Relationship between the coefficients of a polynomial equation and its roots My question is-
Solve: $x^3 - 6x^2 + 3x + 10= 0$  given that the roots are in arithmetic progression.
Any help would be greatly appreciated.
 A: The roots are in arithmetic progression, so we can denote them as $a-d, a, a+d$ where $d$ is the common difference. The sum of these roots is then $3a$ and Viete's formulas shows that the sum of the roots must be $6$, so $a=2.$ So we have found a root, and can get the others by dividing the polynomial by $x-2$ and solving the remaining quadratic.
A: A three degree polynomial would have max 3 roots say p,q,r
The relations between the roots and coefficients in $ax^3 + bx^2 + cx + d$ are:
$\frac{-b}{a} =  p+q+r$
$\frac{c}{a} = pq + qr + pr$
$\frac{-d}{a} = pqr$
Note: where $a \neq 0$. Being 0 simply implies that the order of equation isn't actually 3.
Three equations can solve three unknowns (To use the fact that they are in AP, assume $p,q,r$ to be $q-d$, $q$, $q+d$ where $q$ is the mean and $d$ is the common difference of AP).
Note: The pattern of the roots and coefficients can be the same for n degrees of polynomial as well.
A: Hint: Note that $2$ is a root of the polynomial, how can you get the other two roots ?
A: Let the three roots be $a-d$, $a$, and $a+d$.
The sum of the roots is the negative of the coefficient of $x^2$, so $(a-d)+a+(a+d)=6$, and therefore $a=2$.
The product of the roots is the negative of the constant term, and therefore $(a-d)(a)(a+d)=-10$. Since we know $a$, we can now find $d$.  We get $a^2-d^2=-5$, and therefore $d^2=9$. 
Remarks: $1.$ In general, if we have a cubic equation $px^3+qx^2+rx+s=0$, where $p\ne 0$, and the roots are $\alpha$, $\beta$, and $\gamma$, then 
$$\alpha+\beta+\gamma=-\frac{q}{p}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{r}{p},\qquad \alpha\beta\gamma=-\frac{s}{p}.\tag{$1$}$$
There are analogous relations between coefficients and roots for polynomials of any degree.
$2$. Note that we called the roots $a-d, a, a+d$. We could have called them $a$, $a+d$, and $a+2d$, but the algebra would be a little more complicated. Symmetry is your friend.
$3.$ If the fact we used about roots is not available, there are a couple of things we can do. We can hope that the roots are rational. then they must all divide the constant term $10$. the only candidates are $\pm 1$, $\pm 2$, $\pm 5$, and $\pm 10$, and we quickly get to an answer. 
Or else we can let the roots be $\alpha$, $\beta$, and $\gamma$, and see that the polynomial must be $(x-\alpha)(x-\beta)(x-\gamma)$. Expanding, we find in essence the relationship $(1)$. 
Or else (but this is ugly) we can substitute $a-d$, $a$, and $a+d$ in our polynomial, and expand. Then solve for $a$ and $d$. The equations we get have a fair bit of symmetry, so this is not as difficult as it seems. 
A: Hint $\rm\ \ f(x) = (x+a\!-\!b)(x+a)(x+a\!+\!b)\iff f(x\!-\!a) = x^3 - b^2 x.\:$ But the only shift killing the $\rm\:-6\,x^2\:$ in your cubic is $\rm\:x \to x+2,\:$ yielding $\rm\:f(x\!+\!2) = x^3 - 9\,x,\:$ so $\rm\: b = \pm 3.$
Note  $\rm\,\ \ f(x\!-\!a) = x^3\! -\! b^2x\:\Rightarrow\:b^2 = -f'(x\!-\!a)|_{x=0} =\, -f'(-a)\ \ [\, =  -(3\cdot 2^2\! -\! 12\cdot 2\! +\! 3) = 9\  ].\:$   
Therefore $\rm\ \ f(x) = x^3 + c\: x^2 +\ldots\: \Rightarrow\ a = c/3,\ \ b = \sqrt{-f'(-a)}$
A: First of all the easiest mtd is 
$1)~$ Get one of the roots by hit and trial mtd in your question clearly $~-1~$ is one of the roots i.e when $~x =-1~$ the whole equation results $~0~$ therefore 
$~X=-1 ,~~~ X+1=0~$
$2)~$ apply division method to reduce it to quadratic and then $~u~$ can easily have your roots that are in A. P
