Simple Trinomials
I want to explain some things to you......be patient and you will see how the process pays off.
First, some explanations:
The degree of a term is the sum of the exponents on the variable(s).
e.g. Degree(5x7 )  7
and
Degree(11a 2b4 )  6.
Trinomial means an expression with 3 terms of different degrees.
The number in front of the variable portion of a term is called the coefficient.
So.....for our discussion, a simple trinomial is an expression with 3 terms whose leading
coefficient is 1. e.g. x 2  5x  6 ...........remember: x 2 means 1x 2 .
Expanding and factoring are processes that “undo each other”.
So if you’re asked to expand ( x  2)( x  3) here is the process, as you know already.
( x  2)( x  3)  x2  3x  2 x  6
 x 2  5x  6 (this is called expanding)
FYI: The answer can be derived much more quickly.
So............ ( x  2)( x  3)  x 2  5x  6
23 5
23  6
2
So, in general:  x  a  ( x  b)  x  (a  b) x  ab
Let’s use this idea to factor....i.e. undo expanding.
Let’s look at our first example but let’s undo it.....i.e. go backwards.
Factor x 2  5x  6 : we need to find 2 numbers that add to 5 and multiply to 6.
After a little thought or even trial and error...bingo: 2 &3.
We have x 2  5x  6  ( x  2)( x  3) . You can check the validity of your answer by expanding.
Similarly factor x 2  7 x  12 . We need two numbers with a sum of 7 and a product of 12.
Only one set will work.... 3& 4.
So x2  7 x  12  ( x  3)( x  4).
So far so good......let’s consider a few more examples on the next page.
Factor x 2  2 x  15. We need to find two numbers that have a sum of 2 and a product of 15.
3  5  15 but 3  5  2. This pair, 3&5 , doesn’t quite do it. We need 5&3.
So x2  2 x  15  ( x  5)( x  3).
The order is not important so we could have x2  2 x  15  ( x  3)( x  5).
BUT the signs on the numbers cannot be interchanged.
Here is another one to factor: 3x 2  3x  18. Well this doesn’t follow the rules for simple trinomials...
i.e. the leading coefficient is not 1.....but we can use common factoring and factor out a 3.....
a simple trinomial
3x2  3x  18  3( x2  x  6)
 3( x  3)( x  2) .
Now, I’ll make it a little more challenging.
Factor x 4  13x2  36. This is actually a simple trinomial but the challenge is the exponents....
Here’s the catch........the look at the exponents...the first one is 4, followed by the middle exponent 2,
followed by a constant term, i.e. no variable. Any trinomial that follows the pattern with the middle
term having half the degree of the first term, followed by a constant fits the description for a simple
trinomial.
These are both difference of squares
So, x4  13x 2  36  ( x 2  9)( x 2  4)
 ( x  3)( x  3)( x  2)( x  2)
Now, one last one:
x4  11x3  28x2 .......a trinomial but variables in every term....hmmmmmm!!!
Common factor and presto: x4  11x3  28x2
 x 2 ( x 2  11x  28)
 x 2 ( x  4)( x  7)
So, this is one view of simple trinomials.
The next page has some practice for you.
Factor completely
2) m2  17m  72
3) k 2  10k  9
4) t 2  10t  16
5) y 2  13 y  40
6) h2  12h  36
7) r 2  2r  63
8) h2  5h  66
9) x 2  x  156
1)
x 2  9 x  14
10) 4m2  12m  8
11) 5c2  30c  40
13) x 4  17 x2  16
12) an2  an  110a
14) 2 x4  42 x2  200
Answers
1) ( x  2)( x  7)
2) (m  8)(m  9)
5) ( y  5)( y  8)
6) (h  6)(h  6)....or....(h  6)2
8) (h  11)(h  6)
9) ( x  13)( x  12)
12) a(n  11)(n  10)
3) (k  1)(k  9)
7) (r  9)(r  7)
10) 4(m  1)(m  2)
13) ( x  1)( x  1)( x  4)( x  4)
4) (t  2)(t  8)
11) 5(c  2)(c  4)
14) 2( x  5)( x  5)( x 2  4)