The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation. The given question is:
The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation.
I am confused about how to work out this question, but I feel like it has something to do with using α,β,γ, δ  as roots of the quartic.
I would greatly appreciate all help in a simple and efficient method to solving this question. Please note, according to the answers, there is only one value of $a$ and one value of $b$.
 A: If a polynomial has a $n$-th order repeated root, then that root will also be be a root of the first to the $(n-1)$th derivatives of the polynomial.
So the triple root will also be a root of the first and second derivative, the latter is $12x^2 - 18x - 12$. This quadratic has roots of $2, -\frac 12$.
Based on the first derivative, which is $4x^3 - 9x^2 - 12x + a$, you get $a = 28$ or $a = -\frac {13}4$. The corresponding $b$ values from the original polynomial are $b = -24$ and $b = -\frac 9{16}$ respectively. So $(a,b) = (28,-24)$ or $(-\frac{13}{4}, -\frac {9}{16})$.
If the question stated that $(a,b)$ were integers, then you could discard one set of solutions. But as it stands, both are admissible.
A: We know that the equation has a triple root, so if let $\alpha$ be the triple root and $\beta$ be the single root, we get that
\begin{equation}
x^4 - 3x^3-6x^2 + ax + b = (x-\alpha)^3(x-\beta)
\end{equation}
By expanding the RHS, we see that
\begin{equation}
x^4 - 3x^3-6x^2 + ax + b = x^4 - (3\alpha+\beta)x^3 + (3\alpha\beta + 3\alpha^2)x^2 + \cdots 
\end{equation}
Now equating the coefficients of the $x^2$ and $x^3$ term, we have the simultaneous equations
\begin{equation}
3 = 3\alpha+\beta \quad ; \quad -6 = 3\alpha\beta + 3\alpha^2
\end{equation}
By a simple substitution we conclude $(\alpha,\beta) = (2,-3)$ or $(\alpha,\beta) = (-1/2, 9/2)$. We can check that these indeed satisfy our conditions. It then follows that we have either $(a,b) = (28,-24)$ or  $(a,b) = (-13/4,-9/16)$.
A: the symmetric polynomials of the roots form the coefficients: multiplying out  $\quad (x-r_1)(x-r_2)(x-r_3)(x-r_4)=0\quad$ you see that
the second coefficient is the negative sum of the roots.
And we're given that$\quad r_1=r_2=r_3\quad$ and some other independent $\quad r_4$:
so $\quad 3r_1+r_4=3 \iff r_4=3-3r_1\quad$ and the third coefficient is the six term product-sum:
$$r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=r_1^2+r_1^2+r_1r_4+r_1^2+r_1r_4+r_1r_4\\=3r_1^2+3r_1r_4=-6\iff 3r_1(r_1+r_4)=-6 \iff r_1(r_1+3-3r_1)=-2\\\iff -2r_1^2+3r_1+2=0 \iff (2r_1+1)(r_1-2)=0$$
it's everything you need, the rest is computation.   Then $\quad -a,\quad$ the negative fourth coefficient is
$$r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4$$
and of course the final coefficient is $\quad r_1r_2r_3r_4$
