Download Factoring Trinomials Where a = 1

Factoring Trinomials
Where a = 1
In the last section, we discussed factoring trinomials where a ≠ 1. We
used the expansion (ac) method where we multiplied the first and last
numbers (a and c), and then selected the two factors when added equal
the middle number (b). Those two factors were used to rewrite and
expand the middle term. In this section we will look at trinomials where
a = 1, using a similar procedure. When a = 1, if we multiply the first and
last numbers, we will have a product equal to the last number: 1 · c = c.
Thus, we only need to select two factors of the last number (c) when
added equal the middle number (b).
Example 1: Factor: x2 + 2x – 24
Factors of 24: 1,24; 2,12; 3,8; 4,6
Add to get 2. Since the product is -24, one number must be negative:
-1+24 = 23 or -24+1 = -23 No
-2+12 = 10 or -12+2 = -10 No
-3+8 = 5 or -8+3 = -5
No
-4+6 = 2 or -6+4 = -2
Yes
2
x – 4x + 6x – 24 Rewrite and expand 2x
x2 – 4x + 6x – 24 Split the polynomial in half
x(x – 4) + 6(x – 4) Factor the GCF from both sides
(x – 4)(x + 6)
Notice that we didn’t need to use the expansion method. The final two
binomials consist of the factors that add to the middle number. Once we
find those two numbers, we can simply write the two binomial factors of
the trinomial.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 2: Factor: x2 + 9x + 18
Factors of 18: 1,18; 2,9; 3,6
Add to get 9: 1+18 = 19 No
2+9 = 11 No
3+6 = 9 Yes
(x + 3)(x + 6)
To check the answer, multiply: (x + 3)(x + 6)
x(x + 6) + 3(x + 6)
x2 + 6x + 3x + 18
x2 + 9x + 18 Correct
Example 3: Factor: y2 – 4y + 3
Factors of 3: 1,3
Since the middle term is negative, both numbers must be negative
(y – 1)(y – 3)
To check the answer, multiply: (y – 1)(y – 3)
y(y – 3) – 1(y – 3)
y2 – 3y – 1y + 3
y2 – 4y + 3 Correct
Example 4: Factor: a2 – 8a – 20
Factors of 20: 1,20; 2,10; 4,5
Add to get -8. Since the product is -20, one number must be negative:
-1+20 = 19 or -20+1 = -19 No
-2+10 = 8 or -10+2 = -8 Yes
(a – 10)(a + 2)
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
To check the answer, multiply: (a – 10)(a + 2)
a(a + 2) – 10(a + 2)
a2 + 2a – 10a – 20
a2 – 8a – 20 Correct
Example 5: Factor: x2 – 9xy + 14y2
Factors of 14: 1,14; 2,7
Add to get -9. Since the product is positive, both numbers must be negative:
-1+(-14) = -15 No
-2+(-7) = -9 Yes
(x – 2y)(x – 7y) Don’t forget about the variable y
To check the answer, multiply: x(x – 7y) – 2y(x – 7y)
x2 – 7xy – 2xy + 14y2
x2 – 9xy + 14y2
Example 7: Factor: m2 – mn – 30n2
Factors of 30: 1,30; 2,15; 3,10; 5,6
Add to get -1. Since the product is -30, one number must be negative:
-1+30 = 29 or -30+1 = -29 No
-2+15 = 13 or -15+2 = -13 No
-3+10 = 7 or -10+3 = -7 No
-5+6 = 1 or -6+5 = -1
Yes
(m – 6n)(m + 5n) Don’t forget the variable n
To check the answer, multiply: (m – 6n)(m + 5n)
m(m + 5n) – 6n(m + 5n)
m2 + 5mn – 6mn – 30n2
m2 – 1mn – 30n2
m2 – mn – 30n2 Correct
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 6: Factor: x2 + 5x + 8
Factors of 8: 1,8; 2,4
Add to get 5: 1+8 = 9 No
2+4 = 6 No
Since no set of factors add to 5, the trinomial cannot be factored.
Prime
When factoring, it is important not to forget about the GCF. If all terms
have a common factor, first factor out the GCF before factoring further.
Example 8: Factor: 3z2 – 24z + 45
3(z2 – 8z + 15) Factor out the GCF, 3, first
Factors of 15: 1,15; 3,5
Add to get -8. Since the product is positive, both numbers must be negative:
-1+(-15) = -16 No
-3+(-5) = -8 Yes
(z – 3)(z – 5)
To check the answer, multiply: 3(z – 3)(z – 5)
3[z(z – 5) – 3(z – 5)]
3[z2 – 5z – 3z + 15]
3[z2 – 8z + 15]
3z2 – 24z + 45 Correct
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)