Factoring Polynomials

The two polynomials on the left side are called factors of the polynomial on the right side.

Expressing a given polynomial as a product of other polynomials, that is, finding the factors of a polynomial, is called factoring.

If a polynomial cannot be written as the product of two other polynomials (excluding 1 and -1), then the polynomial is said to be prime. When a polynomial has been written as a product consisting only of prime factors, it is said to be factored completely.

The first factor to look for in a factoring problem is a common monomial factor present in each term of the polynomial. If one is present, use the Distributive Property to factor it out. Continue factoring out monomial factors until none are left.

Factor the Difference of Two Squares and the Sum and Difference of Two Cubes

Difference of Two Squares	x² - a² = (x - a)(x + a)
Perfect Squares	x² - 2ax + a² = (x - a)²
Sum of two cubes	x³ + a³ = (x + a)(x² - ax + a²)
Difference of two cubes	x³ - a³ = (x - a)(x² + ax + a²)

Factor by Grouping

Sometimes a common factor does not occur in every term of the polynomial, but in each of several groups of terms that together make up the polynomial. When this happens, the common factor can be factored out of each group by means of the Distributive Property. This technique is called factoring by grouping.

Distributive Property
(a + b) ⋅ c = a ⋅ c + b ⋅ c

Factor completely by grouping: (x² + 2)⋅x + (x² + 2)⋅3

Notice the common factor x² + 2. By applying the Distributive Property, we have

Let (x² + 2) = c
(x² + 2)⋅x + (x² + 2)⋅3
= c⋅x + c⋅3
= c ⋅ (x² + 3)
Substitute back (x² + 2) for c

(x² + 2)⋅x + (x² + 2)⋅3 = (x² + 2)(x² + 3)

Example 1

Factor completely by grouping: 3(x - 1)²(x + 2)⁴ + 4(x - 1)³(x + 2)³

Example 2

Factor completely by grouping: x³ - 4x² + 2x - 8

To see if factoring by grouping will work, group the first two terms and the last two terms. Then look for a common factor in each group. In this example, we can factor x² from x³ - 4x² and 2 from 2x - 8. The remaining factor in each case is the same, x - 4. This means that factoring by grouping will work, as follows:

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

Factor a Second-Degree Polynomial

The idea behind factoring a second-degree polynomial like x² + Bx + C is to see whether it can be made equal to the product of two, possibly equal, first-degree polynomials.

To factor a second-degree polynomial x² + Bx + C, find integers whose product is C and whose sum is B. That is, if there are numbers a, b, where ab = C and a + b = B,then

x² + Bx + C = (x + a)(x + b)

Example 1

Factoring a Trinomial

Factor completely: x² + 7x + 10

First, determine all pairs of integers whose product is 10 and then compute their sums.

Integers whose product is 10:

10, 1 -10, -1 2, 5 -2, -5

Sum:

11 -11 7 -7

The integers 2 and 5 have a product of 10 and add up to 7.

x² + 7x + 10 = (x + 5)(x + 2)

Example 2

Identifying a Prime Polynomial

Show that x² + 9 is prime.

First, list the pairs of integers whose product is 9 and then compute their sums.

Integers whose product is 9:

9, 1 -9, -1 3, 3 -3, -3

Sum:

10 -10 6 -6

Since the coefficient of the middle term in x² + 9 = x² + 0x + 9 is 0 and none of the sums equals 0, we conclude that x² + 9 is prime.

THEOREM

Any polynomial of the form x² + a², a real, is prime.

Factor a Second-Degree Polynomial: Ax² + Bx + C, A ≠ 1

Find the value of AC.
Find a pair of integers whose product is AC that add up to B. That is, find a and b so that ab = AC and a + b = B.
Write Ax² + Bx + C = Ax² + ax + bx + C

Example 3

Factor completely: 2x² + 5x + 3

Comparing 2x² + 5x + 3 to Ax² + Bx + C , we find that A = 2, B = 5 , and C = 3 .

The value of AC is 2 ⋅ 3 = 6
Determine the pairs of integers whose product is AC = 6 and compute their sums.

Integers whose product is 6:

2, 3 -2, -3 6, 1 -6, -1

Sum:

5 -5 7 -7
The integers whose product is 6 that add up to B = 5 are 2 and 3.
2x² + 5x + 3 = 2x² + 2x + 3x + 3
Factor by grouping:
2x² + 2x + 3x + 3
= 2x(x + 1) + 3(x + 1)
= (2x + 3)(x + 1)

Summary

Type of polynomial

Method

Example

Any polynomial
Look for common monomial factors. (Always do this first!)
6x² + 9x = 3x(2x + 3)

Binomials of degree 2 or higher
Check for a special product.
Difference of two squares, x² - a²
Difference of two cubes, x³ - a³
Sum of two cubes, x³ + a³
x² - 16 = (x - 4)(x + 4)
x³ - 64 = (x - 4)(x² + 4x + 16)
x³ + 27 = (x + 3)(x² - 3x + 9)

Trinomials of degree 2
Check for a perfect square (x ± a)²

Factoring x² + Bx + C
Factoring Ax² + Bx + C, A ≠ 1
x² + 8x + 16 = (x + 4)²
x² - 10x + 25 = (x - 5)²
x² - x - 2 = (x - 2)(x + 1)
6x² + x - 1 = (2x + 1)(3x - 1)

Four or more terms
Grouping
2x³ - 3x² + 4x - 6 = (2x - 3)(x² + 2)

Complete the Square

The idea behind completing the square in one variable is to “adjust” an expression of the form x² + bx to make it a perfect square. Perfect squares are trinomials of the form

x² + 2ax + a² = (x + a)²
or
x² - 2ax + a² = (x - a)²

We do it by adding a number. For example, to make x + 6x a perfect square, add 9. But how do we know to add 9? If we divide the coefficient on the first-degree term, 6, by 2, and then square the result, we obtain 9. This approach works in general.

Completing the Square

Identify the coefficient of the first-degree term. Multiply this coefficient by ½ and then square the result.

That is, determine the value of b in x² + bx and compute

Figure 1

Look at the square in Figure 1. Its area is (y + 4)². The yellow area is y² and each orange area is 4y (for a total area of 8y). The sum of these areas is y² + 8y.

To complete the square, we need to add the area of the green region: 4⋅4 = 16.

As a result, y² + 8y + 16 = (y + 4)².

Example 1

Determine the number that must be added to each expression to complete the square. Then factor the expression.

- y² + 8y
- (½ ⋅ 8)² = 16
- y² + 8y + 16
- (y + 4)²
- p² - 5p
- (½ ⋅ 5)² = 25/4
- p² - 5p + 25/4
- (p + 5/2)²