Solve by factoring: 3x^2 + 5x 1. x^2 + 5x + 6 = 0 2. 2x^2 - x 3. 3x^2 + 5x - 2 = 0

1.  x^2 + 5x + 6 = 0

Begin
with this set up:

(x + __)(x + __) =
0

The missing numbers must have a product of 6 and a sum of
5.

2*3=6    
2+3=5

Therefore...

(x + 2)(x +
3) = 0

Now take each binomial and set it equal to 0.  Solve
for x.

x + 2 = 0     x = -2

x
+ 3 = 0     x = -3

Solution set:  {-2,
-3}

2.  2x^2 - x - 3 =
0

When there are two subtraction signs in the trinomial,
the set up is:

(2x + __)(x - __) = 0 or (2x - __)(x + __) =
0

Now we need two numbers whose product is 3.  Since 3 is
prime, the only options are 1 and 3.  Try substituting 1 and 3 into the set ups above,
use FOIL to see which set up works.

(2x + 1)(x - 3) = 2x^2
- 6x + 1x - 3 = 2x^2 - 5x - 3     NO

(2x + 3)(x - 1) = 2x^2
- 2x + 3x - 3 = 2x^2 + 1x - 3     NO

(2x - 1)(x + 3) = 2x^2
+ 6x - 1x - 3 = 2x^2 + 5x - 3     NO

(2x - 3)(x + 1) = 2x^2
+ 2x - 3x - 3 = 2x^2 - 1x - 3     YES

So now we know
the trinomial can be factored as...

(2x - 3)(x + 1) =
0

Again, set each binomial equal to 0 and solve for
x.

2x - 3 = 0     x = 1.5

x +
1 = 0     x = -1

Solution set:  {1.5,
-1}

3.  3x^2 + 5x - 2 =
0

This one is solved similarly to #2.  Here is the set
up:

(3x + __)(x - __) = 0 or (3x - __)(x + __) =
0

Again, since 2 is prime, the only options for the blanks
are 1 and 2.  Try each combination, use FOIL to see which one
works.

(3x + 1)(x - 2) = 3x^2 - 6x + 1x - 2 = 3x^2 - 5x -
2     NO

(3x + 2)(x - 1) = 3x^2 - 3x + 2x - 2 = 3x^2 + 1x -
2     NO

(3x - 1)(x + 2) = 3x^2 + 6x - 1x - 2 = 3x^2 + 5x -
2     YES

Therefore...

(3x -
1)(x + 2) = 0

Set each binomial equal to 0 and solve for
x.

3x - 1 = 0     x = 1/3

x +
2 = 0     x = -2

Solution set:  {1/3,
-2}

Remember, you can always
check these answers by graphing the equations.  The x-intercepts should equal the
solution set.