# Finding zeros of a polynomial

• Jul 7th 2009, 05:17 AM
hiemmer
Finding zeros of a polynomial
How can I find the zeros(other than 0 itself) of this polynomial?

y = 12x^3 + 12x^2 + 12x

• Jul 7th 2009, 05:22 AM
alexmahone
$\displaystyle 12x^3+12x^2+12x=0$

$\displaystyle 12x(x^2+x+1)=0$

$\displaystyle x=0$ or $\displaystyle x=\frac{-1 \pm \sqrt{1-4}}{2}$

$\displaystyle x=0$ or $\displaystyle x=\frac{-1\pm i\sqrt3}{2}$
• Jul 7th 2009, 05:22 AM
artvandalay11
Quote:

Originally Posted by hiemmer
How can I find the zeros(other than 0 itself) of this polynomial?

y = 12x^3 + 12x^2 + 12x

Factor out 12x, so $\displaystyle y=12x(x^2+x+1)=0$
and now you can see x=0 is a solution and you can use the quadratic formula for $\displaystyle x^2+x+1$
• Jul 7th 2009, 05:25 AM
skeeter
Quote:

Originally Posted by hiemmer
How can I find the zeros(other than 0 itself) of this polynomial?

y = 12x^3 + 12x^2 + 12x

$\displaystyle 12x^3 + 12x^2 + 12x = 0$
$\displaystyle 12x(x^2 + x + 1) = 0$
$\displaystyle x = 0$ , use the quadratic formula to find the other two roots ...
$\displaystyle x = \frac{-1 \pm i\sqrt{3}}{2}$
$\displaystyle x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0$