1. Factoring Quadratic Equations
Miss Liberty
Factoring Quadratic Equations

2. Typo on #11
Should read as x2 -7x -18=0 not as x2 -7x -18 = 2

3. # x2 + 2x -8 = 0
x2 + 2x -8 = 0 we need a=1 b=2 c=-8 product sum a•c b -1• 8 = -8 2

4. # x2 + 2x -8 = 0
x2 + 2x -8 = 0 we need a=1 b=2 c=-8 product sum a•c b -1• 8 = -8 2 tells me the two factors have unlike signs tells me the larger factor is positive

5. # x2 + 2x -8 = 0
x2 + 2x -8 = 0 we need a=1 b=2 c=-8 product sum a•c =-8 b=2 -1• = 7 no -2• = 2 yes x2 + 2x - 8 = 0 x2 + -2x + 4x -8 = 0 replace 2x with -2x+4x

6. # x2 + 2x -8 = 0
x2 + -2x + 4x -8 = 0 replace 2x with -2x+4x Split this into two halves x2 + -2x + 4x -8 = 0 Factor what you can out of each half x( x + -2) + 4( x +-2) =0 Look for a common factor it is (x + -2) record it and make the remaining items into the 2nd factor (x + -2)(x +4) = 0

7. #1 x2 + 2x -8 = 0
(x + -2)(x +4) = 0 Since either factor could equal zero x + -2 = 0 x + 4 = 0 x=2 or x=-4

8. # x2 + 5x +2 = 0
3x2 + 5x + 2 = 0 we need a=3 b=5 c=2 product sum a•c b 3• 2 = 6 5

9. # x2 + 5x +2 = 0
3x2 + 5x +2 = 0 we need a=3 b=5 c=2 product sum a•c b=5 3• 2 = 6 3+2=5 tells me the two factors have like signs tells me the signs are both positive

10. # x2 + 5x +2 = 0
3x2 + 5x +2 = 0 we need a=3 b=5 c=2 product sum a•c =6 b=5 3• 2 = 6 3+2=5 Substitute 3x + 2x in for 5x 3x2 + 3x+ 2x +2 = 0 Split in halves and factor 3x(x +1) + 2(x + 1) = 0

11. 3x(x +1) + 2(x + 1) = 0
Factor out the common (x +1) and record what is left (x + 1)(3x +2) = 0 set each factor =0 x + 1= 0 3x+2=0 x=-1 3x =-2 or x= −2 3