Question

how to solve a cubic polynomial

Answer 1

there is no any formulae for finding roots of cubic polynomial. we can find outs it roots by some fact ,
Let ax³ + bx² +cx +d is a general form of cubic polynomial. which roots alpha, beta and gamma .
then ,
alpha + beta + gamma = -b/a
alpha.beta + beta.gamma + gamma.alpha = c/a
alpha.beta.gamma = -d/a

this property u can use when, at least one root given,

# you can also solve this factorization method .
example :- x³ -3x² + 5x -3 =0
x³ -x² -2x² +2x +3x -3 =0
x²(x -1) -2x(x -1)+3(x -1) =0
(x² -2x +3)(x -1)=0
(x² -2x +3)(x -1) =0

you see here x² -2x +3 is not factorize because D =b² -4ac< 0

hence, x³ -3x² +5x -3 have one real roots e.g 1 and two imaginary roots available

in this way you can find out roots of polynomial .

Answer 2

We have a general cubic formula for all the types of cubics.

Say, a cubic polynomial is of the general form:

$f(x) = ax³+bx²+cx+d=0$

The cubic formula for a general cubic is quite long and it is given by:

$x _{1} = \frac{ - \sqrt[3]{2}b - \sqrt[3]{2 {b}^{3} - 9abc + 27 {a}^{2} {d}^{2} + 3a \sqrt{3} \sqrt{ - Δ} } - \sqrt[3]{2 {b}^{3} - 9abc + 27 {a}^{2} {d}^{2} - 3a \sqrt{3} \sqrt{ - Δ} } }{3 \sqrt[3]{2} a}$

$x _{2,3} = \frac{ - 2 \sqrt[3]{2}b + (1±\sqrt{3}i) \sqrt[ 3 ]{2 {b}^{3} - 9abc + 27 {a}^{2} {d}^{2} + 3a \sqrt{3} \sqrt{ - Δ} } + (1∓ \sqrt{3}i) \sqrt[3]{2 {b}^{3} - 9abc + 27 {a}^{2} {d}^{2} - 3a \sqrt{3} \sqrt{ - Δ}} }{6 \sqrt[3]{2} a}$

Where,

Δ is the cubic discriminant given by:

Δ = –27a²d²+18abcd–4ac³–4b³d+b²c²

However, Every cubic is writable in its depressed form:

$f(x)≡x ^{3} + px + q = 0$

where;

$p = \frac{3ac - {b}^{2} }{3 {a}^{2} } \: and , \: q = \frac{27 {a}^{2}d - 9abc + 2 {b}^{3} }{27 {a}^{3} }$

So, its solutions are pretty simple:

$x_{1} = \sqrt[3]{ - \frac{q }{2} + \sqrt{( \frac{q}{2} ) ^{2} + ( \frac{p}{3} ) ^{3} } } + \sqrt[3]{ - \frac{q }{2} - \sqrt{( \frac{q}{2} ) ^{2} + ( \frac{p}{3} ) ^{3} } }$

$x_{2,3} = ( \frac{ - 1±  \sqrt{3}i }{2}) \sqrt[3]{ - \frac{q}{2} + \sqrt{( \frac{q}{2} )^{2} + {( \frac{p}{3} )}^{3} } } + ( \frac{ - 1∓ \sqrt{3} i}{2} )\sqrt[3]{ - \frac{q}{2} - \sqrt{( \frac{q}{2} )^{2} + {( \frac{p}{3} )}^{3} } }$

So, the depressed cubic discriminant is very simple:

Δ= –27q²–4p³

If you'll notice this closely; you will see that the same solutions are just multiplied by the three cube roots of unity.

If you want the derivation of this cubic formula, i can give you the proof if you need it. :)