Question

How do I find the zeros of 3x^3-x^2-2x?

Answer 1

\[3x ^{3}-x ^{2}-2x\]

Answer 2

x(3x^2-x-2) ? I'm stuck :(

Answer 3

ok... so you now need to factor the quadratic...

so looking at

\[3x^2 - x - 2\]

multiply 3 and -2

then find the factors that add to -1

the larger factor is negative....

then its

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

remove a common factor from a binomial, and you'll find
it cancels with the denominator...

whats left is the factored form...

hope it helps

Answer 4

Review: "zeros" are values of x at which the given function is zero.

Fanduekisses has done a great job of typing out this expression in proper math symbols.

How would you factor \[3x^2 -x -2?\]

The first thing I'd do would be to multiply the first and last coefficients of this polynomial together: 3*2=-6.

Then I'd list all integer factors of the result (6), to obtain

-6*1
-3*2
-2*3

and so on.

Next, I'd look for the pair of factors, if any, whose sum is -1, the coeff. of the x term.

Answer 5

Which pair would that be?

Answer 6

-3 and 2

Answer 7

Campbell and I have presented essentially the same approach.
According to rules of mathematical logic, does that mean that he and I are both right? :)

Answer 8

If you choose -3 and 2, which i would choose myself,

then I'd "factor by grouping:"

3x^2 -x - 2 becomes
3x^2 -3x + 2x - 2, which follows your choice of -3 and 2 as factors of -6

Factor by grouping:

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

x-1 is the factor common to both sides, so factor that out and determine what remains; what remains is the 2nd factor.

Can you finish this?

Answer 9

ohh yes :) so zeros are 1 and 1?

Answer 10

Campbell: What would you do here to determine the zeros?

Answer 11

lol... oh well

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

remove a common factor from the 1st binomial

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

cancel the common factor leaves

(x -1)(3x +2)

so the factored form is

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

the reason why its confusing is teachers make students split the middle term and group in pairs... which can be extremely difficult for students...

Answer 12

oooohhhhh I see, yeah at first I found it confusing lol yeah some teachers make us do it differently which is actually more confusing :) Thanks guys :)

Answer 13

to find the zeros set each factor to zero

x = 0

x - 1 = 0

3x + 2 = 0

and solve each for x

Answer 14

Hope that you, Fondue, will check your answers. Substitute each supposed zero back into the original expression. If the expression is then zero, great; you have that zero correct. Check the other supposed zero also.

Answer 15

Thanks so much guys <3

Answer 16

Very happy to be of help. Thanks again, Campbell.