# find the zeros

• May 10th 2008, 05:10 PM
gumi
find the zeros
Suppose you know that 1/3 and -2 are zeros of 3x^4-x^3-21x^2-11x+6
Find all the zeros!
• May 10th 2008, 05:28 PM
TKHunny
Hint: 1/3 and -2 are zero of (3x-1)(x+2).

You need the Remainder Theorem.

How's your Synthetic Division? Long division?
• May 10th 2008, 05:31 PM
gumi
not really good:(
• May 10th 2008, 05:38 PM
TKHunny
It's time to practice.

Every time you find a zero, use it to deflate the polynomial to one degree lower. Eventually, you'll get it down to quadratic and the last two are easy.
• May 10th 2008, 06:03 PM
sleepingcat
You could also use Vieta's Formulas and get the system
$\displaystyle \frac13-2+r_3+r_4=\frac13$
$\displaystyle -\frac23r_3r_4=2$
and solve it simultaneously for the other roots. You should get the same quadratic (I'm bad at synthetic division too).
• May 11th 2008, 03:16 AM
Moo
Hello,

Quote:

Originally Posted by gumi
Suppose you know that 1/3 and -2 are zeros of 3x^4-x^3-21x^2-11x+6
Find all the zeros!

I'd do it this way :

$\displaystyle \frac 13$ and $\displaystyle -2$ are zeros of $\displaystyle 3x^4-x^3-21x^2-11x+6$

Hence $\displaystyle 3x^4-x^3-21x^2-11x+6=(x-\frac 13)(x+2)P(x)$

where P is a polynomial of degree 2.

We can rewrite this way :

\displaystyle \begin{aligned} 3x^4-x^3-21x^2-11x+6 &=3(x-\frac 13)(x+2)Q(x) \\ &=(3x-1)(x+2)Q(x) \end{aligned}

$\displaystyle Q(x)=\frac{P(x)}{3}$, but we don't care.

$\displaystyle Q(x)=ax^2+bx+c$

Since the first term is $\displaystyle 3x^4$, a=1 because we would have 3x*x*ax˛ as the dominating term.

Hence $\displaystyle Q(x)=x^2+bx+c$

Let's study the constant term :
-1*2*c=6

-> $\displaystyle c=-3$

Thus $\displaystyle Q(x)=x^2+bx-3$

Now, you can develop...