How Do You Know if a Polynomial Has No Complex 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 10 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 ane 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 manus, iii is non a root of this polynomial considering it is non a value that cancels the polynomial, in other words, if nosotros substitute ten 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 at present you empathize 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 come across in the next section.

How to find all the roots of a polynomial

To discover all the roots of a polynomial, yous must exercise the post-obit steps:

First, find all the divisors (or factors) of the constant term of the polynomial.
Second, evaluate the polynomial at all the values institute in the previous pace.
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 case stride past step then 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, considering any root of a polynomial is also a factor of the abiding term. So, the factors of vi are:

Factors of half-dozen: +1, -one, +2, -2, +3, -iii

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

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

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

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

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

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

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

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

Then the polynomial equals to nothing only when the variable 10 is +ii or +3, and then these are the roots of the polynomial:

Roots or zeros of the polynomial: +ii and +three

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

Backdrop of polynomial roots

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

As we take seen before, the integer roots (or zeros) of a polynomial are divisors of the abiding term of the polynomial.

If we know all the roots of a polynomial, nosotros tin express the polynomial in the grade 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, nosotros can rewrite the polynomial in the form of three multiplications of factors, each one formed by the variable $x$ and a root changed sign:

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

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

A polynomial has as many complex roots every bit its degree indicates. So a 2d-caste polynomial will have 2 roots, a third-caste polynomial volition take three roots, a 4th-degree polynomial will have four roots, and then on.

If a polynomial does not have a abiding term, it means that at to the lowest degree 1 of its roots is 0. Then, the rest of the polynomial roots are divisors of the coefficient of the lowest degree monomial.

For instance, the following polynomial has no constant term:

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

So a root of the polynomial must necessarily exist 0. And the residuum of the roots are factors of the coefficient of the term of lowest caste, 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

When the roots of a polynomial cannot exist 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 -ane, that is, -1 and +1. Then 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 upshot in 0, therefore it does not accept integer roots and information technology is an irreducible polynomial over the integers.

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.

Every bit an case, 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

Make up one's mind whether $x=-4$ is a root of the post-obit polynomial:

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

To know if x=-4 is a root of the polynomial we must evaluate information technology 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 zero, then it is a root of the polynomial.

Problem two

Find all the roots of the post-obit quadratic polynomial:

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

Beginning, to find the possible roots of the polynomial nosotros 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, nosotros must evaluate the polynomial at all these values:

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

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

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

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

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

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

Problem three

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 besides a cistron of its constant term. The factors of four are:

Divisors of iv: +one, -1, +2, -two, +4, -4

So the possible polynomial roots or zeros are ±ane, ±ii 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

Trouble 4

Find all the zeros of the post-obit polynomial:

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

In this instance, the polynomial has no abiding term. Therefore, according to the 4th property of the polynomial roots explained above, nosotros know that 1 of the roots of the polynomial must exist 0.

Roots of the polynomial: $x=0$

Moreover, in this case the possible roots are non the factors of the constant term, but of the coefficient of the everyman caste term, that is 8:

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

Then 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 +two or +4, and so these values are roots of the polynomial. However, we also have to add the root 0 that we plant at the offset of the trouble. In conclusion, all the roots of the polynomial are:

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

Trouble 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 accept seen in the sixth holding of the polynomial roots, when the polynomial is formed past 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 2d property of the polynomial roots, nosotros can deduce that the root of the commencement factor is +1 and the root of the 2nd cistron is -3.

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

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

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

Divisors of -2: +1, -ane, +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 ii.

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

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

riveraageres.blogspot.com

Source: https://www.algebrapracticeproblems.com/polynomial-roots-zeros/

How Do You Know if a Polynomial Has No Complex Zeros

What are the roots of a polynomial?

How to find all the roots of a polynomial

Example

Backdrop of polynomial roots

Practice problems on finding polynomial roots

Problem 1

Problem two

Problem three

Trouble 4

Trouble 5

0 Response to "How Do You Know if a Polynomial Has No Complex Zeros"

Postar um comentário

Iklan Atas Artikel

Iklan Tengah Artikel 1

Iklan Tengah Artikel 2

Iklan Bawah Artikel