How do you factor #3x ^ { 5} + 9x ^ { 4} - 6x# completely?
1 Answer
Explanation:
=
There is a method to solve a quartic equation in general by hand (and calculator) on paper.
This method is called 'the method of Descartes'.
Substituting x=y+p in
if we take 4p+a=0 or p=-a/4, the coefficient of y³ becomes zero, and we get :
(with p = -3/4)
The method of Descartes writes
elimination of m and n results in the cubic equation
with x = k², so once we have found positive x, we know k,m,n and the solutions of the quartic equation as solutions of two quadratic equations.
In our example this cubic equation is :
There is a method to solve a cubic equation in general by hand (and simple calculator) on paper.
Dividing by the first coefficient yields :
Substituting x=y+p in
if we take 3p+a=0 or p=-a/3, the first coefficient becomes zero, and we get :
(with p = 9/4)
Substituting y=qz in
if we take q = sqrt(|b|/3), the coefficient of z becomes 3 or -3, and we get :
(here q = 1.63299316)
Substituting z = t - 1/t, yields :
Substituting u = t³, yields the quadratic equation :
A root of this quadratic equation is u=0.22921437.
Substituting the variables back, yields :
t = cuberoot(u) = 0.61199416.
z = -1.02200836.
y = -1.66893267.
x = 0.58106733.
The other roots of the cubic can be found by dividing and solving the remaining quadratic equation.
Filling in this value for k² yields as roots :
So the factorization is
Note that here there is a shortcut possible if we use y=1/x as substitution then we have a quartic with first coefficient already zero and yields in a cubic resolvent with first coefficient also zero already. I did not use the shortcut to show you the general method of Descartes.