Suppose you know that 1/3 and -2 are zeros of 3x^4-x^3-21x^2-11x+6
Find all the zeros!
Hello,
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...