Factoring problem: Factor completely showing all work: 3x^2- 2x -8?

Question

I need to show all the work for this problem,: factor completely
3x^2-2x-8

Answer 1

3x² - 2x - 8

Factors of 3:
3 x 1

Factors of -8:
1 x -8
2 x -4
4 x -2
8 x -1

Your answer will have the following form:
(3x )(x )

Let's try the various factors of -8
(3x + 1)(x - 8) --> this results in -23x in the middle
(3x + 2)(x - 4) --> this results in -10x in the middle
(3x + 4)(x - 2) --> this results in -2x in the middle (done!)

Or if you have to show the factoring step by step, you know where you need to get to, you need a 4x and -6x, so split the expression as:

3x² - 6x + 4x - 8

Factor out 3x from the first two terms:
3x(x - 2) + 4x - 8

Factor a 4 out of the second two terms:
3x(x - 2) + 4(x - 2)

Now you have a common (x - 2):
(3x + 4)(x - 2)

Answer 2

My personal favorite method for such factoring is the expanding and grouping method. You take the first coefficient and multiply it in the last coefficient (when in the order given above).
3*-8=-24
Now you want to find 2 factors of -24 that add up to -2
-6 and -4 work, so we use those instead of -2
you go from 3x^2-2x-8 to 3x^2-6x+4x-8
Now, in the first part of the equation, you have 3x^2-6x, which can be written as 3x(x-2)
And the second part, 4x-8, can be written as 4(x-2). Notice the (x-2) factor is the same. The fact that these are the same serves in itself as a check to let you know that this is factorable and your math so far is correct.

Now you have the first part 3x(x-2) and 2nd part, 4(x-2), and you are adding them, so you have
3x(x-2)+4(x-2)
You have a common factor of (x-2), so you use distributive property and get (3x+4)(x-2) That is your final answer. If you were to solve the equation, and were told it is equal to zero, you would set each factor equal to zero and get x=-4/3 and x=2 as solutions. Since the problem is asking simply for factoring, however, you leave it as (3x+4)(x-2).

Keep in mind this only works for trinomials in which the second term (x) has half the power of the first term (x^2)

Answer 3

Factor 3x² - 2x -8
since in the form ax² + bx + c where a= 3, b= -2 and c= -8 we follow some basic steps
1. find ac
ac = (3)(-8) = -24
2. find factors o ac that when added gives b and multiplied gives ac
factors are 4 and -6 ( 4 + -6)= -2 and (4)(-6)= -24
3. replace the term with b using these factors
3x² - 6x + 4x -8
4. factor by grouping
3x(x-2) +4 (x-2)
then factor completely
(x-2)(3x+4)

you can check your answer by multiplying out it should give you the original equation

Answer 4

1. 5x^2 - 20 = 5 ( x^2 - 4 ) = 5 ( x -2 ) ( x + 2 ) 2. 3x^2 + 12x + 9 = 3 ( x^2 + 4x + 3 ) = 3 [ x ( x + 4 ) + 3 ] 3. 2x^2 - 50 = 2 ( x^2 - 25 ) = 2 ( x -5 ) ( x + 5 ) 4. 2x^2 - 12x - 32 = 2 ( x^2 - 6x - 16 ) = 2 ( x - 8 ) ( x + 2 ) Good Luck - hope all is clear.

Answer 5

factors of 3x^2 are 3x and x
facotors of -8 are 1*-8 or -1*8 or -2*4 or 2 * -4

Trying one set and FOIL to check
(3x +1)(x - 8) = 3x^2 -24x + x - 8 = 3x^2 - 23x - 8 not right
(3x + 8)(x - 1) = 3x^2 + 5x - 8 not right either

(3x +4)(x - 2) = 3x^2 -6x + 4x - 8 = 3x^2 -2x - 8 CORRECT

(3x +4)(x - 2) is the combination that works.

Answer 6

okay, lets see... break it up first.

3x^2=3*2*x*2=6x2
combine like terms:
6x2-2x-8=4x2-8=4x-6

I think 4x-6 is the correct answer, but I could be totatlly wrong!!! Good luck and sorry if I am.....

Answer 7

3x^2-2x-8 can be written as
3x^2-6x+4x-8
3x(x-2)+4(x-2)
(x-2)(3x+4)

Answer 8

(3x + 4)(x - 2)