factoring math prob?

Question

3x^2 + 5x +2=0
solve by factoring, please help, show me so I can do the other factoring probs I have.

Answer 1

Multiply 3 and 2. You get 6.
What factors of 6 add up to 5? You get 3 and 2.
Insert these into the equation in place of 5x.
3x^2 + 3x + 2x + 2
Factor by grouping
= 3x(x+1) + 2(x+1)
= (3x+2)(x+1)

(3x+2)(x+1) = 0
Set each factor equal to zero.
3x+2 = 0 or x+1 = 0
x = -2/3 or x = -1

Answer 2

3x² + 5x + 2 = 0

The middle term is + 5x

Find the sum of the middle term

Multiply the first term 3 times the last term 2 equals 6 and factor

factors of 6

1 x 6
2 x 3. . . .<=. .Use these factors

+ 3 and + 2 satisfy the sum of the middle term

Insert + 3x and + 2x into the equation

3x² + 5x + 2 = 0

Gropup factor

3x² + 3x + 2x + 2 = 0

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

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

- - - - - - - -

Roots

3x + 2 = 0

3x + 2 - 2 = 0 - 2

3x = - 2

3x / 3 = - 2 / 3

x = - 2/3

- - - - - - - -

Roots

x + 1 = 0

x + 1 - 1 = 0 - 1

x = - 1

- - - - - - - - s-

Answer 3

First, let's look at the 3 in front of the x^2. The only way to factor a 3 is 3*1 (or -3 * -1, but we can ignore that for the first term), so you know any factored solution must start with:
(3x ......... ) * (x ......... )

Next, look at the 2.
The only ways to factor a 2 are 2*1 and -2 * -1.
This leaves us with four possibilities:
(3x + 2)*(x + 1)
(3x + 1)*(x+2)
(3x - 2)*(x - 1)
(3x - 1)*(x - 2)

The last two possibilities can be thrown out, since they would lead to a negative coefficient for the middle term, and that coefficient needs to be 5.

Next, you FOIL out the two possibilities you have left, and see if either one works.
(3x + 2)*(x + 1) = 3x^2 + 5x + 2
(3x + 1)*(x + 2) = 3x^2 + 7x + 2

Looks like the first option is a winner!

Answer 4

First: multiply the 1st & 3rd term to get 6. find two numbers that give you 6 when multiplied & 5 (2nd term) when added/subtracted. the numbers are (2 & 3). rewrite the expression with the new middle terms.

3x^2 + 2x +3x + 2 = 0

Sec: group "like" terms & factor both sets of parenthesis.

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

Third: set both parenthesis to "0" & solve the "x" variables.

a. x+1 = 0
subtract 1 from both sides - when you move a term to the opposite side, always use the opposite sign.

x+1-1 = 0-1
x = -1

b. 3x+2 = 0
subtract 2 from both sides.

3x+2-2 = 0-2
3x = -2 (divide both sides by 3).
3x/3 = -2/3
x = -2/3

Solutions: -1 and -2/3

P.S. We solve the "x" variables because, the problem is an equation set to "0"

Answer 5

(3x + 2)(x + 1)

I look at it this way:

The first term I look at is 3x^2. 3 happens to be prime, so you know that the first terms in the solution are some combination of 3x and 1x. (It also could be -3x and -x....)

Then I look at the last term of the polynomial: the 2. 2 is also prime, so the solution is a combination of 2 and 1 or -2 and -1.

To get the final solution, you have to solve for that 5x.
3x(a) + x(b) = 5x where a, b are either (-)1 or (-)2 (from the factors of the last term, 2.)
Since 3x(1) + x(2) = 5x, we get the final solution that the 3x and the 1 are in different factors.
(3x + 2)(x +1)

Answer 6

The factors of 2 are 1,2...

You know you need a 3x^2 as the first term in the polynomial.

(3x + )( x + )

Now you need the middle term in the poly to add to 5 so you need to use the 2 to multiply with the x and not the 3x.

(3x + 2)(x + 1)

Test the result using FOIL (First outer inner last)
3x*x + 3x*1 + 2*x + 2*1 =
3x^2 + 3x + 2x + 2 =
3x^2 + 5x + 2 (check)

Hope this helps.

Answer 7

3x^2 + 5x +2=0
=(3x + ?)(x + ?) = 0
Since ? times ? = 2 one ? 2 and the other =1.
So = (3x +2)(x+1) = 0

Answer 8

( 3 x +2 )(x+1)

Answer 9

3x^2+5x+2=3x^2+3x+2x+2
3x(x+1)+2(x+1)
=(x+1)(3x+2)answer