# Finding zeros of a polynomial

• Jul 7th 2009, 06:17 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, 06:22 AM
alexmahone
$12x^3+12x^2+12x=0$

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

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

$x=0$ or $x=\frac{-1\pm i\sqrt3}{2}$
• Jul 7th 2009, 06:22 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 $y=12x(x^2+x+1)=0$
and now you can see x=0 is a solution and you can use the quadratic formula for $x^2+x+1$
• Jul 7th 2009, 06:25 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

$12x^3 + 12x^2 + 12x = 0$

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

$x = 0$ , use the quadratic formula to find the other two roots ...

$x = \frac{-1 \pm i\sqrt{3}}{2}$
• Jul 7th 2009, 09:13 AM
Sean12345
Or if you're not a fan of the quadratic formula (I never was), you could complete the square on the quadratic giving,

$
x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0
$