help with factoring?

Question

can any one plz factor this and show me the steps how 2 do it--

3x^2 + x - 2


thank youuuuuu!!
<3

Answer 1

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

Answer 2

You know that you will have (3x + _)(x+_) because that will give you the first term 3x^2.

Next we need to fill in the blanks. You know that that two numbers multiplied together will give you -2 (the last term). So now you need to think about, what numbers added together will give you -2 and multipled will give you 1 (for the 1 in front of the x).

This is an easy case because there are only two combinations that will give you -2 (-1 times 2 or -2 times 1). Try filling in both in the blanks above then multiply it out to see if you get the same thing you began with.

Here are all the steps put together

Start with: 3x^2 + x -1
(3x + _)(x+_)
Try: (3x - 1)(x + 2)
Multiply it out: 3x^2 +6x -x -2 = 3x^2 -5x -2 (Not what we're looking for)
Try again: (3x - 2) (x+1)
Multiply it out: 3x^2 +3x -2x -2 = 3x^2 +x -2 (what we originally started with)
Finished answer: (3x-2)(x+1)

Hope this helps!

Answer 3

OK here see u have 3x^2+x-2,
first you will have to multiply first n last digits like 3 x -2 = -6,
now u have six right now look for any two factors that gives you -6 when multiplied....... and in the middle " 1 " when added or subtracted....... soo then possibility is 3 x -2 which r only factors that can give +1 when subtracted ......
soo now it will be
3x^2 + 3x - 2x - 2 right to make the middle 1x......k cuz +ve 3 and -2 when added gives +1 i.e +3+(-2) then just find the common from first two digits and last two digits that is
== 3x^2 + 3x - 2x - 2
== 3x( x + 1 ) - 2 ( x + 1)
make sure first bracket is same as second one that is most important......they are always same.......always......only then its right step...........
then write the brackets first and the remaining two digits in second bracket........ that is.........

== ( x + 1 ) ( 3x - 2 ) ===== answer..............

now you can check always by multiplying the two brackets and you must get
3x^2 + x - 2 always..........

OK............

Answer 4

The usual way. Look for terms that do the trick. If your factors are (ax+b)(cx+d), bd=-2, ac=3, and (bc+ad)=1. Unfortunately, you can solve this by trial and arror.
For the x, your a and c in factors would be 3 and 1 OR -3 and -1. For the constant term, your b d would be +2 and -2 and your d -1 and +1 respectively. You have 3x-2 and x+1 as factors in the end.

Answer 5

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

I just looked at what I needed: 3x^2 factors into 3x and x. Now I need to find two factors of -2 that I will multiply by 3x and x. Once I have these numbers, I will add them together and they need to equal the middle term, x. So, 3x(1) + x(-2) = 3x-2x=x, so 1 and -2 work because their product is -2. Then I write the equation in terms of the two factors. I have to criss-cross the numbers (3x and 1 and x and -2) because of FOIL.

Answer 6

to factor this out u need to conjugate binomials!
3x^2+x-2
1) well first you put the x's in quantative form! (3x- )(x+ )
NOTE: plus and minus equals a negative when it is a -2
2) next thing is to find factors of -2 that has the sum of x which equal 1! (3x-1)(x+2)
3) so ur answer is : (3x-1)(x+2) if you don't believe me do the FOIL system! [F=first, O=outer, I=insides, L=last]

Answer 7

-2*3=-6 (take the outsides and times them)

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

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

answer: (x+1)(3x-2)=0, x= -1,.2/3