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The word polynomial literally means "many terms". Polynomials are made up of constants and variables connected with algebraic operations, whereas the exponent of a variable must be a non-negative integer. For example, $x+a$, $x^{2}-a^{2}+bx+c$, $x^{3}+3x^{2}+2x+1$, $z^{3}-y^{3}-7yz-6$ etc. are polynomials.
The degree of polynomial $2x+5$ is 1. The degree of polynomial $3x^{2}+7x-6$ is 2.The degree of $6x^{5}-3x^{2}y^{3}+7xy^{2}$ is 5. The general form of the polynomial is ![]() |
A function which contains polynomial is called polynomial function. So, we can write that function of the form:

For example, $f(x)=x^{3}+x^{2}y^{2}+y^{3}$ is a polynomial function.
Polynomials are classified into various types. There are two categories on the basis of which polynomials are classified.
Classification on the Basis of Degree:
- Linear Polynomial: A polynomial of degree "one" is known as linear polynomial such as $3x+5$.
- Quadratic Polynomial: A polynomial of degree "two" is known as quadratic polynomial. For example, $3x^{2}-7x-8$.
- Cubic Polynomial: A polynomial of degree "three" is known as cubic polynomial such as $x^{3}+2x^{2}-x+2$.
- Bi-quadratic Polynomial: A polynomial of degree "four" is known as bi-quadratic polynomial, like $x^{4}-3x^{3}+2x-3$.
Classification on the Basis of Number of Terms:
- Monomial: A polynomial containing one non-zero term is called monomial. For example, 5, 3x etc.
- Binomial: A polynomial containing two non-zero terms is called binomial. For example, 3 + 6x, x - 5y etc.
- Trinomial: A polynomial containing three non-zero terms. For example, $x^{5}-x^{2}y^{3}+y^{5}$, $x^{2}-7x+9$ etc.
- Zero Polynomial: A polynomial containing only one term that is zero is called zero polynomial.
- Constant Polynomial: A polynomial containing only one constant term is called constant polynomial. For example, 7, 28, -4 etc.
$x^{2}+3x+2=(x+1)(x+2)$.
Such polynomials are known as reducible polynomial.
In contrast with reducible polynomials, irreducible polynomial are those which cannot (by any means) be factorized. For example, $x^{2}+x+1$ and $x^{2}+1$ are irreducible polynomials. Such polynomials are also known as prime polynomials.
Polynomials are factorized by different methods. Few important methods are listed below:
1. Finding common factor: Polynomials can be factorized by finding common terms.
Consider $x^{2}+2x$
Here, x is common in both the terms. So, we can write it in the following way:
$x^{2}+2x$ = $x(x+2)$
2. Grouping and Finding Common Factor: Sometimes, we need to group like terms and then take the common factor out.
Consider $x^{3}+x^{2}y+y^{3}+xy^{2}$
$x^{3}+x^{2}y+y^{3}+xy^{2}$ = $(x^{3}+x^{2}y)+(y^{3}+xy^{2})$
= $x^{2}(x+y)+y^{2}(y+x)$
= $x^{2}(x+y)+y^{2}(x+y)$
= $(x+y)(x^{2}+y^{2})$
3. Decomposition Method: This method is used to factorize a quadratic polynomial. A quadratic polynomial of the form $ax^{2}+bx+c$ is to be factorized in the following way:
- Multiply the values of a and c (along with their signs).
- Determine the value of b (along with the sign).
- Split the product ac in two terms so that, after adding or subtracting those splits parts, we must get the product ac.
- Write those split parts in place of b.
- Factorize by grouping and taking common factor out.
- $a^{2}-b^{2}=(a+b)(a-b)$
- $(a+b)^{2}=a^{2}+2ab+b^{2}$
- $(a-b)^{2}=a^{2}-2ab+b^{2}$
- $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$
- $(a-b)^{3}=a^{3}-b^{3}-3ab(a-b)$
- $a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)$
- $a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)$
- $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac$
- $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)$
Consider $x^{2}-6x+5$
$x^{2}-6x+5$ = $x^{2}-6x+9-9+5$
= $(x^{2}-6x+9)-4$
= $(x-3)^{2}-2^{2}$
Using the identity, $a^{2}-b^{2}=(a+b)(a-b)$, we get
$(x-3+2)(x-3-2)$
$(x-1)(x-5)$
Roots of a polynomials are also known as zeros of polynomials. Roots or zeros of a polynomial are the points at which the value of a polynomial is zero. In other words,
The number "a" is said to be the root or zero of a polynomial P(x), if P(a) = 0
To find the roots or zeros of a polynomial, we simply need to equate the polynomial to zero and determine the values of variable contained in it.
The number of roots or zeros of a polynomial are equal to its degree.
Consider $P(x)=x^{2}-3x$
Since the degree of given polynomial is 2, hence it must have 2 roots.
To find the zeros of P(x), we take P(x) = 0
$x^{2}-3x=0$
$x(x-3)=0$
$x=0\ and\ x=3$
Therefore, 0 and 3 are two zeros of polynomial P(x).
The graph of a polynomial function,$a_nx^n + a_{n-1}x^{n-1} + ....+ a_1x + a_0$ depends upon two factors:
- Whether 'a' is positive or negative.
- Whether 'n' is even or odd.
Case 1: $a> 0$ and n is even
In this case, graph rises at both left and right endpoints.

Case 2: $a> 0$ and n is odd
In this case, graph falls at left endpoint and rises at right endpoint.

Case 3: $a< 0$ and n is even
In this case, graph falls at both left and right endpoints.

Case 4: $a< 0$ and n is odd
In this case, graph rises at left endpoint and falls at right endpoint.

Basic algebraic operations on polynomials are performed in the following manner:
1. Addition: Combine like terms and then proceed.
2. Subtraction: Flip the signs of the terms of polynomial which is to be subtracted, then combine like terms and proceed.
3. Multiplication: While multiplying, following rules of exponents should be kept in mind:
- $a^{m} \times a^{n}=a^{mn}$
- $a^{m} \times b^{m}=(ab)^{m}$
- $(a^{m})^{n}=a^{mn}$
4. Division: While dividing, divide each term separately and follow the following rules of exponents:
- $\frac{a^{m}}{a^{n}}$ = $(a)$$^{m-n}$
- $\frac{a^{m}}{b^{m}}=(\frac{a}{b})^{m}$
- $\frac{1}{a^{m}}$ = $a$$^{-m}$
Let us simplify $\frac{x^{2}-4}{x+2}$
$\frac{x^{2}-4}{x+2}$ = $\frac{(x+2)(x-2)}{x+2}$
$(x+2)$ in numerator is cancelled out from $(x+2)$ in denominator and we are left with $(x-2)$
=> $\frac{x^{2}-4}{x+2}$ = x - 2
Few word problems based on polynomials are given below:
Solved Examples
Question 1: The length and breadth of a rectangle are two successive integers. If the perimeter of the rectangle is 26 meters, find the measure of both the sides.
Solution:
Solution:
Let us consider that length of rectangle = $l$.
Then, breadth of rectangle = $l$ + 1
Perimeter of rectangle = 2 (length + breadth)
$2(l+l+1)=26$
$l+l+1$ = $\frac{26}{2}$
$2l+1=13$
$2l=12$
$l=6$
Therefore, length = 6 meters
Breadth = 6 + 1 = 7 meters.
Then, breadth of rectangle = $l$ + 1
Perimeter of rectangle = 2 (length + breadth)
$2(l+l+1)=26$
$l+l+1$ = $\frac{26}{2}$
$2l+1=13$
$2l=12$
$l=6$
Therefore, length = 6 meters
Breadth = 6 + 1 = 7 meters.
Question 2: A train ticket for an adult is $\$$5 and for a child is $\$$2. Find the general polynomial for the cost of journey for a family of x adults and y children, if it costs $\$$16 to family.
Solution:
Solution:
Cost for one adult = $\$$5
Cost for x adults = $\$$5 * x
Cost for one child = $\$$2
Cost for y children = $\$$2 * y
Cost of journey = 5x + 2y
According to the question, the polynomial is given by:
5x + 2y = 16
Cost for x adults = $\$$5 * x
Cost for one child = $\$$2
Cost for y children = $\$$2 * y
Cost of journey = 5x + 2y
According to the question, the polynomial is given by:
5x + 2y = 16
Solved Examples
Question 1: Determine whether 0, 1, -1, 2 and 3 are the zeros of polynomial $P(x)=x^{3}-2x^{2}-3x$.
Solution:
Solution:
Recall that:
The number "a" is said to be the root or zero of a polynomial P(x), if P(a) = 0
Substituting x = 0,
$P(0)=0^{3}-2 \timesĀ 0^{2}-3 \times 0$
= 0
Therefore, 0 is a zero of the polynomial P(x).
Substituting x = 1,
$P(1)=1^{3}-2 \times 1^{2}-3 \times 1$
= 1 - 2 - 3 $\neq $ 0
Therefore, 1 is not a zero of the polynomial P(x).
Substituting x = -1,
$P(1)=(-1)^{3}-2 \times (-1)^{2}-3 \times (-1)$
= -1 - 2 + 3 = 0
Therefore, -1 is a zero of the polynomial P(x).
Substituting x = 2,
$P(1)=2^{3}-2 \times 2^{2}-3 \times 2$
= 8 - 8 - 6 $\neq $ 0
Therefore, 2 is not a zero of the polynomial P(x).
Substituting x = 3,
$P(3)=3^{3}-2 \times 3^{2}-3 \times3$
= 27 - 18 - 9 = 0
Therefore, 3 is a zero of the polynomial P(x).
The number "a" is said to be the root or zero of a polynomial P(x), if P(a) = 0
Substituting x = 0,
$P(0)=0^{3}-2 \timesĀ 0^{2}-3 \times 0$
= 0
Therefore, 0 is a zero of the polynomial P(x).
Substituting x = 1,
$P(1)=1^{3}-2 \times 1^{2}-3 \times 1$
= 1 - 2 - 3 $\neq $ 0
Therefore, 1 is not a zero of the polynomial P(x).
Substituting x = -1,
$P(1)=(-1)^{3}-2 \times (-1)^{2}-3 \times (-1)$
= -1 - 2 + 3 = 0
Therefore, -1 is a zero of the polynomial P(x).
Substituting x = 2,
$P(1)=2^{3}-2 \times 2^{2}-3 \times 2$
= 8 - 8 - 6 $\neq $ 0
Therefore, 2 is not a zero of the polynomial P(x).
Substituting x = 3,
$P(3)=3^{3}-2 \times 3^{2}-3 \times3$
= 27 - 18 - 9 = 0
Therefore, 3 is a zero of the polynomial P(x).
Question 2: Factorize the polynomial $27x^{3}-8$.
Solution:
Solution:
$27x^{3}-8$ = $(3x)^{3}-2^{3}$
Using the identity $a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)$, we obtain,
= $(3x-2)\left \{ (3x)^{2}+2^{2}+3x \times 2 \right \}$
$(3x-2)(9x^{2}+4+6x)$
Using the identity $a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)$, we obtain,
= $(3x-2)\left \{ (3x)^{2}+2^{2}+3x \times 2 \right \}$
$(3x-2)(9x^{2}+4+6x)$

