The word polynomial literally means
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 Above polynomial is known as polynomial of degree n, provided that $a_{n}\neq 0$ and every exponent of x is a non-negative integer. |

A function which contains polynomial is called polynomial function. So, we can write that function of the form:

Above polynomial is known as polynomial function.

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:

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:

**Polynomials can be factorized by finding common terms.**

1. Finding common factor:

1. Finding common factor:

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.

**4. Using Formulas:**There are different formulas for factorizing polynomials:

- $a^{2}-b^{2}=(a+b)(a-b)$
- $(a+b)^{2}=a^{2}+2ab+b^{2}$
- $(a-b)^{2}=a^{2}-2ab+b^{2}$
- $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$
- $(a-b)^{3}=a^{3}-b^{3}-3ab(a-b)$
- $a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)$
- $a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)$
- $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac$
- $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)$

**5. Completing Perfect Square:**Another way to factorize a polynomial is to complete the perfect square. After completing perfect square, we may use formulas described above as per the requirement of question.

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:**

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:**

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:**

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:**

$27x^{3}-8$ = $(3x)^{3}-2^{3}$

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,

= $(3x-2)\left \{ (3x)^{2}+2^{2}+3x \times 2 \right \}$

$(3x-2)(9x^{2}+4+6x)$