Factorization of polynomials

An expression of the form

P(x)=a₀xⁿ + a₁x^n-1 + ... + a_n-1x + a_n

is called a polynomial of x of degree n with real coefficients if the coefficients a_i - real numbers, and a₀ ≠ 0. The coefficient a_n is called the free member of the polynomial.
As is known, if a complex number

z₁ = g + ih

is a root of a polynomial, then its complex conjugate number

z₂^* = g - ih

is also a root of the polynomial. Therefore, their product
is a quadratic expression.
Thus, any polynomial with real coefficients can always be represented as a product of linear and quadratic factors

P(x)=a0(x-x1)k1*...*(x-xs)ks * (x2 + b1x + c1)r1 *...* (x2 + bpx + cp)rp,

where x₁, ..., x_s - real roots of the polynomial and That is, if all the roots of a polynomial with real coefficients are known, then we can immediately write its factorization.

Selecting the roots of a polynomial

In general, finding the roots of a polynomial of degree n is a rather difficult task, but you can try to find at least one root x₀. By dividing the original polynomial by the binomial x-x₀, we get a polynomial of degree n-1. Thus, we have simplified the original problem, since now we need to factor a polynomial of degree n-1. For example, for a polynomial of degree three, after dividing by x-x₀, we get a polynomial of degree two, the roots of which we find by simply solving a quadratic equation. The following theorem can be of significant help in selecting rational roots of a polynomial.
Theorem. If the polynomial P(x)=a₀xⁿ + a₁x^n-1 + ... + a_n-1x + a_n, a₀ ≠ 0, with integer coefficients has a rational root x₀ = (and this fraction is irreducible), then p – divisor of the free term a_n, and q – divisor of the leading coefficient a₀.
It follows from this theorem that if the leading coefficient is equal to one, then the integer roots of the polynomial should be sought only among the divisors of the free term.

Let's try to apply this theorem to factorize a polynomial.

Examples of factorization of polynomials

Example 1.
Factorize a polynomial 3x⁴ - 5x³ - 17x² + 13x + 6.
Solution.
A fourth degree polynomial can have at most 4 different roots. Let's try to find these roots.
Divisors of the free member: ±1, ±;2, ±3, ±6.
Divisors of the leading coefficient: ±1, ±3.
Therefore, rational roots of a polynomial can be sought among the numbers ±1, ±;2, ±3, ±6, .
Substituting x = 1 into the polynomial, we see that x = 1 is the root of the polynomial. This means that the original polynomial must be divided by x-1. Let's use Horner's scheme.



So, 3x⁴ - 5x³ - 17x² + 13x + 6=(x - 1)(3x³ - 2x² - 19x - 6).
Now we need to factorize the polynomial 3x³ - 2x² - 19x - 6.
Divisors of the free member: ±1, ±;2, ±3, ±6.
Divisors of the leading coefficient: ±1, ±3.
That is, its rational roots can be sought among the same numbers as the roots of the original polynomial:
±1, ±;2, ±3, ±6, .
Substituting x = -2 into the polynomial, we see that x = -2 is the root of the polynomial. This means that the original polynomial must be divided by x + 2.
Let's use Horner's scheme again.


Therefore, 3x³ - 2x² - 19x - 6 = (x + 2)(3x² - 8x - 3). Let's try to lower the degree of the polynomial 3x² - 8x - 3 in the same way, although we can simply solve the quadratic equation.
Divisors of the free member: ±1, ±;2, ±3.
Divisors of the leading coefficient: ±1, ±3.
That is, its rational roots can be sought among numbers ±1, ±;2, ±3, . We check by substitution that the numbers are suitable and 3. All roots have been found, which means the factorization has also been found:
3x⁴ - 5x³ - 17x² + 13x + 6 = 3(x - 1)(x + 2)(x - 3)(x + 1/3).
Answer: 3x⁴ - 5x³ - 17x² + 13x + 6 = (x - 1)(x + 2)(x - 3)(3x + 1).

Example 2.
Factorize a polynomial x⁵ - 2x⁴ - 10x³ + 20x² + 9x - 18.
Solution.
A fifth-degree polynomial can have at most 5 different roots. Let's try to find these roots. Since the divisor of the leading coefficient is 1, rational roots of the polynomial can be sought among the divisors of the free term: ±1, ±;2, ±3, ±6, ±9, ±18. Substituting x = 1 into the polynomial, we see that x = 1 is a root of the polynomial. This means that the original polynomial must be divided by x-1. Let's use Horner's scheme.



So, x⁵ - 2x⁴ - 10x³ + 20x² + 9x - 18 = (x - 1)(x⁴ - x³ - 11x² + 9x + 18). Similarly, the rational roots of the polynomial x⁴- x³- 11x² + 9x + 18 can be sought among the same divisors of the free term: ±1, ±;2, ±3, ±6, ±9, ±18. Substituting x=2 into the polynomial, we see that x=2 is a root of the polynomial. This means that this polynomial must be divided by x-2.



Thus, x⁴ - x³ - 11x² + 9x + 18 = (x - 2)(x³ + x² - 9x - 9). The roots of a polynomial x³ + x² - 9x - 9 can be found among the divisors of the free term 9:±1, ±3, ±9. It is easy to see that x = 3 is a root of the polynomial. Let us lower the degree of this polynomial by dividing it by x-3.



Hence, x³ + x² - 9x - 9 = (x - 3)(x² + 4x + 3). The roots of the square trinomial x²+ 4x + 3 are the numbers -1 and -3. The process of finding the roots of the original polynomial is complete.
Answer: x⁵ - 2x⁴ - 10x³ + 20x² + 9x - 18 = (x - 1)(x + 1)(x - 2)(x - 3)(x + 3).