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 $x$.

$p(x)=x^5+2x^4-3x^3+x^2-1$ has as a root $1$, $$p(1)=1^5+2\cdot1^4-3\cdot1^3+1^2-1=0$$ and $1$ divides the independent term $-1$.

The polynomial $p(x)=2x^5+5x^4+4x^3-x^2+x$ 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 $x-a_i$.

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

The polynomial $p(x)=x^2+5x+6$ 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.

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 $x^2-16$.

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

We can see that the previous polynomial is a square of a difference: $$x^2-6x+9=x^2-2\cdot3\cdot x+3^2=(x-3)^2$$ Therefore, the polynomial has the root $x=3$.

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

We can see that the previous polynomial corresponds to the cube of a sum: $$x^3+12x^2+48x+64=x^3+3\cdot4\cdot x^2+3\cdot4^2x+4^3=(x+4)^3$$ 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 $p(x)$.

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 $$x=\dfrac{ -(-1)\pm\sqrt{ (-1)^2-4\cdot1\cdot(-2) } }{2\cdot1}=\dfrac{1\pm\sqrt{9}}{2}= \left\{\begin{array}{c} \dfrac{1+3}{2}=2 \\\\ \dfrac{1-3}{2}=-1 \end{array} \right.$$

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

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 $$x=\dfrac{ -1\pm\sqrt{ 1^2-4\cdot1\cdot(-6) } }{2\cdot1}=\dfrac{1\pm\sqrt{25}}{2}= \left\{\begin{array}{c} \dfrac{1+5}{2}=3 \\\\ \dfrac{1-5}{2}=-2 \end{array} \right.$$

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 $p(x)$.

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

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

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

We apply the formula to find the roots of a quadratic equation $$x=\dfrac{ -(-5)\pm\sqrt{ (-5)^2-4\cdot1\cdot4 } }{2\cdot1}=\dfrac{5\pm\sqrt{9}}{2}= \left\{\begin{array}{c} \dfrac{5+3}{2}=4 \\\\ \dfrac{5-3}{2}=1 \end{array} \right.$$

Now we undo the change: $$x^2=4 \Rightarrow \left\{\begin{array}{c} x=2 \\\\ x=-2 \end{array} \right.$$ $$x^2=1 \Rightarrow \left\{\begin{array}{c} x=1 \\\\ x=-1 \end{array} \right.$$

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

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

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

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

We apply the formula to find the roots of a quadratic equation $$x=\dfrac{ -(-10)\pm\sqrt{ (-10)^2-4\cdot1\cdot9 } }{2\cdot1}=\dfrac{10\pm\sqrt{64}}{2}= \left\{\begin{array}{c} \dfrac{10+8}{2}=9 \\\\ \dfrac{10-8}{2}=1 \end{array} \right.$$

Now we undo the change: $$x^2=9 \Rightarrow \left\{\begin{array}{c} x=3 \\\\ x=-3 \end{array} \right.$$ $$x^2=1 \Rightarrow \left\{\begin{array}{c} x=1 \\\\ x=-1 \end{array} \right.$$

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:

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

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:

$$p(1)=1^2-3\cdot1+2=0$$

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

$$p(2)=2^2-3\cdot2+2=0$$

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

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

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

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:

$$p(1)=1^2+6\cdot1-7=0$$

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

$$p(7)=7^2+6\cdot7-7=84$$

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

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