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

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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.

You are allowed to ask only one question at a time. I am solving one of the problems, you can apply the same method for the others.

To solve 3x^2 + 5x - 2 = 0, we express 5 in terms of two numbers that add up to 5 and the product of the numbers is 3*-2 = -6. The numbers that satisfy this are 6 and -1.

3x^2 + 5x - 2 = 0

=> 3x^2 + 6x - x - 2 = 0

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

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

3x - 1 = 0

=> x = 1/3

x + 2 = 0

=> x = -2

**The solution of 3x^2 + 5x - 2 = 0 is x = 1/3 and x = -2**

Going to do #2

2x^2 - x - 3 = 0

a b c

multiply a by c

2 x -3 = -6

find factors of -6 that minus to b (-1) which would be -3 and 2

plug those numbers in as b

2x^2 + 2x - 3x - 3

group

(2x^2 + 2x)( - 3x - 3)

factor out

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

group the numbers outside together:

(2x - 3 ) (x + 1)

that's the answer but you can go further to find the solutions by setting the parentheses = 0

2x - 3 =0

2x = 3

**x = 3 / 2 or 1.5**

x + 1 = 0

**x = 0**