Download The X-Games

Math 4
Factoring Notes
December 16, 2012
The X-Games
Each X below is a puzzle. Find two numbers whose product is the top number and whose sum
is the bottom number (This is similar to the snowman method, although I believe the numbers are
switched top to bottom).
Example:
Multiply to
-15
?
-15
-5
?
-2
3
-2
Add to
−5 + 3 = −2 and −5 · 3 = −15
Solve each of the puzzles below. They get more challenging as you go.
32
108
56
12
21
15
-96
-64
-121
-4
12
0
46
132
343
-25
-23
-56
The puzzles you have just completed will help you factor basic quadratics. A basic quadratic is
considered a polynomial of degree two with a front coefficient of 1.
Let’s take a look at factoring this one:
x2 + 9x + 18 = (x + 3) (x + 6)
18
6
3
9
Try this one:
x2 + 2x − 24 = (x
) (x
)
-24
2
Okay, now why does this work? Multiply the two binomials below, but don’t combine like terms.
(x + 6)(x − 4) =
What do you notice about the puzzle above and your multiplication? What connections can
you make?
Consider this:
x2 + ax + bx + ab = (x + a)(x + b)
a·b
a+b
What is common about all of the following polynomials?
a. x2 − 16
c. x2 − 144
e. 16y 2 − 81
g. 25z 2 − w2
b. x2 − 25
d. 4x2 − 36
f. y 2 z 2 − 4
h. (x + 2)2 − (y − z)2
Now remember from our FOILing, what happens when you expand a set of binomials that looks
like this?
(X − Y )(X + Y )
Now factor the polynomials above using difference of squares:
a.
c.
e.
g.
b.
d.
f.
h.
What do you think you should do when all three terms share a factor? How would you factor
the following?
2x2 + 20x + 42
Using what you’ve learned, factor the following polynomials. Remember that you can always
check your answer by foiling the binomials.
1. x2 + 10x + 21
10. x2 − 36x + 180
2. x2 + 15x + 56
11. x2 − 12x + 36
3. x2 + 13x + 42
12. x2 − 24x + 144
4. x2 − x − 30
13. x2 − 49
5. x2 + 7x − 18
14. x2 − 121
6. x2 − 5x − 84
15. 4x2 − 9
7. x2 + 2x − 143
16. 2x2 + 2x − 24
8. x2 + 14x − 51
17. 3x2 + 30x − 33
9. x2 − 48x − 1024
18. x2 + xy − 6y 2
Now that you feel more comfortable with basic factoring, we need to address what to do with
quadratics when the front coefficient is not 1. Within this category, there are two types: Quadratics
with multiple prime coefficients and quadratics with composite (non-prime) coefficients.
Type 1: Notice that A and C are primes. This makes it MUCH easier.
3x2 − 20x − 7
Now look at the possible combinations:
1. The coefficents of x HAVE TO BE 3 and 1.
2. The constants HAVE TO BE 1 and 7.
(3x 1)(x 7)
or
(3x 7)(x 1)
Now you have to decide on the signs. Keep in mind the sign of the middle term and how large
the middle coefficient is. In this case it is large and negative so the following makes sense:
(3x − 1)(x + 7)
Work an Example:
−5x2 − 16x − 3
Type 2: Notice that A and C are not primes and C has lots of factors. This can be done
intuitively like Type 1, but you can also use a method called ”Factoring by Grouping.”
10x2 + 11x − 840
This is like the X-games or Snowman method you have learned before. Only this time the top
number is A times C. (Which if you think about it, has how it has always been, just previously A
was 1!).
Multiply to
-840
-840
?
-24
?
11
35
11
Add to
35 − 24 = 11 and −24 · 35 = −840*
Now here’s where it changes. Take the two side numbers and split the middle term (it does not
matter which order they are in!). Notice that the second equation is equal to the first one! Also,
when one of the split terms is negative it is best to do it as below: add the negative. See example:
10x2 + 11x − 84
10x2 + 35x + (−24)x − 84
Now Group the first two terms and the last two terms. Make sure that if the second middle
term is negative the negative sign is inside the parenthesis:
(10x2 + 35x) + (−24x − 84)
Then factor out the common terms. First from the two groups, and then the binomials!
5x(2x + 7) − 12(2x + 7)
Factor out common terms!
(2x + 7)(5x − 12)
Thusly: 10x2 + 11x − 84 = (2x + 7)(5x − 12)
THIS ALWAYS WORKS. (As long as the quadratic is not prime).
*Remember that to find the side numbers (the arms of the snowman from last class), you can
use the prime factorization of the big numbers:
840 = 84 · 10 = (7 · 12) · (5 · 2) = (7 · 2 · 2 · 3) · (5 · 2) = 2 · 2 · 2 · 3 · 5 · 7
And in this case, you can see it in the second step: 7 × 5 = 35 and 12 × 2 = 24. Figuring out
which factors to use depends on what you are trying to add to. If the number they need to sum to
is small then both factors will be around the same size, the larger the sum number the bigger the
difference between the factors.
Practice
Factor each of the following using whatever method you so choose. Although if A and C are
prime, try to do it using the first method (Type 1).
1. 12x2 + 5x − 2
2. 3x2 − 3x − 90
3. 6x2 − 11x + 4
4. 15x2 + 14x − 8
5. 24x2 y + 34xy + 12y
6. 9x2 + 12x + 4
7. 3x2 + 14x − 5
8. 7x2 + 8x + 1
9. 2x2 − 15x − 17