Factoring: Special Cases
Learning Objective(s)
· Factor trinomials that are perfect squares.
· Factor binomials in the form of the difference of squares.
Introduction
One of the keys to factoring is finding patterns between the trinomial and the factors of the trinomial. Learning to recognize a few common polynomial types will lessen the amount of time it takes to factor them. Knowing the characteristic patterns of special products—trinomials that come from squaring binomials, for example—provides a shortcut to finding their factors.
Perfect squares are numbers that are the result of a whole number multiplied by itself or squared. For example 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 are all perfect squares—they come from squaring each of the numbers from 1 to 10. Notice that these perfect squares can also come from squaring the negative numbers from −1 to −10, as (−1)( −1) = 1, (−2)( −2) = 4, (−3)( −3) = 9, and so on.
A perfect square trinomial is a trinomial that is the result of a binomial multiplied by itself or squared. For example, (x + 3)2 = (x + 3)(x + 3) = x2 + 6x + 9. The trinomial x2 + 6x + 9 is a perfect square trinomial. Let’s factor this trinomial using the methods you have already seen.
| Example | ||
| Problem | Factor x2 + 6x + 9. | |
| x2 + 3x + 3x + 9 | Rewrite 6x as 3x + 3x, as 3 • 3 = 9, the last term, and 3 + 3 = 6, the middle term. | |
| (x2 + 3x) + (3x + 9) | Group pairs of terms. | |
| x(x + 3) + 3(x + 3) | Factor x out of the first pair, and factor 3 out of the second pair. | |
| (x + 3)(x + 3) or (x + 3)2 | Factor out x + 3. (x + 3)(x + 3) can also be written as (x + 3)2. | |
| Answer | (x + 3)(x + 3) or (x + 3)2 | |
Notice that in the trinomial x2 + 6x + 9, the a and c terms are each a perfect square, as x2 = x • x, and 9 = 3 • 3. Also the middle term is twice the product of the x and 3 terms, 2(3)x = 6x.
Let’s look at a slightly different example next. The above example shows how (x + 3)2 = x2 + 6x + 9. What do you suppose (x – 3)2 equals? Using what you know about multiplying binomials, you see the following.
| (x – 3)2 |
| (x – 3)(x – 3) |
| x2 – 3x – 3x + 9 |
| x2 – 6x + 9 |
Look: (x + 3)2 = x2 + 6x + 9, and (x – 3)2 = x2 – 6x + 9! Here 9 can be written as (−3)2, so the middle term is 2(−3)x = −6x. So when the sign of the middle term is negative, the trinomial may be factored as (a – b)2.
Let’s try one more example: 9x2 – 24x + 16. Notice that 9x2 is a perfect square, as (3x)2 = 9x2 and that 16 is a perfect square, as 42 = 16. However, the middle term, –24x is negative, so try 16 = (−4)2. In this case, the middle term is 2(3x)( −4) = −24x. So the trinomial 9x2 – 24x + 16 is a perfect square and factors as (3x – 4)2.
You can also continue to factor using grouping as shown below.
| Example | ||
| Problem | Factor 9x2 – 24x + 16. | |
| 9x2 – 12x – 12x + 16 | Rewrite −24x as −12x – 12x. | |
| (9x2 – 12x) + (-12x + 16) | Group pairs of terms. (Keep the negative sign with the 12.) | |
| 3x(3x – 4) – 4(3x – 4) | Factor 3x out of the first group, and factor out −4 from the second group. | |
| (3x – 4)(3x – 4) or (3x – 4)2 | Factor out (3x – 4). (3x – 4)(3x – 4) can also be written as (3x – 4)2. | |
| Answer | (3x – 4)2 | |
Notice that if you had factored out 4 rather than −4, the 3x – 4 factor would have been −3x + 4, which is the opposite of 3x – 4. By factoring out the −4, the factors from the grouping come out the same, both as 3x – 4. We need that to happen if we are going to pull a common grouping factor out for our next step.
The pattern for factoring perfect square trinomials lead to this general rule.
Perfect Square Trinomials
A trinomial in the form a2 + 2ab + b2 can be factored as (a + b)2.
A trinomial in the form a2 – 2ab + b2 can be factored as (a – b)2.
Examples:
The factored form of 4x2 + 20x + 25 is (2x + 5)2.
The factored form of x2 – 10x + 25 is (x – 5)2.
Let’s factor a trinomial using the rule above. Once you have determined that the trinomial is indeed a perfect square, the rest is easy. Notice that the c term is always positive in a perfect trinomial square.
| Example | ||
| Problem | Factor x2 – 14x + 49. | |
| x2 – 14x + 49 | Determine if this is a perfect square trinomial. The first term is a square, as x2 = x • x. The last term is a square as 7 • 7 = 49. Also −7 • −7 = 49. So, a = x and b = 7 or −7. | |
| −14x = −7x + −7x | The middle term is −2ab if we use b = 7, because −2x(7) = −14x. It is a perfect square trinomial. | |
| (x – 7)2 | Factor as (a – b)2. | |
| Answer | (x – 7)2 | |
You can, and should, always multiply to check the answer. (x – 7)2 = (x – 7)(x – 7) = x2 – 7x – 7x + 49 = x2 – 14x + 49.
Factor x2 – 12x + 36.
A) (x – 4)(x – 9)
B) (x + 6)2
C) (x – 6)2
D) (x + 6)(x – 6)
Show/Hide Answer
Factoring a Difference of Squares
The difference of two squares, a2 – b2, is also a special product that factors into the product of two binomials.
Let’s factor 9x2 – 4 by writing it as a trinomial, 9x2 + 0x – 4. Now you can factor this trinomial just as you have been doing.
9x2 + 0x – 4 fits the standard form of a trinomial, ax2 + bx + c. Let’s factor this trinomial the same way you would any other trinomial. Find the factors of ac (9 • −4 = −36) whose sum is b, in this case, 0.
| Factors of −36 | Sum of the factors |
| 1 • -36 = −36 | 1 + (−36) −35 |
| 2 • −18 = −36 | 2 + (−18) = −16 |
| 3 • −12 = −36 | 3 + (−12) = −9 |
| 4 • −9 = −36 | 4 + (−9) = −5 |
| 6 • −6 = −36 | 6 + (−6) = 0 |
| 9 • −4 = −36 | 9 + (−4) = 5 |
| … | … |
There are more factors, but you have found the pair that has a sum of 0, 6 and −6. You can use these to factor 9x2 – 4.
| Example | ||
| Problem | Factor 9x2 – 4. | |
| 9x2 + 0x – 4 9x2 – 6x + 6x – 4 | Rewrite 0x as −6x + 6x. | |
| (9x2 – 6x) + (6x – 4) | Group pairs. | |
| 3x(3x – 2) + 2(3x – 2) | Factor 3x out of the first group. Factor 2 out of the second group. | |
| (3x – 2)(3x + 2) | Factor out (3x – 2). | |
| Answer | (3x – 2)(3x + 2) | |
Since multiplication is commutative, the answer can also be written as (3x + 2)(3x – 2).
You can check the answer by multiplying (3x – 2)(3x + 2) = 9x2 + 6x – 6x – 4 = 9x2 – 4.
Factoring a Difference of Squares
A binomial in the form a2 – b2 can be factored as (a + b)(a – b).
Examples
The factored form of x2 – 100 is (x + 10)(x – 10).
The factored form of 49y2 – 25 is (7y + 5)(7y – 5).
Let’s factor the difference of two squares using the above rule. Once you have determined that you have the difference of two squares, you just follow the pattern.
| Example | ||
| Problem | Factor 4x2 – 36. | |
| 4x2 – 36 | 4x2 = (2x)2, so a = 2x 36 = 62, so b = 6 And 4x2 – 36 is the difference of two squares. | |
| (2x + 6)(2x – 6) | Factor as (a + b)(a – b). | |
| Answer | (2x + 6)(2x – 6) | |
Check the answer by multiplying: (2x + 6)(2x – 6) = 4x2 – 12x + 12x – 36 = 4x2 – 36.
Factor 4b2 – 25.
A) (2b – 25)(2b + 1)
B) (2b + 5)2
C) (2b – 5)2
D) (2b + 5)(2b – 5)
Show/Hide Answer
Notice that you cannot factor the sum of two squares, a2 + b2. You might be tempted to factor this as (a + b)2, but check it by multiplying: (a + b)2 = (a + b)(a + b) =
a2 + ab + ab + b2 = a2 + 2ab + b2, NOT a2 + b2.
Summary
Learning to identify certain patterns in polynomials helps you factor some “special cases” of polynomials quickly. The special cases are:
- trinomials that are perfect squares, a2 + 2ab + b2 and a2 – 2ab + b2, which factor as (a+ b)2 and (a – b)2, respectively;
- binomials that are the difference of two squares, a2 – b2, which factors as (a + b)(a – b).
For some polynomials, you may need to combine techniques (looking for common factors, grouping, and using special products) to factor the polynomial completely.
