Question

Solve: 2x^3+x^2+x=0
any help?

Answer 1

factor out an x and what do you have?

Answer 2

^

Answer 3

you can help bloop

Answer 4

I'll let you have this one.

Answer 5

x(x^2+x+1)=0?

Answer 6

Divide both sides of the equation by x, and you have a quadratic equation. Solve.

Answer 7

keep the x, its a 0

Answer 8

you will have 3 answers

Answer 9

0 is one of them, now solve the quadratic

Answer 10

zzr0ck3r made mention to 3 solutions. Keep in mind that this is a mathematical law: The degree of a polynomial is equal to its roots.

Answer 11

So should my final answer be (x=0) (x=-1) (x=-1)? This answer just doesn't seem right though...

Answer 12

\[0=2x^3+x^2+x=x(2x^2+x+1)\]
ok so x =0 is one solution so lets look at
\[2x^2+x+1=0\\2x^2+x=-1\\
2(x^2+\frac{x}{2})=-1\\2(x+\frac{1}{4})^2=-1+2(\frac{1}{4})^2=-1+\frac{1}{8}=\frac{-7}{8}\\so\\2(x+\frac{1}{4})^2=\frac{-7}{8}\\x=\pm\sqrt{\frac{-7}{16}}-\frac{1}{4}=\pm\frac{\pm\sqrt{7}i-1}{4}\]

Answer 13

COMPLEX ROOTS FTW.

Answer 14

the method I used to solve the quadratic was "completing the square"
you could have used \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 15

thanks guys much love!