# Find all Zeros of the Function

• Oct 15th 2009, 06:35 PM
Porogoro
Find all Zeros of the Function
Hallooo
Ok this problem has been giving me troubles on my Take Home Test(we are allowed to use all resources available to use to do it :P)

Given that 1-sqrt(3)i is a zero of h(x)=3x^3-4x^2+8x+8, find all the zeros of h.

I put this into a graphing calculator online and the graph intersects the x-axis at (-.667,0) which is one of the possible rational zeros -(2/3) but when I try to figure out the other zeros, I end up at a dead end. : /
Any help is appreciated!
• Oct 15th 2009, 06:39 PM
GamesonPlanes
You have one of the zeros, 1-sqrt(3)i. Since they always come in groups, (some theorem i forgot that states if (a+bi) is an factor, then (a-bi) is also one) then another zero is 1+sqrt(3)i. Then you times them together and use long division with the answer you just got and 3x^3-4x^2+8x+8 to get the answer.
• Oct 15th 2009, 06:43 PM
apcalculus
Quote:

Originally Posted by Porogoro
Hallooo
Ok this problem has been giving me troubles on my Take Home Test(we are allowed to use all resources available to use to do it :P)

Given that 1-sqrt(3)i is a zero of h(x)=3x^3-4x^2+8x+8, find all the zeros of h.

I put this into a graphing calculator online and the graph intersects the x-axis at (-.667,0) which is one of the possible rational zeros -(2/3) but when I try to figure out the other zeros, I end up at a dead end. : /
Any help is appreciated!

complex roots come in pairs so $\displaystyle 1 + \sqrt{3} i$ is also a zero.

$\displaystyle (x - 1 - \sqrt{3} i) (x - 1 + \sqrt{3} i) =(x-1)^2 + 3$ is one of the factor.

Perform long division to find the remaining factor or factors. Good luck!
• Oct 16th 2009, 05:36 AM
stapel
Complex-valued and square-root zeroes come in pairs because that's how the Quadratic Formula spits them out: in pairs, because of the "plus / minus" in the Formula.

So any time they give you a zero with a square root or an imaginary (or both), another zero will be the conjugate of the zero they've given you. (Wink)