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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.

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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.

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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.

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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.

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where $a\neq 0$ – Belgi Jun 14 '12 at 14:39
Thanks. I have edited the answer – Saurabh Agarwal Jun 14 '12 at 14:39

Hint: Note that $2$ is a root of the polynomial, how can you get the other two roots ?

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I did not use the fact that the roots are in arithmetic progression – Belgi Jun 14 '12 at 14:36
That fact could be used to ascertain that 6/3 = 2 is one of the roots. – hardmath Feb 1 '14 at 4:40

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 $\rm\ \ f(x) = x^3 + c\: x^2 +\ldots\: \Rightarrow\ a = c/3,\ \ b = \sqrt{-f'(-a)}$

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