Question

can someone help me with the problem Im about to post below please

Answer 1

Find all the solutions to \[x^3+3x^2+2x \le0\]

Answer 2

Ignore the "smaller than or equal to " symbol for now, and then find the roots of

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

Answer 3

Hint: Factor out x from all four terms.

Answer 4

I don't get it

Answer 5

If you don't get it, then ask some questions so that the problem will make more sense.
If what I've suggested is not clear, point out what is not clear and guess what it means.
You need to be involved in the solution of this problem.

Answer 6

how do I factor out the x ?

Answer 7

That's a lot better. Thank you.

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

Can you reduce each term? Note that 0/x is still 0.

Answer 8

Yes

Answer 9

Great. Reduce each term, one by one.

Answer 10

ok

Answer 11

Wouldn't it be 3+3^2+2=0

Answer 12

In the first term, x^3 (read "x cubed" can be re-written as x*x*x, where * means "multiply."

What is

x^3
--- ? It's not 3.
x

Answer 13

x^2?

Answer 14

or just 2

Answer 15

Remember, I explained that x^3 means x*x*x.

What is

x*x*x
----- ? Hint: It's not 3, it's not 2. Here's the rule of exponents you need:
x

\[\frac{ x^m }{ x^m }=x ^{m-n}\]

Answer 16

Hint: x^3 is \[x^m\]

Answer 17

and x is\[x^n\]

Answer 18

so, using m=3 and n=1, find\[\frac{ x^m }{ x^n }\]

Answer 19

So it would be \[x^{3-1}\]

Answer 20

which would give you x^2

Answer 21

Yes, that's right.

Answer 22

ok

Answer 23

Please finish:

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

Answer 24

For 3x^2 what would happen with the 3

Answer 25

Keep it. 3x^2 3x
---------- = --- = ?
x^1 1

Answer 26

2x

Answer 27

Actually, that'd be 3x. It's the next term that would be 2 (not 2x).

Answer 28

How did you get the 3x?

Answer 29

So you've gotten this far in the factoring process:

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

Hold the first x. Factor x^2 + 3x + 2.

Answer 30

(x+2)(x+1)

Answer 31

You should get two binomial factors: (x + ? ) (x + ? )

Answer 32

so 2 and 1?

Answer 33

(x+1)(x+2), yes.

So, now you have x(x+1)(x+2)=0, in completely factored form.

Answer 34

yes

Answer 35

On the left side you have 3 factors. Set each one to 0 separately and solve each equation for x. The 3 values you find are your solution set.

Answer 36

ok

Answer 37

So it would be -2 for the first one

Answer 38

-1 and 0

Answer 39

Yes, very good. I'd write your solution set as {-2, -1, 0}.