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 $$x=-7$$.