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Identify the best method to use to factor each polynomial:

  1. a b + a + 4 b + 4
  2. 3 k 2 + 15
  3. p 2 + 9 p + 8

factor using grouping no method undo using FOIL

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Factor trinomials of the form ax 2 + bx + c With a gcf

Now that we have organized what we’ve covered so far, we are ready to factor trinomials whose leading coefficient is not 1, trinomials of the form a x 2 + b x + c .

Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods in the last section. Let’s do a few examples to see how this works.

Watch out for the signs in the next two examples.

Factor completely: 2 n 2 8 n 42 .

Solution

Use the preliminary strategy.

Is there a greatest common factor? 2 n 2 8 n 42 Yes, GCF = 2. Factor it out. 2 ( n 2 4 n 21 ) Inside the parentheses, is it a binomial, trinomial, or are there more than three terms? It is a trinomial whose coefficient is 1, so undo FOIL. 2 ( n ) ( n ) Use 3 and −7 as the last terms of the binomials. 2 ( n + 3 ) ( n 7 )

Factors of −21 Sum of factors
1 , −21 1 + ( −21 ) = −20
3 , −7 3 + ( −7 ) = −4 *

Check.

2 ( n + 3 ) ( n 7 )

2 ( n 2 7 n + 3 n 21 )

2 ( n 2 4 n 21 )

2 n 2 8 n 42

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Factor completely: 4 m 2 4 m 8 .

4 ( m + 1 ) ( m 2 )

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Factor completely: 5 k 2 15 k 50 .

5 ( k + 2 ) ( k 5 )

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Factor completely: 4 y 2 36 y + 56 .

Solution

Use the preliminary strategy.

Is there a greatest common factor? 4 y 2 36 y + 56 Yes, GCF = 4. Factor it. 4 ( y 2 9 y + 14 ) Inside the parentheses, is it a binomial, trinomial, or are there more than three terms? It is a trinomial whose coefficient is 1. So undo FOIL. 4 ( y ) ( y ) Use a table like the one below to find two numbers that multiply to 14 and add to −9 . Both factors of 14 must be negative. 4 ( y 2 ) ( y 7 )

Factors of 14 Sum of factors
−1 , −14 −1 + ( −14 ) = −15
−2 , −7 −2 + ( −7 ) = −9 *

Check.

4 ( y 2 ) ( y 7 )

4 ( y 2 7 y 2 y + 14 )

4 ( y 2 9 y + 14 )

4 y 2 36 y + 42

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Factor completely: 3 r 2 9 r + 6 .

3 ( r 1 ) ( r 2 )

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Factor completely: 2 t 2 10 t + 12 .

2 ( t 2 ) ( t 3 )

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In the next example the GCF will include a variable.

Factor completely: 4 u 3 + 16 u 2 20 u .

Solution

Use the preliminary strategy.

Is there a greatest common factor? 4 u 3 + 16 u 2 20 u Yes, GCF = 4 u . Factor it. 4 u ( u 2 + 4 u 5 ) Binomial, trinomial, or more than three terms? It is a trinomial. So “undo FOIL.” 4 u ( u ) ( u ) Use a table like the table below to find two numbers that 4 u ( u 1 ) ( u + 5 ) multiply to −5 and add to 4 .

Factors of −5 Sum of factors
−1 , 5 −1 + 5 = 4 *
1 , −5 1 + ( −5 ) = −4

Check.

4 u ( u 1 ) ( u + 5 )

4 u ( u 2 + 5 u u 5 )

4 u ( u 2 + 4 u 5 )

4 u 3 + 16 u 2 20 u

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Factor completely: 5 x 3 + 15 x 2 20 x .

5 x ( x 1 ) ( x + 4 )

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Factor completely: 6 y 3 + 18 y 2 60 y .

6 y ( y 2 ) ( y + 5 )

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Factor trinomials using trial and error

What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Let’s factor the trinomial 3 x 2 + 5 x + 2 .

From our earlier work we expect this will factor into two binomials.

3 x 2 + 5 x + 2 ( ) ( )

We know the first terms of the binomial factors will multiply to give us 3 x 2 . The only factors of 3 x 2 are 1 x , 3 x . We can place them in the binomials.

This figure has the polynomial 3 x^ 2 +5 x +2. Underneath there are two terms, 1 x, and 3 x. Below these are the two factors x and (3 x) being shown multiplied.

Check. Does 1 x · 3 x = 3 x 2 ?

We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1 and 2. But we now have two cases to consider as it will make a difference if we write 1, 2, or 2, 1.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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