Equations of Higher Degrees
Consider equation of the form P(x)=0, where P(x) is polynomial of degree higher than 2.
To solve such equations method of factoring and method of entering new variable are used.
Factoring of polynomial of n-th degree is based on the following fact.
Fact. Suppose we are given polynomial `P(x)=a_0x^n+a_1x^(n-1)+...+a_(n-1)x+a_n`, all coefficients of which are integer numbers (`a_0!=0`). Then if integer number x=b is root of the polynomial P(x), then it is divider of free member `a_n`.
Example 1. Solve equation `x^3+4x^2-24=0`.
Let's try to find integer root of the equation. For this we write out all dividers of free member -24: `+-1;+-2;+-3;+-4;+-6;+-12;+-24`.
If `x=1` then `1^3+4*1^2-24=-19!=0`, i.e. `x=1` is not root.
If `x=-1` then `(-1)^3+4*(-1)^2-24=-21!=0`, i.e. `x=-1` is not root.
If `x=2` then `2^3+4*2^2-24=0`, i.e. `x=2` is root.
Now, we divide polynomial `x^3+4x^2-24` by `x-2`: `x^3+4x^2-24=(x-2)(x^2+6x+12)`.
Quadratic equation `x^2+6x+12=0` doesn't have roots, therefore initial equation has only one root: x=2.
Example 2. Solve equation `21x^3+x^2-5x-1=0`.
Divide both sides of equation by `x^3`: `21+1/x-5/x^2-1/x^3=0`.
Let `1/x=y` then `21+y-y^2-y^3=0` or `y^3+5y^2-y-21=0`.
Let's try to find integer root of the equation. For this we write out all dividers of free member -21: `+-1;+-3+-7`.
After checking we find that `y=-3` is root of the equation. Dividing `y^3+5y^2-y-1` by `y+3` gives: `y^3+5y^2-y-21=(y+3)(y^2+2y-7)`.
Quadratic equation `y^2+2y-7=0` has two roots: `-1+-2sqrt(2)`.
Now, since `y=1/x` then `x=1/y`.
Therefore, initial equation has 3 roots: `-1/3,1/(-1+2sqrt(2)),1/(-1-2sqrt(2))`.