Root and factorization of a polynomial

Concept of root

The root or zero of a polynomial $p (x)$ is that value $a$ that $p (a) = 0$

Mathematicians, throughout history, have always been fascinated by finding the roots of any polynomial. In general, this is a very complicated problem.

So, using the remainder theorem and the factor theorem, we can deduce some properties of the roots of a polynomial:

1) The roots of a polynomial are divisors of the independent term. If it does not have an independent term, it means that it is divisible by $x - a$ , where $a = 0$ , this is, it is divisible by $x$ .

Example

$p (x) = x^{5} + 2 x^{4} - 3 x^{3} + x^{2} - 1$ has as a root $1$ , $p (1) = 1^{5} + 2 \cdot 1^{4} - 3 \cdot 1^{3} + 1^{2} - 1 = 0$ and $1$ divides the independent term $- 1$ .

Example

The polynomial $p (x) = 2 x^{5} + 5 x^{4} + 4 x^{3} - x^{2} + x$ has the independent term equal to $0$ .

Then, using the factor theorem, $0$ is a root of $p (x)$ and therefore $x - 0 = x$ divides the polynomial $p (x)$ exactly.

2) Being $a_{i}$ the $i$ roots of a polynomial, we can express this polynomial as product of polynomials like $x - a_{i}$ .

Example

The polynomial $p (x) = x^{2} - 3 x + 2$ has roots $x = 2$ and $x = 1$ . Therefore, it can be expressed as $p (x) = (x - 2) \cdot (x - 1)$

Example

The polynomial $p (x) = x^{2} + 5 x + 6$ has root $x = - 2$ and $x = - 3$ . Therefore, it can be expressed as $p (x) = (x + 2) \cdot (x + 3)$

3) A polynomial is called irreducible or prime if it does not have any rational number that is a root.

Example

The polynomials $p (x) = x^{2} + x + 1$ and $q (x) = x^{2} + 1$ do not have any root in the rational numbers.

Factorization of a polynomial

The process of factorization of a polynomial consists in finding all of its roots.

There are different techniques used to find the roots of a polynomial. Next, we will explain the most outstanding ones:

Using notable products

The idea is to use the notable products but in the opposite sense. For example, if we know that: $(a - b) \cdot (a + b) = a^{2} - b^{2}$ It is clear that we can apply it the other way around: $a^{2} - b^{2} = (a - b) \cdot (a + b)$ Therefore, if we have a polynomial like the following one $x^{2} - 16$ which is $x^{2} - 4^{2}$ Applying the formula $x^{2} - 16 = x^{2} - 4^{2} = (x - 4) (x + 4)$ This is, $4$ and $- 4$ are the roots of the polynomial $x^{2} - 16$ .

Example

Factorize the following polynomial $x^{2} - 6 x + 9$ .

We can see that the previous polynomial is a square of a difference: $x^{2} - 6 x + 9 = x^{2} - 2 \cdot 3 \cdot x + 3^{2} = (x - 3)^{2}$ Therefore, the polynomial has the root $x = 3$ .

Example

Factorize the following polynomial $x^{3} + 12 x^{2} + 48 x + 64$ .

We can see that the previous polynomial corresponds to the cube of a sum: $x^{3} + 12 x^{2} + 48 x + 64 = x^{3} + 3 \cdot 4 \cdot x^{2} + 3 \cdot 4^{2} x + 4^{3} = (x + 4)^{3}$ Therefore, the polynomial has $x = - 4$ as a root.

Using the formula to solve quadratic equations

If we have a polynomial $p (x)$ of degree $2$ , we can equal it to $0$ and find the solution of the quadratic equation $p (x) = 0$ .

These values solution will be the roots of the polynomial $p (x)$ .

Example

Factorize the following polynomial $x^{2} - x - 2$ .

We must solve the following equation $x^{2} - x - 2 = 0$ .

We apply the formula to find the roots of a quadratic equation $x = \frac{- (- 1) \pm \sqrt{(- 1)^{2} - 4 \cdot 1 \cdot (- 2)}}{2 \cdot 1} = \frac{1 \pm \sqrt{9}}{2} = {\begin{matrix} \frac{1 + 3}{2} = 2 \\ \frac{1 - 3}{2} = - 1 \end{matrix}$

Therefore, the polynomial has $x = 2$ and $x = - 1$ as roots.

Example

Factorize the following polynomial $x^{2} + x - 6$ .

We must solve the following equation $x^{2} + x - 6 = 0$ .

We apply the formula to find the roots of a quadratic equation $x = \frac{- 1 \pm \sqrt{1^{2} - 4 \cdot 1 \cdot (- 6)}}{2 \cdot 1} = \frac{1 \pm \sqrt{25}}{2} = {\begin{matrix} \frac{1 + 5}{2} = 3 \\ \frac{1 - 5}{2} = - 2 \end{matrix}$

Therefore, the polynomial has $x = 3$ and $x = - 2$ as its roots.

Using the formula to solve biquadratic equations

If we have a polynomial $p (x)$ of degree $4$ and even exponents, we can equal it to $0$ and find the solution of the biquadratic equation $p (x) = 0$ .

These value solution will be the roots of the polynomial $p (x)$ .

Example

Factorize the following polynomial $x^{4} - 5 x^{2} + 4$ .

We must solve the following equation $x^{4} - 5 x^{2} + 4 = 0$ .

We change the variable $x^{2} = t$ $t^{2} - 5 t + 4 = 0$

We apply the formula to find the roots of a quadratic equation $x = \frac{- (- 5) \pm \sqrt{(- 5)^{2} - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{5 \pm \sqrt{9}}{2} = {\begin{matrix} \frac{5 + 3}{2} = 4 \\ \frac{5 - 3}{2} = 1 \end{matrix}$

Now we undo the change: $x^{2} = 4 \Rightarrow {\begin{matrix} x = 2 \\ x = - 2 \end{matrix}$ $x^{2} = 1 \Rightarrow {\begin{matrix} x = 1 \\ x = - 1 \end{matrix}$

Therefore, the polynomial has $x = 2, x = - 2, x = 1$ and $x = - 1$ as its roots.

Example

Factorize the following polynomial $x^{4} - 10 x^{2} + 9$ .

We must solve the following equation $x^{4} - 10 x^{2} + 9 = 0$ .

We change the variable $x^{2} = t$ $t^{2} - 10 t + 9 = 0$

We apply the formula to find the roots of a quadratic equation $x = \frac{- (- 10) \pm \sqrt{(- 10)^{2} - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{10 \pm \sqrt{64}}{2} = {\begin{matrix} \frac{10 + 8}{2} = 9 \\ \frac{10 - 8}{2} = 1 \end{matrix}$

Now we undo the change: $x^{2} = 9 \Rightarrow {\begin{matrix} x = 3 \\ x = - 3 \end{matrix}$ $x^{2} = 1 \Rightarrow {\begin{matrix} x = 1 \\ x = - 1 \end{matrix}$

Therefore, the polynomial has $x = 3, x = - 3, x = 1$ and $x = - 1$ as its roots.

Using the factor theorem

For polynomials of a larger degree, our only tool is to use the factor theorem.

This way, to find the roots of a polynomial, it will only be necessary to evaluate the polynomial for the values of $x$ that are divisors of the independent term, and for the values in which the expression turns out to be empty, these values will be the roots of the polynomial. With a few examples, we will visualize the procedure:

Example

Factorize the following polynomial $p (x) = x^{2} - 3 x + 2$ .

As the properties of the factor theorem show, if $a$ is a root of $p (x)$ , then $p (a) = 0$ .

But, what value does $a$ have? There is a property that will be extremely useful:

If $a$ is a root of $p (x)$ , $a$ is a divisor of the independent term of $p (x)$ .

In our case, the divisors of the independent term of the polynomial (of value $2$ ) are $1, - 1, 2, - 2$ Therefore, it is only necessary to evaluate these values in the polynomial and apply the factor theorem:

$p (1) = 1^{2} - 3 \cdot 1 + 2 = 0$

$p (- 1) = (- 1)^{2} - 3 \cdot (- 1) + 2 = 6$

$p (2) = 2^{2} - 3 \cdot 2 + 2 = 0$

$p (- 2) = (- 2)^{2} - 3 \cdot (- 2) + 2 = 12$

And so, the roots of $p (x)$ are $x = 1$ and $x = 2$ .

Example

Factorize the following polynomial $p (x) = x^{2} + 6 x - 7$ .

The divisors of the independent term of the polynomial (of value $7$ ) are: $1, - 1, 7, - 7$ Therefore, it is only necessary to evaluate these values in the polynomial and apply the factor theorem:

$p (1) = 1^{2} + 6 \cdot 1 - 7 = 0$

$p (- 1) = (- 1)^{2} + 6 \cdot (- 1) - 7 = - 12$

$p (7) = 7^{2} + 6 \cdot 7 - 7 = 84$

$p (- 7) = (- 7)^{2} + 6 \cdot (- 7) - 7 = 0$

And so, the roots of $p (x)$ are $x = 1$ and $x = - 7$ .