content that was edited to the style and standards of the LibreTexts platform.

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