# THIS IS WORTH 50 POINTS!!I’ve been stuck on this for awhile now, I don’t know if anyone will understand this, but I’m praying that

THIS IS WORTH 50 POINTS!! I’ve been stuck on this for awhile now, I don’t know if anyone will understand this, but I’m praying that someone does, I’ll send more pictures of the problem!

$THIS IS WORTH 50 POINTS!! I’ve been stuck on this for awhile now, I don’t know if anyone will und$

## This Post Has 8 Comments

1. Expert says:

step-by-step explanation:

2. claydale1659 says:

I don’t know...I have to make my answer 20 characters long

3. bigmandavis says:

OK U

Step-by-step explanation:

4. jaxondbagley says:

5. mojr1000 says:

I think B is the answer

6. lexas2894 says:

Notice that when $x=-1$, you have

$(-1)^3+2(-1)+3=-1-2+3=0$

so $x=-1$ is a solution, and the cubic has $x+1$ as a factor. This means you can divide to get another quadratic factor and thus two more solutions.

$\dfrac{x^3+2x+3}{x+1}=x^2-x+3$

So,

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

You can use the quadratic formula here:

$x=\dfrac{1\pm\sqrt{1-12}}2=\dfrac12\pm i\dfrac{\sqrt{11}}2$

7. Expert says:

ab: 2 n83b8eb7ebi. 8e yen in

8. gracianoa says:

it's 10

hope that helped lowk