Factorization of cubic polynomials

Factorization of Cubic Polynomial: To factorize a cubic polynomial, we need to find its zeros, which are the values of the variable that make the polynomial equal to zero; but for that, we need at least one zero or factor of the polynomial. Once we have one zero or factor of the cubic polynomial, then we can use the synthetic division to find the other factors. And how do get the first factor? Well, in this regard, the remainder theorem is very helpful. If the cubic polynomial has an integer root, then one of the roots must be a factor of the constant term.

Example: Find the factors of $x^3 - 6x^2 + 11x -6$.

Solution:

Let $p(x) = x^3 - 6x^2 + 11x -6$. In order to find one factor or zero of the above polynomial, check the factors of constant term $“-6”$, which are -1, 1, -2, 2, -3, and 3. The leading coefficient is 1, so any rational root must have a denominator of 1.

Therefore, the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6$. We can use the Remainder Theorem to test each of these possible roots. For example, testing the root 1, we have:

$p(1) = (1)^3 - 6(1)^2 + 11(1) -6$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= 1 - 6 + 11 - 6$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= 0$

Hence, $1$ is a zero of the polynomial, therefore $x-1$ is a factor of the given polynomial. Therefore, now using the synthetic division to get the other factors, we get:

Therefore, the quotient is $x^2 -5x + 6$, which can be factorized, i.e.

$x^2 - 5x + 6 = x^2 - 2x-3x+6$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= x(x - 2)-3(x-2)$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= (x - 2)(x-3)$

Hence, $(x-1), (x-2),$ and $(x-3)$ are the factors of the given polynomial.