# 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: $-1/3,1/(-1+2sqrt(2)),1/(-1-2sqrt(2))$.