How to find the roots of a polynomial of $n^{t h}$ $n^(th)$ degree?

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How to find the roots of a polynomial of $n^{t h}$ $n^(th)$ degree?

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Answer 1 · 2015-12-23T21:09:35

It gets more complicated fast as $n$ $n$ increases...

Explanation:

Linear

$a x + b = 0$ $ax+b = 0$

Solution $x = - \frac{b}{a}$ $x = -b/a$

Quadratic

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$

Solutions $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

Cubic

$a x^{3} + b x^{2} + c x + d = 0$ $ax^3+bx^2+cx+d = 0$

Substitute $x = t - \frac{b}{3 a}$ $x = t-b/(3a)$ to get a cubic in the form:

$a t^{3} + b_{1} t + c_{1} = 0$ $at^3+b_1t+c_1 = 0$

If this has exactly one Real root, then I like to use Cardano's method, substituting $t = u + v$ $t = u+v$ to get:

$a u^{3} + a v^{3} + (3 a u v + b_{1}) (u + v) + c_{1} = 0$ $au^3+av^3+(3auv+b_1)(u+v)+c_1 = 0$
then adding the constraint $v = - \frac{b_{1}}{3 a u}$ $v = -b_1/(3au)$ to eliminate the $(u + v)$ $(u+v)$ term. Multiplying through by $u^{3}$ $u^3$ results in a quadratic in $u^{3}$ $u^3$ , whose two roots $r_{1}$ $r_1$ and $r_{2}$ $r_2$ are $u^{3}$ $u^3$ and $v^{3}$ $v^3$ .

Hence:

$x_{1} = - \frac{b}{3 a} + \sqrt[3]{r_{1}} + \sqrt[3]{r_{2}}$ $x_1 = -b/(3a)+root(3)(r_1)+root(3)(r_2)$

is the Real root, and the Complex roots are given by:

$x_{2} = - \frac{b}{3 a} + ω \sqrt[3]{r_{1}} + ω^{2} \sqrt[3]{r_{2}}$ $x_2 = -b/(3a)+omega root(3)(r_1)+omega^2 root(3)(r_2)$

$x_{3} = - \frac{b}{3 a} + ω^{2} \sqrt[3]{r_{1}} + ω \sqrt[3]{r_{2}}$ $x_3 = -b/(3a)+omega^2 root(3)(r_1) + omega root(3)(r_2)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2 i$ is the primitive Complex root of $1$ $1$ .

There are other methods for solving cubics. If the cubic has $3$ $3$ Real roots then there is a neat trigonometric solution in terms of $cos$ $cos$ and $arccos$ $arccos$ that you can reach by shoehorning the cubic into a form in terms of $cos (3 θ)$ $cos(3 theta)$ using

$cos (3 θ) = 4 {cos}^{3} (θ) - 3 cos (θ)$ $cos(3 theta) = 4cos^3(theta)-3cos(theta)$ .

Of course this is not quite as algebraically nice as expressions in terms of $n$ $n$ th roots.

Quartic

$a x^{4} + b x^{3} + c x^{2} + d x + e = 0$ $ax^4+bx^3+cx^2+dx+e = 0$

Substitute $x = t - \frac{b}{4 a}$ $x = t-b/(4a)$ and divide through by $a$ $a$ to convert to the form:

$t^{4} + a_{1} t^{2} + b_{1} t + c_{1} = 0$ $t^4+a_1t^2+b_1t+c_1 = 0$

Next consider possible quadratic factors of this:

$t^{4} + a_{1} t^{2} + b_{1} t + c_{1} = (t^{2} + A t + B) (t^{2} - A t + C)$ $t^4+a_1t^2+b_1t+c_1 = (t^2+At+B)(t^2-At+C)$

$= t^{4} + (B + C - A^{2}) t^{2} + A (C - B) t + B C$ $=t^4+(B+C-A^2)t^2+A(C-B)t+BC$

Hence:

$B + C = a_{1} + A^{2}$ $B+C = a_1+A^2$

$C - B = \frac{b_{1}}{A}$ $C-B = b_1/A$

$B C = c_{1}$ $BC= c_1$

So:

${(a_{1} + A^{2})}_{1}^{2} = {(B + C)}^{2} = {(C - B)}^{2} + 4 B C = \frac{b_{1}^{2}}{A^{2}} + 4 c_{1}$ $(a_1+A^2)^2 = (B+C)^2 = (C-B)^2 + 4BC = b_1^2/(A^2) + 4c_1$

Hence:

$A^{2} {(a_{1} + A^{2})}_{1}^{2} - 4 c_{1} A^{2} - b_{1}^{2} = 0$ $A^2(a_1+A^2)^2-4c_1A^2-b_1^2 = 0$

If you multiply this out, you get a cubic in $A^{2}$ $A^2$ , hence $3$ $3$ possible ways to factor the quartic as two quadratics.

Quintic

$a x^{5} + b x^{4} + c x^{3} + {d x}^{2} + e x + f = 0$ $ax^5+bx^4+cx^3+dx^2+ex+f = 0$

In general this has no algebraic solution in terms of $n$ $n$ th roots.

For example, the roots of $x^{5} + 4 x + 2 = 0$ $x^5+4x+2 = 0$ are not expressible in terms of $n$ $n$ th roots.

Answer 2 · 2015-12-23T21:13:54

You can find approximations for the roots of $n$ $n$ th degree polynomials using Newton's method.

Explanation:

Let $f (x)$ $f(x)$ be a polynomial, or indeed any nicely behaved function, whose zeros we want to find.

Given a first approximation $a_{0}$ $a_0$ , you can iteratively apply the formula:

$a_(i+1) = a_i - f(a_i)/(f'(a_i))$

to find successively better approximations.

This works with both Real and Complex zeros.

To find different roots, start with different initial approximations.

To find reasonable initial approximations, look at the behaviour of the function a little, perhaps evaluating $f(x)$ for a few different values first.

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