1. ## 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!

2. Hint: 1/3 and -2 are zero of (3x-1)(x+2).

You need the Remainder Theorem.

How's your Synthetic Division? Long division?

3. not really good

4. 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.

5. You could also use Vieta's Formulas and get the system
$\frac13-2+r_3+r_4=\frac13$
$-\frac23r_3r_4=2$
and solve it simultaneously for the other roots. You should get the same quadratic (I'm bad at synthetic division too).

6. Hello,

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 :

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

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

where P is a polynomial of degree 2.

We can rewrite this way :

\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}

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

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

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

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

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

-> $c=-3$

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

Now, you can develop...