Polynomials

Vachmi

What is a Polynomial?

A Polynomial is an algebraic expression or a function with non-negative integral powers of the variables in the equation.

Polynomial Functions

A Polynomial is also known as polynomial expression or polynomial function.

For example, $4x+9$ is a polynomial that has power equal to 1.

Let us understand this in more detail.

As you may be aware, an algebraic expression is made up of constants and variables.

A symbol having a fixed numerical value is called a constant
For example 2, -7, 0.05, $\frac{3}{7}$ are all constants. These are generally denoted as $a,b,c$, etc.
A symbol which can take any real value is called a variable
These are generally denoted by letters such as $x,y,z$, etc. An algebraic expression is made up by combining constants and variables
Example $5x + \frac{5}{6}y - 9$ is an algebraic expression.

Different parts of an algebraic expression are joined by $+$ or $-$ operators. These parts are called terms.

Notice that in the definition of algebraic expression, we said that it should contain non-negative integral powers. Read this as the terms of polynomials should have variables with their powers as whole numbers.

Examples of Polynomials

$4x^2 - 15x + 9$ is a polynomial in one variable

$5x + 2y - 120$ is a polynomial in two variables

$5x + 2y^{-1} - 120$ is not a polynomial as the power of variable y is -1

$y + 2z - \dfrac{1}{2}$ is a polynomial

$y^{\frac{6}{7}} + 2z - \dfrac{1}{2}$ is not a polynomial as power of y is $\dfrac{6}{7} $ which is not a whole number or a non-negative integer.

$5x^2$ is a monomial

Given an expression $5x - 2y - 120$, please note that in this expression, $5$ and $-2$ are coefficients of $x$ and $y$ respectively and -120 is a constant.

Difference between a Monomial, Binomial, Trinomial and a Polynomial

The parts of an algebraic expression that are separated by addition or subtraction, are called terms. The number of terms decides the type of expression, whether it is a monomial, binomial, trinomial or polynomial.

A Monomial is an algebraic expression that has only one term.

$4x^2$
$xy$
$24y$
$5z$

A Binomial is an algebraic expression that has two unlike terms.

$x + y$
$xy + 6$
$3p - 2q$
$m + 6$

A Trinomial is an algebraic expression that has three terms in it.

$4x + y + 5$
$4x^2 + 6x + 7$
$m + n + 10$
$ab + a + b$

The term poly means many. An expression that has more than one term is called polynomial, non-negative integral exponents of a variable..

$ax + by + ca$
$4x^3 + 6x + 120$
$1.2ab - 2.4b + 3.6a
$4 + x^2 + xy$

Polynomial Solutions

Refer Polynomials Solved Examples here

Degree of a Polynomial

In case of a polynomial in one variable, the degree of a polynomial is the highest power of the variable.
Examples

$5x + 2$ is a polynomial in $x$ of degree 1 as the highest power of $x$ is 1.
$y^3 + 2y^2 + 5y + 4$ is a polynomial in $y$ of degree 3 as the highest power of $y$ is 3.

In case of a polynomial in two or more variables, the degree of a polynomial is found by taking the sum of the powers of the variables in each term and then taking the highest sum so obtained.
Examples

$5x^2y^3 + 2x^2y^2 - 8xy$ is a polynomial in $x$ and $y$ of degree 5.
$y^3 + 2x^2y^2 + 5xy + 4$ is a polynomial in $x$ and $y$ of degree 4.

Factoring Polynomials

The process by which we find the constituent factors of a higher-degree polynomial is called factoring polynomials.
For example, by multiplying $x+2$ and $x-1$, we get $x^2+x-2$; where $x+2$ and $x-1$ are the factors of the expression $x^2+x-2$.
By the fundamental theorem of algebra, we know that any polynomial of degree n has n roots, either real or complex. Thus, it also has n factors as well. as every unique root gives a unique factor to the provided expression.

Steps for Factoring a Polynomial

For factorizing any polynomial, we need to follow three steps as follows:

Factor out the common factor of all the terms of the given polynomial.
Choose the best technique for factoring polynomials.
Write all the factors as the product, as it gives the provided polynomial.

To factor polynomials, we generally make use of the following techniques.

Common Factors

The first method is the common factors method, in which if there is a common factor for each term in the polynomial, we factor that common term out and write the remaining polynomial.

Example
What are the factors of $3x^2 + 6x +12?$

Solution
Let $f(x)$ = $3x^2 + 6x +12$
As 3 is present in each term as expression can be rewritten as
$f(x)$ = $3(x2+2x+4)$

Thus, factors of the $3x^2 + 6x +12$ are $3$ and $x^2 + 2x +4$

Grouping Method

This method factors polynomials by grouping terms with two or more terms together and finding the greatest common factor for each grouping.

Example
Factorize the expression $ax^2+7abx+ax+7ab$ if a and b are some real numbers.

Solution
Let $f(x)$ = $ax^2+7abx+ax+7ab$
Grouping two elements at a time,
$f(x)$ = $ax^2+ax+7abx+7ab$
$f(x)$ = $ax(x+1)+7ab(x+1)$
$f(x)$ = $(ax+7ab)(x+1)$

Thus, factors of the $ax^2+7abx+ax+7ab$ are $(ax+7ab)$ and $(x+1)$

Splitting the middle term

In this method, the middle term is factored in to two factors.

Example
Factorize $6x^2+19x+10$

Solution
Let $f(x)$ = $6x^2+19x+10$
Step 1. Find the product of 1st and last term $( a \times c)$
$6 \times 10 = 60$

Step 2. Find the factors of this product ($60$ in this case) such that addition of those factors is the middle term. These factors can be positive or negative integers.
$60 = (15 \times 4$) and $15 + 4 = 19$.
$19$ is the middle term in the above expression.

Step 3. Replace the centre term by these two terms.
$f(x)$ = $6x^2+19x+10$ = $6x^2+15x+4x+10$

Step 4. Now using the Grouping method, form pairs by finding common factors.
$6x^2+15x+4x+10$ = $3x(2x+5)+2(2x+5)$ = $(3x+2)(2x+5)$

Thus, factors of the $6x^2+19x+10$ are $(3x+2)$ and $(2x+5)$

Remainder Theorem

If a polynomial $f(x)$ of degree n >= 1, is divided by $(x - a)$, then the remainder is $f(a)$.
$f(a)$ is obtained by putting $a$ as value of variable $x$ in polynomial $f(x)$.

Example
Find the remainder when $f(x)$ = $x^3 - 2x^2 + 5$ is divided by $(x - 2)$

Solution:
$(x - 2) = 0$ ⇒ $x$ = 2
By the Remainder Theorem, we know that if a polynomial $f(x)$ is divided by $(x - 2)$, the remainder is $f(2)$.
Hence finiding $f(2)$ = $2^3 - 2*2^2 + 5$ = $5$.

Factor Theorem

In Remainder Theorem, we saw that if a polynomial $f(x)$ of degree n >= 1, is divided by $(x - a)$, then the remainder is $f(a)$.
If $f(a)$ so obtained is 0, then $(x - a)$ is a factor of $f(x)$.

The reverse is also true.
If $(x - a)$ is a factor of $f(x)$, then $f(a)$ = 0

Example
Show that $(x - 2)$ is a factor of $f(x)$ = $x^3 - 2x^2$

Solution
By the Factor Theorem, $(x - 2)$ is a factor if $f(2)$ = 0
Now, $f(x)$ = $x^3 - 2x^2$
⇒ $f(2)$ = $2^3 - 2*2^2$ = $0$.
Hence, $(x - 2)$ is a factor of $f(x)$ = $x^3 - 2x^2$

Algebraic or Polynomial Identities

An equation that is true for every value of the variable is called an identity.
For factorization of polynomials, we use the following identities.

$(x + y)^2 = x^2 + 2xy + y^2$

$(x - y)^2 = x^2 - 2xy + y^2$

$(x + a)(x + b)$ $= x^2 + (a + b)x + ab$

$(x + y)^3 = x^3 + y^3 + 3xy(x + y)$

$(x - y)^3 = x^3 - y^3 - 3xy(x - y)$

$(x + y + z)^2$ $= x^2 + y^2 + z^2 + 2xy + 2yz + 2xz$

$(x^2 - y^2) = (x + y)(x - y)$
⇒ This also means that if $(x + y) = 0$, then $(x^2 - y^2) = 0$

$(x^3 + y^3) = (x + y)(x^2 - xy +y^2)$
⇒ This also means that if $(x + y) = 0$, then $(x^3 + y^3) = 0$

$(x^3 - y^3) = (x - y)(x^2 + xy + y^2)$
⇒ This also means that if $(x - y) = 0$, then $(x^3 - y^3) = 0$

$(x^3 + y^3 + z^3)$ $= (x + y + z)(x^2 + y^2 + z^2 - xy - yz - xz)$ $+ 3xyz$
⇒ This also means that if $(x + y + z) = 0$, then $(x^3 + y^3 + z^3) = 0$