# Solving cubic or higher degree polynomials by hand

1. Mar 4, 2006

### dak246

How does one find the roots of an equation, for instance:
r^4 + r^3 - 7r^2 - r + 6 = 0
...completely by hand. Is there some type of non-lengthy process or trick to find special circumstances for easy solving? Thanks.

2. Mar 4, 2006

### rindech

Such equations, up to the third degree generally, may be solved by hand, but the method of solution is quite complex and involved. For equations above the third degree, unless you can come up with a factoring which brings it down to equations of the third degree or less, it is generally impossible to solve for the roots of such equations completely by hand. One would be better off utilizing a computer driven equation solving program like MatLab or Maple.
If such an equation is given by an instructor as an assignment, and one is not going to use the computer, it is a good bet that it can be factored. Look for quadratic factoring in your example. Then once factored, use the known solution technique of quadratic equations to find the four roots, 2 roots for each quadratic equation. (i.e. look for a,b,c,d, where (r^2+ar+b)(r^2+cr+d)=r^4+r^3-7r^2-r+6. So a+c=1, b+d+ac=-7, ad+bc=-1, bd=6).

3. Mar 4, 2006

### benorin

The use of the rational root theorem can be handy (if your polynomial has integer coeffs).

4. Mar 4, 2006

### dak246

Both those posts helped me greatly. Thanks alot.

5. Mar 4, 2006

### robert Ihnot

You can try a few small values to get the "feel" of it. In this case, r=1 is a root, and so by the factor theorem we have: (r-1)(r^3+2r^2-5r-6). In the second poloynominal we have the constant value 6, so that integers that might divide it are of the form plus and minus: 1, 2,3,6.

6. Mar 5, 2006

7. Mar 5, 2006

### mathwonk

an extract from a famous work of algebra:

i got bored with fixing all the superscripts that reproduced badly after a while:

For the solutions x1, x2 of the quadratic equation x^2+px+q = 0, we have (since at least 825 AD) the formula x = (1/2){-p + D^(1/2)}, in terms of the coefficients p, q, where D = (x1-x2)^2 = p^2-4q. The two solutions are obtained by taking the two square roots of D. For the cubic equation x^3+px+q = 0, years of toil and some intrigue led to the publication, by Cardano in 1545, of the following formula. x =

(1/3)[{(-27q/2)+(3/2)(-3D)^(1/2)}^(1/3) - 3p/{(-27q/2) + (3/2)(-3D)^(1/2)}^(1/3)].

Fixing a value of (-3D)^(1/2), where
D = (x1- x2)^2(x1-x3)^2(x2-x3)^2 = -4p^3-27q^2, and varying the cube root gives all three solutions. Eg., in the equation x^3-1 = 0, p = 0, q = -1, D = -27, so we get x = (1/3){27/2 + 27/2}^(1/3) =
{1/2 + 1/2}^(1/3) = 11/3, as hoped.

Similarly, for x^3-a = 0, we have p = 0, q = -a, D = -27a^2, hence
x = (1/3){27a/2 + (3/2)(81a^2)^(1/2)}^(1/3) = (1/3){27a}^(1/3) = a^(1/3).

For x^3 - 4x = 0, we get p = -4, q = 0, D = 256, and so

x = (1/3) [{(3/2)(-768)^(1/2)}^(1/3) + 12/{(3/2)(-768)^(1/2)}^(1/3)]

= (1/3) [ {(-27)(64)}^(1/6) + 12/{(-27)(64)}^(1/6) ]

= (1/3) [ 2¯3 i1/3 + 12/{2¯3 i1/3} ] = (2/¯3 )( i1/3 + i-1/3)

= (4/¯3 )Re(i1/3).

Varying the cube roots of i in this formula gives
(4/¯3 )(cos(