Polynomial roots (or zeros)

On this post you will find what the roots (or zeros) of a polynomial are and how to calculate all the roots of a polynomial. Also, you will see examples and exercises solved step by step of polynomial roots.

What are the roots of a polynomial?

The roots (or zeros) of a polynomial are the values of x for which the polynomial is equal to zero, that is, x=a is a polynomial root if P(a)=0.

For example, let P(x) be a polynomial:

 P(x)=x^2-3x+2

We can check that one of the roots of the polynomial is x=1 evaluating the polynomial at that point:

P(1)=1^2-3\cdot1+2=1-3+2\color{orange}\bm{= 0}

On the other hand, 3 is not a root of this polynomial because it is not a value that cancels the polynomial, in other words, if we substitute x for 3 the polynomial is nonzero:

 P(3)=3^2-3\cdot 3+2 = 9-9+2=2 \color{orange} \bm{\neq 0}

I hope that now you understand better what the root of a polynomial is, but wouldn’t you like to know how many roots a polynomial has? Or how to find all the roots of a polynomial? Well, this is precisely what we are going to see in the next section.

How to find all the roots of a polynomial

To find all the roots of a polynomial, you must do the following steps:

  1. First, find all the divisors (or factors) of the constant term of the polynomial.
  2. Second, evaluate the polynomial at all the values found in the previous step.
  3. Third, if the evaluation of a number results in zero, this number is a root of the polynomial. Otherwise, that number is not a root of the polynomial.

Example

Next we are going to solve an example step by step so that you can better understand how to find the zeros of a polynomial.

  • What are all the roots (or zeros) of the following polynomial?

 P(x) = x^2-5x+6

First of all, we must find the factors of the constant term of the polynomial, because any root of a polynomial is also a factor of the constant term. So, the factors of 6 are:

Factors of 6: +1, -1, +2, -2, +3, -3

Remember that if a number is a factor, its negative is also. Since a number is divisible by positive and negative numbers.

So the possible roots or zeros of the polynomial are: ±1, ±2, ±3. Therefore, we must evaluate the polynomial at all these values. For that, we substitute all these values in the expression of the polynomial for x:

 P(1) = 1^2 -5\cdot 1 +6= 1 -5 +6 =2

 P(-1) = (-1)^2 -5\cdot (-1) +6 =1+5+6 = 12

 P(2) = 2^2 -5\cdot 2 +6 =4-10+6= \color{blue} \bm{0}

 P(-2) = (-2)^2 -5\cdot (-2) +6 =4+10+6 =20

 P(3) = 3^2 -5\cdot 3 +6 =9-15+6=\color{blue} \bm{0}

 P(-3) = (-3)^2 -5\cdot (-3) +6 =9+15+6 =30

So the polynomial equals to zero only when the variable x is +2 or +3, so these are the roots of the polynomial:

Roots or zeros of the polynomial: +2 and +3

On the other hand, note that the polynomial has as many roots as its degree. In the section of the properties of the roots of a polynomial (below), we will see why this characteristic always holds for any polynomial.

Properties of polynomial roots

The polynomial roots (or zeros) have the following characteristics:

  1. As we have seen before, the integer roots (or zeros) of a polynomial are divisors of the constant term of the polynomial.
  1. If we know all the roots of a polynomial, we can express the polynomial in the form of products of binomials of the type (x-a).

For example, the polynomial P(x)=x^3+3x^2-x-3 has three roots, which are x=+1,x= -1, and x = -3. Thus, we can rewrite the polynomial in the form of three multiplications of factors, each one formed by the variable x and a root changed sign:

\displaystyle\definecolor{vermell}{HTML}{F44336}\definecolor{blau}{HTML}{2196F3}\definecolor{verd}{HTML}{27AE60} P(x) =x^3+3x^2-x-3 \ \longrightarrow \ \text{roots} \begin{cases} x=\color{verd}\bm{+1} \\[2ex] x=\color{vermell}\bm{-1} \\[2ex] x=\color{blau}\bm{-3}\end{cases}

\definecolor{vermell}{HTML}{F44336}\definecolor{blau}{HTML}{2196F3}\definecolor{verd}{HTML}{27AE60}P(x) =x^3+3x^2-x-3 = (x\color{verd}\bm{-1}\color{black})\cdot (x\color{vermell}\bm{+1}\color{black}) \cdot (x\color{blau}\bm{+3}\color{black})

  1. A polynomial has as many complex roots as its degree indicates. So a second-degree polynomial will have 2 roots, a third-degree polynomial will have 3 roots, a fourth-degree polynomial will have 4 roots, and so on.
  1. If a polynomial does not have a constant term, it means that at least one of its roots is 0. Then, the rest of the polynomial roots are divisors of the coefficient of the lowest degree monomial.

For example, the following polynomial has no constant term:

 P(x) =x^3+x^2-2x

So a root of the polynomial must necessarily be 0. And the rest of the roots are factors of the coefficient of the term of lowest degree, that is, -2. In particular, the other polynomial zeros are x=+1 and x=-2.

Roots or zeros of the polynomial: 0, +1 and -2

  1. When the roots of a polynomial cannot be determined, we say that it is an irreducible polynomial.

For example, we are going to try to calculate the roots of the following polynomial:

 P(x) =x^2+3x-1

The only possible roots of the polynomial are the factors of -1, that is, -1 and +1. So we evaluate the polynomial at these values:

 P(1) = 1^2 +3\cdot 1 -1= 1 +3 -1 =3 \neq 0

 P(-1) = (-1)^2 +3\cdot (-1)-1 =1-3-1 =-3 \neq 0

In no case does the polynomial result in 0, therefore it does not have integer roots and it is an irreducible polynomial over the integers.

  1. When the polynomial is composed of the product of several polynomials, it is not necessary to compute the product to calculate the roots, but the roots of the polynomial are the roots of each factor.

As an example, if we have the following polynomial:

 P(x) = (x-2) \cdot (x+1)

From the second property of the polynomial roots, we can deduce that the root of the polynomial on the left is +2 and the root of the polynomial on the right is -1.

 \displaystyle (x-2) \ \longrightarrow \ \text{root} \ x=+2

 \displaystyle (x+1) \ \longrightarrow \ \text{root} \ x=-1

Therefore, the roots of the polynomial resulting from the multiplication of the two factors are their respective roots, that is, +2 and -1.

 \displaystyle  P(x) = (x-2) \cdot (x+1) \ \longrightarrow \ \text{roots} \ \begin{cases}x=+2 \\[2ex] x=-1 \end{cases}

Practice problems on finding polynomial roots

Problem 1

Determine whether x=-4 is a root of the following polynomial:

P(x)=x^3+2x^2-11x-12

To know if x=-4 is a root of the polynomial we must evaluate it at that value. So:

 \begin{aligned}P(-4)& =(-4)^3+2\cdot (-4)^2-11\cdot (-4) -12 \\[2ex] & = -64+2\cdot 16 +44 -12 \\[2ex] & = -64+32+44 -12 \\[2ex] & = 0 \end{aligned}

The polynomial evaluated at x=-4 is null, so it is a root of the polynomial.

 

Problem 2

Find all the roots of the following quadratic polynomial:

 P(x)=x^2-3x+2

First, to find the possible roots of the polynomial we have to find the divisors of the constant term. Thus, the divisors of 2 are:

Divisors of 2: +1, -1, +2, -2

So the possible polynomial roots or zeros are ±1 and ± 2. Therefore, we must evaluate the polynomial at all these values:

 P(1)=1^2-3\cdot 1+2 =1-3+2=0

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

 P(2)=2^2-3\cdot 2+2 =4-6+2=0

 P(-2)=(-2)^2-3\cdot (-2)+2 =4+6+2=12

So the polynomial is zero when x is +1 or +2, so these are the polynomial roots:

Roots or zeros of the polynomial: +1 and +2

 

Problem 3

Find the roots of the following cubic polynomial:

 P(x)=x^3-x^2-4x+4

First we must find all the factors of the constant term, since the root of a polynomial is also a factor of its constant term. The factors of 4 are:

Divisors of 4: +1, -1, +2, -2, +4, -4

So the possible polynomial roots or zeros are ±1, ±2 and ±4. Now we have to evaluate the polynomial at all these values:

 P(1)=1^3-1^2-4\cdot 1+4  =1-1-4+4=0

 P(-1)=(-1)^3-(-1)^2-4\cdot (-1)+4 =-1-1+4+4=6

 P(2)=2^3-2^2-4\cdot 2+4 =8-4-8+4=0

 P(-2)=(-2)^3-(-2)^2-4\cdot (-2)+4 =-8-4+8+4=0

 P(3)=3^3-3^2-4\cdot 3+4 =27-9-12+4=10

 P(-3)=(-3)^3-(-3)^2-4\cdot (-3)+4 =-27-9+12+4=20

So the polynomial roots are:

Roots or zeros of the polynomial: +1, +2 and -2

 

Problem 4

Find all the zeros of the following polynomial:

 P(x)=x^3-6x^2+8x

In this case, the polynomial has no constant term. Therefore, according to the fourth property of the polynomial roots explained above, we know that one of the roots of the polynomial must be 0.

Roots of the polynomial: x=0

Moreover, in this case the possible roots are not the factors of the constant term, but of the coefficient of the lowest degree term, that is 8:

Divisors of 8: +1, -1, +2, -2, +4, -4, +8, -8

So the possible roots or zeros of the polynomial are ±1, ±2, ±4 and ±8.

 P(1)=1^3-6\cdot 1^2+8\cdot 1 = 1-6+8=3

 P(-1)=(-1)^3-6\cdot (-1)^2+8\cdot (-1) = -1-6-8=-15

 P(2)=2^3-6\cdot 2^2+8\cdot 2 = 8-24+16=0

 P(-2)=(-2)^3-6\cdot (-2)^2+8\cdot (-2) = -8-24-16=-48

 P(4)=4^3-6\cdot 4^2+8\cdot 4 = 64-96+32=0

 P(-4)=(-4)^3-6\cdot (-4)^2+8\cdot (-4) = -64-96-32=-192

 P(8)=8^3-6\cdot 8^2+8\cdot 8 = 512-384+64=192

 P(-8)=(-8)^3-6\cdot (-8)^2+8\cdot (-8) = -512-384-64=-960

So the polynomial equals to zero if x is +2 or +4, so these values are roots of the polynomial. However, we also have to add the root 0 that we found at the beginning of the problem. In conclusion, all the roots of the polynomial are:

Roots or zeros of the polynomial: 0, +2 and +4

 

Problem 5

Use the properties of the polynomial roots to calculate all the zeros of the following polynomial:

 P(x)=(x-1)(x+3)(x^2-x-2)

As we have seen in the sixth property of the polynomial roots, when the polynomial is formed by the product of factors, it is not necessary to calculate all the roots, since the roots of the entire polynomial are the roots of each factor.

Furthermore, from the second property of the polynomial roots, we can deduce that the root of the first factor is +1 and the root of the second factor is -3.

 \displaystyle (x-1) \ \longrightarrow \ \text{root} \ x=+1

 \displaystyle (x+3) \ \longrightarrow \ \text{root} \ x=-3

Therefore, we only need to find the roots of the last factor. To do this, we find the divisors of the constant term (-2):

Divisors of -2: +1, -1, +2, -2

And evaluate the polynomial at all these values:

 q(x)= x^2-x-2

 q(1)=1^2-1-2=1-1-2=-2

 q(-1)=(-1)^2-(-1)-2=1+1-2=0

 q(2)=2^2-2-2=4-2-2=0

 q(-2)=(-2)^2-(-2)-2=4+2-2=4

So the roots of the polynomial on the right are -1 and 2.

Therefore, the roots of the polynomial are all the roots found:

Roots or zeros of the polynomial : +1, -1, +2, -3

 

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