How to solve polynomials

Question

Solve the equation $64x^3-240x^2+284x-105=0$ given that the roots are in an arithmetic . I tried having the roots as $a, (a+d), (a+2d)$ Factorising out $a$, $a(1+d+2d)$

Answer 1

You could try $a-d, a,$ and $a+d$.

Then you would have $a(a^2-d^2)=\frac{105}{64}$, but more importantly you would have $3a=\frac{240}{64}$. Both by Vieta's formulas.

Answer 2

Using the rational root theorem, the roots may involve the factors of the coefficients of the first and last terms. We can see that $64=4^3$ and $105=3\times5\times7$ and we can factor this polynomial easily.

$$64x^3-240x^2+284x-105=0=(4 x - 3) (4 x - 5) (4 x - 7)$$ so we have:

$$x=\frac{3}{4}, \frac{5}{4}, \frac{7}{4}$$

Answer 3

This can be solved using the cubic formula. Since the discriminant is negative, you have $3$ real solutions, and should use the trig form, $$ax^{3}+bx^{2}+cx+d=0,\ x=\frac{2\sqrt{b^{2}-3ac}\cos\left(\frac{1}{3}\cos^{-1}\left(-\frac{b^{3}-4.5abc+13.5a^{2}d}{\sqrt{\left(b^{2}-3ac\right)^{3}}}\right)\right)-b}{3a}$$ or use the rational root theorem to get $1$ solution, then solve for the other $2$ in terms of the $1$st. $$x_{2,\ 3}=\frac{\sqrt{b^{2}-4ac-3a^{2}x_{1}^{2}-2abx_{1}}-b-ax_{1}}{2a}$$