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)).