Section 5: Factoring Polynomials

Factor a polynomial by factoring out a common factor.

In this lesson we summarize and apply the factoring methods presented in the previous sections. It is important to make sure that a problem is completely factored. This means that in some cases you may need to factor more than once or apply more than one factoring method.

Recall that a polynomial is completely factored when:

1. the polynomial is written as a product of factors which are prime polynomials with integer coefficients
2. none of the polynomial factors can be factored further

In order to help you organize your factoring methods, the following steps may help:

Factoring A Polynomial

1. Factor out any common factors (GCF).
2. If the polynomial is a binomial, check to see if it is the difference of squares, the difference of cubes, or the sum of cubes.
   If the polynomial is a trinomial, check to see if it is a perfect square trinomial. If it is not, then try factoring using the AC Method.
   If the polynomial has more than three terms, try to factor by grouping.
3. Check the factored form by multiplying.

Example 1. Factor 9p + 45

Solution.

Notice in this example that there is a GCF of 9. This means we can factor out (or divide) 9 from each term in the polynomial.

9 p + 45 = 9⋅p + 9⋅5 = 9(p + 5)

Therefore, 9p + 45 = 9 (p + 5).

Example 2. Factor 8m²p² + 4mp

Solution.

Notice in this example that there is a GCF of 4mp. This means we can factor out (or divide) 4mp from each term in the polynomial.

8m²p² + 4mp = (4mp)(2mp) + (4mp)(1) = 4mp(2mp + 1)

Therefore, 8m²p² + 4mp = 4mp( 2mp + 1).

Factor a Binomial

Example 3. Factor 64m² – 9n²

Solution.

Notice in this example that we have the subtraction of two terms where both terms are perfect squares:
64p² = ( 8p )² = (8p)(8p) and 9n² = (3n)²= (3n)(3n).

This means we can use the difference of squares formula to factor this problem.

In order to do so, we open two sets of parenthesis and include in each parenthesis one factor of each squared term. Next we include in one parenthesis a “+” sign and a “–” sign in the other.

64p² – 9n2 = ( 8p )² – (3n)² = ( 8m + 3n )(8m – 3n) = ( 8m + 3n)(8m – 3n).
Therefore, Therefore, 64m² – 9n² = (8m + 3n )(8m – 3n) .

Example 4. Factor 8p³ – 27

Solution.

Notice in this example that we have the subtraction of two terms where both terms are perfect cubes: 8p³ = (2p)³ and 27 = (3)³ .

This means we can use the difference of cubes formula to factor this problem.

In order to do so, we open two sets of parentheses. In the first set of parentheses, we include one factor of each cubed term and the sign will be the same as that of the original problem. In the second parentheses, the first term will be the same as the first term in the first parentheses squared; the last term will be the same as the last term in the first parentheses squared; the middle term will be the product
of the terms in the first parentheses and the sign will be the opposite of that in the first parentheses.

8p³ – 27 = ( 2p )³ – (3)³ = ( 2p – 3 )[( 2p ) 2 + ( 2 p • 3 ) + 3² ] = ( 2p – 3 )(4p² + 6p + 9 )
Therefore, 8p³ – 27 = (2p – 3)(4p² + 6p + 9).

Factor a Trinomial (No Fuss Factoring Method)

Example 5. Factor 2k2 – k – 6

Solution.

In this example, a = 2, b = 1 and c = 6.
So we factor by using the AC Method.

Step 1: a • c = 2 • 6 = 12

Step 2: We need to find two numbers that multiply (product) to give you –12 and add (sum) to give you –1.
We can make a table to list of all the ways we can multiply to get –12 and check to see if the sum is –1:

Notice that the pair of numbers, 3 and –4, will multiply to give you –12 and add to give you –1. So these are the correct two numbers.
 

Step 3: We need to open two sets of parenthesis and fill in the variables and the numbers found in step 2. (Notice in the trinomial you have 2k² , this means 2•k•k. So we put one of each variable in each parenthesis.Also, for now, we will include 2 in each parenthesis (we will simplify it later). We don’t need to worry about the variable in –k because this term will appear if you multiply out the answer)

(2k + 3)(2k 4)

Step 4: Notice the expression in the second parenthesis is not completely simplified. Therefore, we need to reduce further. The terms in the second parenthesis can be divided by 2

(2k + 3) (2k– 4) ⇒ Divide out 2 from (2k – 4 ) ⇒ (k – 2)
= (2k + 3) (k – 2)

Therefore, 2k2 – k – 6 = ( 2k + 3 )( k – 2 ) .

Example 6. Factor 28z2 + 6z – 10

Solution.

Notice in this example that there is a GCF of 2. This means we can factor out (or divide) 2 from each term in the polynomial.

Factor 28z² + 6z – 10 = 2(14z² + 3z – 5)

Notice that we can still factor further because (14z2 + 3z – 5 ) can be factored using the AC Method. In this example, a = 14, b = 3 and c = 5.

Step 1: a • c = 14 • 5 = 70

Step 2: We need to find two numbers that multiply (product) to give you –70 and add (sum) to give you 3.
We can make a table to list of all the ways we can multiply to get –70 and check to see if the sum is 3:

Notice that the pair of numbers, –7 and 10, will multiply to give you –70 and add to give you 3. So these are the correct two numbers.

Step 3: We need to open two sets of parenthesis and fill in the variables and the numbers found in step 2.

((Notice in the trinomial you have 14z2, this means 14 •z •z. So we put one of each variable in each parenthesis. Also, for now, we will include 14 in each parenthesis (we will simplify it later in step 4).

We don’t need to worry about the variable in 3z because this term will appear if you multiply out the answer)

(14z − 7)(14z + 10)

Step 4: Notice the expression in the second parenthesis is not completely simplified. Therefore, we need to reduce further. The terms in the first parenthesis can be divided by 7 and the terms in the second parenthesis can be divided by 2

(14z –7) (14z +10) ⇒Divide out 7 from (14z –7)⇒ (2z– 1)
⇒Divide out 2 from (14z +10 )⇒ (7z + 5)

= (2z – 1)(7z + 5)

Therefore, 28z² + 6z – 10 = 2 ( 2z – 1) (7z + 5) .

Test Your Knowledge by opening up the Test Yourself Activity.