Factoring Trinomials
When using FOIL, we can express the product (k – 3)(k + 1) as k 2  2k  3 .
How can we obtain such a product from k 2  2k  3 ? We can undo the multiplying by
factoring the trinomial.
Example: x 2  5 x  6
We must find 2 numbers that when multiplied equal 6, but
when added, equal 5.
Factors of 6: 1, 2, 3, 6
Sum of 5: 2 + 3 = 5
Therefore, we express our two binomials as (x + 2)(x + 3)
FOIL to check: (x + 2)(x + 3)
x 2  2 x  3x  6
x 2  5x  6
****This method only works when the leading coefficient is 1!!!****
Sign Changes
 When all signs of the polynomial are positive (+), the sign in each binomial factor
will be positive (+).
 When the middle term is negative (-), both binomial signs will be negative (-).
Ex: x 2  9 x  20  ( x  5)( x  4)
Note: The positive last term tells us both signs will be the same. The
middle term then tells us they will both be negative, since the middle term is
negative.
Similarly, as in the first example, the last term was positive, so both
binomials had the same sign. Since the middle term was also positive, both of
those signs were positive.
 When the last term is negative (-), one binomial will be positive and one binomial
will be negative.
Ex: p 2  2 p  15  ( p  5)( p  3)
Prime Polynomials
x 2  5 x  12
Since the third term is positive, both signs will be the same.
Since the second term is negative, both of those signs will be
negative.
Factors of 12: 1* 12, 2*6, 3*4
Sum of -5: None of our 3 pairs have a sum of -5.
Therefore, our polynomial cannot be factored using ONLY integers, thus it is prime.
Examples:
1) y 2  12 y  20
3) t 2  12t  32
5) a 2  9a  22
7) r 2  3r  4
9) k 2  8k  11
2) x 2  9 x  18
4) y 2  10 y  24
6) r 2  6r  16
8) m 2  2m  5
10) m 2  9m  14
Factoring Trinomials with Two Variables
z 2  2bz  3b 2
Since the leading term is a z, we know the coefficient of our
middle term z is -2b.
Goal: Find 2 numbers that multiply to  3b 2 and add to -2b.
Factors of 3: 1*3
Factors of b 2 : b*b
Third term is negative, so one binomial is positive and one is negative.
Factors that add to -2: -3 + 1
Therefore: (z – 3b)(z + b)
Trinomials with a GCF
4 x 5  28 x 4  40 x 3
3
GCF: 4x -------------------
Ex:
1) b 2  3ab  4a 2
3) 2 p 3  6 p 2  8 p
4 x 3 ( x 2  7 x  10)
4 x 3 ( x  5)( x  2)
2) r 2  6rs  8s 2
4) 3 x 4  15 x 3  18 x 2