# Thread: Finding Zeros of Cubic Polynomial

1. ## Finding Zeros of Cubic Polynomial

I am having problems factoring this cubic polynomial. I can't find how they got to the factored form on the right.

However i was able to get to the same answer but first factored, then found possible zeros using p/q, then did long division, then used the quadratic formula. Is there a faster way to do this?

Thanks.

2. ## Re: Finding Zeros of Cubic Polynomial

Hey petenice.

Factoring is a weird thing: in root finding a wide range of approaches are taken.

One that is commonly taught is to "guess" one root and then factor the polynomial by dividing P(X) by (x-a) where a is the root. It's not a mechanical systematic thing unless you have previous experience or can "see" something in the polynomial itself (i.e. an easy solution).

There is however a formula for obtaining the roots to a cubic polynomial given a cubic equation and it is more complex than the standard quadratic.

You also have other results in algebra that give ways to factorize equations of a particular form.

The cubic formula gives you a systematic way to get the roots and the proof shows you how the routine is constructed, but when you don't have these and try to "guess" roots then that's what happens: you make a few guesses and if you get one right then you do the long division with a polynomial and you factorize out to solve something simpler.

3. ## Re: Finding Zeros of Cubic Polynomial

I would use the rational roots theorem to see if a rational root exists as you did, but I would first factor out a 2 to make things a little simpler.

4. ## Re: Finding Zeros of Cubic Polynomial

hello! if u intressted i can show u how to solve all the roots with rational root theorem and polynomal division!

5. ## Re: Finding Zeros of Cubic Polynomial

Hi, please take a look at: Cubic function - Wikipedia, the free encyclopedia, you can use some of the simpler methods there.

Salahuddin
Maths online

6. ## Re: Finding Zeros of Cubic Polynomial

Hello, petenice!

I don't think there is a faster way . . .

We want to solve: . $32x^3 - 48x^2 + 12x + 2 \:=\:0$

Divide by 2: . $16x^3 - 24x^2 + 6x + 1 \:=\:0$

According to the Rational Roots Theorem,
. . the only possible rational roots are: . $\pm1,\:\pm\tfrac{1}{2},\:\pm\tfrac{1}{4},\:\pm \tfrac{1}{8},\:\pm\tfrac{1}{16}$

We find that $x = \tfrac{1}{2}$ is a zero of the polynomial.
. . Hence, $2x-1$ is a factor.

Using long division: . $16x^3 - 24x^2 + 6x + 1 \;=\;(2x-1)(8x^2-8x-1)$

The other two roots arise from: . $8x^2 - 8x - 1 \:=\:0$

. . $x \;=\;\frac{\text{-}(\text{-}8) \pm \sqrt{(\text{-}8)^2 - 4(8)(\text{-}1)}}{2(8)} \;=\;\frac{8 \pm\sqrt{96}}{16} \;=\; \frac{8 \pm4\sqrt{6}}{16} \;=\;\frac{2\pm\sqrt{6}}{4}$