Tuesday, 10/14/14
Today’s Character Counts Quote of the Day:
Respect
When someone does something
good, applaud! You will make two
people happy.
– Samuel Goldwyn
Tuesday, 10/14/14
Today’s agenda:
Check your work (in 5 minutes)
–Problem Set 2.01 (1 – 12)
Notes
• Factoring “any” quadratic
–Quadratic coefficient ≠ 1
–Standard & not standard forms
Assignment
• Problem Set 2.02 (1 – 12)
Problem Set 2.01 (1 – 12) solutions
2 x  2  x  2   0
1)
20
No Solution
2)
or
or
x20
x  2
or
or
x20
x2
x  44 x  5  0
x40
x4
or
or
4x  5  0
5
x
4
Problem Set 2.01 (1 – 12) solutions
 p  1 p  3  0
3)
p 1  0
p  1
4)
or
or
p 3  0
p3
33 x  1 x  3  0
30
No Solution
or
or
3x  1  0
or
x3 0
1
x
3
or
x  3
Problem Set 2.01 (1 – 12) solutions
x  2 x  10 x  20
5)
2
x  12 x  20  0 2
2
x  2  0 or
20
12
10
x  10  0
6)
x x  2  10 x  2 
x  2x  10  0
x  2 or
x  10
x  9 x  10 x  90
2
x  19 x  90  0 -9
2
x  9  0 or
90
-10
-19
x  10  0
x x  9    10 x  9 
x  9x  10  0
x  9 or
x  10
Problem Set 2.01 (1 – 12) solutions
a  3a  6a  18
7)
2
a  9a  18  0
2
18
3
9
6
a  3  0 or a  6  0
8)
aa  3  6a  3
a  3a  6  0
a  3 or a  6
r  10r  10r  100
2
r  100  0
2
-100
10
-10 r r  10   10r  10
0
r  10r  10  0
r  10  0 or r  10  0
r  10 or r  10
Problem Set 2.01 (1 – 12) solutions
9)
11m  63  3m  10m  7
2
2
 10m  3m  7  10m  3m  7
2
2
m  3m  70  0
2
m  7m  10m  70
2
-70
7
-10 mm  7    10m  7 
-3
m  7  0 or m  10  0
m  7 m  10  0
m  7 or m  10
Problem Set 2.01 (1 – 12) solutions
10)
 4 x  11x  36  9  7 x  5 x
2
2
5x  7 x  9  9  7 x  5x
2
2
x  4 x  45  0
2
x  9 x  5 x  45
2
-45
-9 5
-4
x  9  0 or
x5  0
x x  9   5 x  9 
x  9x  5  0
x  9 or
x  5
Problem Set 2.01 (1 – 12) solutions
11)
 7 x  12 x  40  8 x  9 x
2
2
 8x  9 x  8x  9 x
2
2
x  3x  40  0
2
x  8 x  5 x  40
2
-40
-8 5
-3
x  8  0 or
x5  0
x x  8  5 x  8
x  8x  5  0
x  8 or
x  5
Problem Set 2.01 (1 – 12) solutions
12)
 2n  n  95  3n  5
2
2
 3n  5  3n  5
2
2
n  n  90  0
2
n  9n  10n  90
2
-90
9 -10
-1
n  9  0 or n  10  0
nn  9   10n  9 
n  9n  10  0
n  9 or n  10
What if the leading coefficient isn’t 1?
1. Write the quadratic in standard
form ax 2  bx  c and use the method…
What about this one?
2 x  3x  4  0
2
2 x  4 x  1x  4  0
2
-4
-4
1
-3
2 x x  2  1 x  4
No Common Factors!!!
…because the “a” term is not 1
So, we need to modify our method
What if the leading coefficient isn’t 1?
1. Write the quadratic in std. form
2. Multiply the leading coefficient
times the constant term, and put
that in the top of the diamond. Put
the linear coefficient in the bottom.
3. Solve the diamond puzzle
4. Use the left and right numbers from
the puzzle to rewrite the linear term
5. Factor by grouping (look for
common factors twice)
What if the leading coefficient isn’t 1?
Example:
2
6 x  23x  20  0
2
6 x  15 x  8 x  20  0
6
120
20
-15 -8
3 x2 x  5   42 x  5  0
-23
2 x  53x  4  0
2x  5  0
2x  5
5
x
2
3x  4  0
3x  4
4
x
3