Find all the factors of 72.
If you missed this problem, review Example 1.2.
Find the product: ( 3 y + 4 ) ( 2 y + 5 ) . ( 3 y + 4 ) ( 2 y + 5 ) .
If you missed this problem, review Example 5.28.
Simplify: −9 ( 6 ) ; −9 ( 6 ) ; −9 ( −6 ) . −9 ( −6 ) .
If you missed this problem, review Example 1.18.
You have already learned how to multiply binomials using FOIL . Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.
To figure out how we would factor a trinomial of the form x 2 + b x + c , x 2 + b x + c , such as x 2 + 5 x + 6 x 2 + 5 x + 6 and factor it to ( x + 2 ) ( x + 3 ) , ( x + 2 ) ( x + 3 ) , let’s start with two general binomials of the form ( x + m ) ( x + m ) and ( x + n ) . ( x + n ) .
| Foil to find the product. |
| Factor the GCF from the middle terms. |
| Our trinomial is of the form x 2 + b x + c . x 2 + b x + c . |
This tells us that to factor a trinomial of the form x 2 + b x + c , x 2 + b x + c , we need two factors ( x + m ) ( x + m ) and ( x + n ) ( x + n ) where the two numbers m and n multiply to c and add to b.
Factor: x 2 + 11 x + 24 . x 2 + 11 x + 24 .
Factor: q 2 + 10 q + 24 . q 2 + 10 q + 24 .
Factor: t 2 + 14 t + 24 . t 2 + 14 t + 24 .
Let’s summarize the steps we used to find the factors.
In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.
How do you get a positive product and a negative sum? We use two negative numbers.
Factor: y 2 − 11 y + 28. y 2 − 11 y + 28.
Again, with the positive last term, 28, and the negative middle term, −11 y , −11 y , we need two negative factors. Find two numbers that multiply 28 and add to −11 . −11 .
| y 2 − 11 y + 28 y 2 − 11 y + 28 | |
| Write the factors as two binomials with first terms y . y . | ( y ) ( y ) ( y ) ( y ) |
| Find two numbers that: multiply to 28 and add to −11. |
| Factors of 28 28 | Sum of factors |
|---|---|
| − 1 , −28 − 1 , −28 − 2 , −14 − 2 , −14 − 4 , −7 − 4 , −7 | − 1 + ( − 28 ) = −29 − 1 + ( − 28 ) = −29 − 2 + ( − 14 ) = −16 − 2 + ( − 14 ) = −16 − 4 + ( − 7 ) = −11 * − 4 + ( − 7 ) = −11 * |
| Use −4 , −7 −4 , −7 as the last terms of the binomials. | ( y − 4 ) ( y − 7 ) ( y − 4 ) ( y − 7 ) |
| Check: ( y − 4 ) ( y − 7 ) y 2 − 7 y − 4 y + 28 y 2 − 11 y + 28 ✓ ( y − 4 ) ( y − 7 ) y 2 − 7 y − 4 y + 28 y 2 − 11 y + 28 ✓ |
Factor: u 2 − 9 u + 18 . u 2 − 9 u + 18 .
Factor: y 2 − 16 y + 63 . y 2 − 16 y + 63 .
Now, what if the last term in the trinomial is negative? Think about FOIL . The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.
How do you get a negative product and a positive sum? We use one positive and one negative number.
When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.
Factor: 2 x + x 2 − 48 . 2 x + x 2 − 48 .
| 2 x + x 2 − 48 2 x + x 2 − 48 | |
| First we put the terms in decreasing degree order. | x 2 + 2 x − 48 x 2 + 2 x − 48 |
| Factors will be two binomials with first terms x . x . | ( x ) ( x ) ( x ) ( x ) |
| Factors of −48 −48 | Sum of factors |
|---|---|
| −1 , 48 −1 , 48 −2 , 24 −2 , 24 −3 , 16 −3 , 16 −4 , 12 −4 , 12 −6 , 8 −6 , 8 | −1 + 48 = 47 −1 + 48 = 47 −2 + 24 = 22 −2 + 24 = 22 −3 + 16 = 13 −3 + 16 = 13 −4 + 12 = 8 −4 + 12 = 8 −6 + 8 = 2 * −6 + 8 = 2 * |
| Use −6 , 8 as the last terms of the binomials. Use −6 , 8 as the last terms of the binomials. | ( x − 6 ) ( x + 8 ) ( x − 6 ) ( x + 8 ) |
| Check: ( x − 6 ) ( x + 8 ) x 2 − 6 q + 8 q − 48 x 2 + 2 x − 48 ✓ ( x − 6 ) ( x + 8 ) x 2 − 6 q + 8 q − 48 x 2 + 2 x − 48 ✓ |
Factor: 9 m + m 2 + 18 . 9 m + m 2 + 18 .
Factor: −7 n + 12 + n 2 . −7 n + 12 + n 2 .
Sometimes you’ll need to factor trinomials of the form x 2 + b x y + c y 2 x 2 + b x y + c y 2 with two variables, such as x 2 + 12 x y + 36 y 2 . x 2 + 12 x y + 36 y 2 . The first term, x 2 , x 2 , is the product of the first terms of the binomial factors, x · x . x · x . The y 2 y 2 in the last term means that the second terms of the binomial factors must each contain y. To get the coefficients b and c, you use the same process summarized in How To Factor trinomials.
Factor: r 2 − 8 r s − 9 s 2 . r 2 − 8 r s − 9 s 2 .
We need r in the first term of each binomial and s in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
| r 2 − 8 r s − 9 s 2 r 2 − 8 r s − 9 s 2 | |
| Note that the first terms are r , r , last terms contain s . s . | ( r s ) ( r s ) ( r s ) ( r s ) |
| Find the numbers that multiply to −9 and add to −8. |
| Factors of −9 −9 | Sum of factors |
|---|---|
| 1 , −9 1 , −9 | −1 + 9 = 8 −1 + 9 = 8 |
| −1 , 9 −1 , 9 | 1 + ( −9 ) = − 8 * 1 + ( −9 ) = − 8 * |
| 3 , −3 3 , −3 | 3 + ( −3 ) = 0 3 + ( −3 ) = 0 |
| Use 1 , −9 as coefficients of the last terms. Use 1 , −9 as coefficients of the last terms. | ( r + s ) ( r − 9 s ) ( r + s ) ( r − 9 s ) |
| Check: ( r − 9 s ) ( r + s ) r 2 + r s − 9 r s − 9 s 2 r 2 − 8 r s − 9 s 2 ✓ ( r − 9 s ) ( r + s ) r 2 + r s − 9 r s − 9 s 2 r 2 − 8 r s − 9 s 2 ✓ |
Factor: a 2 − 11 a b + 10 b 2 . a 2 − 11 a b + 10 b 2 .
Factor: m 2 − 13 m n + 12 n 2 . m 2 − 13 m n + 12 n 2 .
Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.
Factor: u 2 − 9 u v − 12 v 2 . u 2 − 9 u v − 12 v 2 .
We need u in the first term of each binomial and v in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
| u 2 − 9 u v − 12 v 2 u 2 − 9 u v − 12 v 2 | |
| Note that the first terms are u , u , last terms contain v . v . | ( u v ) ( u v ) ( u v ) ( u v ) |
| Find the numbers that multiply to −12 and add to −9. |
| Factors of − 12 − 12 | Sum of factors |
|---|---|
| 1 , −12 1 , −12 −1 , 12 −1 , 12 2 , −6 2 , −6 −2 , 6 −2 , 6 3 , −4 3 , −4 −3 , 4 −3 , 4 | 1 + ( −12 ) = −11 1 + ( −12 ) = −11 −1 + 12 = 11 −1 + 12 = 11 2 + ( −6 ) = −4 2 + ( −6 ) = −4 −2 + 6 = 4 −2 + 6 = 4 3 + ( −4 ) = −1 3 + ( −4 ) = −1 −3 + 4 = 1 −3 + 4 = 1 |
Note there are no factor pairs that give us −9 −9 as a sum. The trinomial is prime.
Factor: x 2 − 7 x y − 10 y 2 . x 2 − 7 x y − 10 y 2 .
Factor: p 2 + 15 p q + 20 q 2 . p 2 + 15 p q + 20 q 2 .
Let’s summarize the method we just developed to factor trinomials of the form x 2 + b x + c . x 2 + b x + c .
When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.
x 2 + b x + c ( x + m ) ( x + n ) When c is positive, m and n have the same sign. b positive b negative m , n positive m , n negative x 2 + 5 x + 6 x 2 − 6 x + 8 ( x + 2 ) ( x + 3 ) ( x − 4 ) ( x − 2 ) same signs same signs When c is negative, m and n have opposite signs. x 2 + x − 12 x 2 − 2 x − 15 ( x + 4 ) ( x − 3 ) ( x − 5 ) ( x + 3 ) opposite signs opposite signs x 2 + b x + c ( x + m ) ( x + n ) When c is positive, m and n have the same sign. b positive b negative m , n positive m , n negative x 2 + 5 x + 6 x 2 − 6 x + 8 ( x + 2 ) ( x + 3 ) ( x − 4 ) ( x − 2 ) same signs same signs When c is negative, m and n have opposite signs. x 2 + x − 12 x 2 − 2 x − 15 ( x + 4 ) ( x − 3 ) ( x − 5 ) ( x + 3 ) opposite signs opposite signs
Notice that, in the case when m and n have opposite signs, the sign of the one with the larger absolute value matches the sign of b.
Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form a x 2 + b x + c . 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 we’ve used so far. Let’s do an example to see how this works.
Factor completely: 4 x 3 + 16 x 2 − 20 x . 4 x 3 + 16 x 2 − 20 x .
| Is there a greatest common factor? | 4 x 3 + 16 x 2 − 20 x 4 x 3 + 16 x 2 − 20 x |
| Yes, GCF = 4 x . GCF = 4 x . Factor it. | 4 x ( x 2 + 4 x − 5 ) 4 x ( x 2 + 4 x − 5 ) |
| Binomial, trinomial, or more than three terms? | |
| It is a trinomial. So “undo FOIL.” | 4 x ( x ) ( x ) 4 x ( x ) ( x ) |
| Use a table like the one shown to find two numbers that multiply to −5 and add to 4. | 4 x ( x − 1 ) ( x + 5 ) 4 x ( x − 1 ) ( x + 5 ) |
| Factors of − 5 − 5 | Sum of factors |
|---|---|
| − 1 , 5 − 1 , 5 1 , −5 1 , −5 | − 1 + 5 = 4 * − 1 + 5 = 4 * 1 + ( − 5 ) = −4 1 + ( − 5 ) = −4 |
| Check: 4 x ( x − 1 ) ( x + 5 ) 4 x ( x 2 + 5 x − x − 5 ) 4 x ( x 2 + 4 x − 5 ) 4 x 3 + 16 x 2 − 20 x ✓ 4 x ( x − 1 ) ( x + 5 ) 4 x ( x 2 + 5 x − x − 5 ) 4 x ( x 2 + 4 x − 5 ) 4 x 3 + 16 x 2 − 20 x ✓ |
Factor completely: 5 x 3 + 15 x 2 − 20 x . 5 x 3 + 15 x 2 − 20 x .
Factor completely: 6 y 3 + 18 y 2 − 60 y . 6 y 3 + 18 y 2 − 60 y .
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 . 3 x 2 + 5 x + 2 .
From our earlier work, we expect this will factor into two binomials.
3 x 2 + 5 x + 2 ( ) ( ) 3 x 2 + 5 x + 2 ( ) ( )We know the first terms of the binomial factors will multiply to give us 3 x 2 . 3 x 2 . The only factors of 3 x 2 3 x 2 are 1 x , 3 x . 1 x , 3 x . We can place them in the binomials.
Check: Does 1 x · 3 x = 3 x 2 ? 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, 2. But we now have two cases to consider as it will make a difference if we write 1, 2 or 2, 1.
Which factors are correct? To decide that, we multiply the inner and outer terms.
Since the middle term of the trinomial is 5 x , 5 x , the factors in the first case will work. Let’s use FOIL to check.
( x + 1 ) ( 3 x + 2 ) 3 x 2 + 2 x + 3 x + 2 3 x 2 + 5 x + 2 ✓ ( x + 1 ) ( 3 x + 2 ) 3 x 2 + 2 x + 3 x + 2 3 x 2 + 5 x + 2 ✓
Our result of the factoring is:
3 x 2 + 5 x + 2 ( x + 1 ) ( 3 x + 2 ) 3 x 2 + 5 x + 2 ( x + 1 ) ( 3 x + 2 )Factor completely using trial and error: 3 y 2 + 22 y + 7 . 3 y 2 + 22 y + 7 .