## Chapter 2 Polynomials Answers | NCERT Solutions

Chapter 2 Polynomials Solution polynomials are polynomials that describe the roots or solutions of a given polynomial.

When you have a polynomial, finding polynomials solutions is often a crucial step in solving problems in various fields, including mathematics, physics, engineering, and economics.

For example, consider the polynomial

𝑓(𝑥)=𝑎𝑥^{2}+𝑏𝑥+𝑐=0 where a, b, and c are constants and x is the variable.

The solutions to this quadratic polynomial, also known as the roots of the equation, can be found using various methods such as factoring, completing the square, or using the quadratic formula.

The solution polynomial in this case would be a polynomial representing these roots.

In general, the solution polynomial can have multiple terms depending on the original polynomial equation’s degree and its roots’ multiplicities.

Solution polynomials are important because they provide a concise way to represent the solutions of polynomial equations, making it easier to analyze and understand the behavior of the equations in various contexts.

They are beneficial in fields like control theory, where understanding the roots of polynomial equations helps in designing stable systems.

When teaching polynomial solutions to students, it’s important to start with the basics and gradually increase the complexity as they become more comfortable with the concepts. Here’s a suggested approach:

### Chapter 2 Polynomials **Introduction:**

Define what a polynomial is: an expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents.

Give examples of polynomials and non-polynomials.

Introduce key terms like degree, leading coefficient, and standard form.

#####
**Degree of Polynomials:**

Explain how to determine the degree of a polynomial.

Practice identifying the degree of various polynomials.

#####
**Operations with Polynomials:**

Addition and subtraction: Teach how to add and subtract polynomials by combining like terms.

Multiplication: Show how to multiply polynomials using the distributive property and FOIL method.

Division: Introduce polynomial long division and synthetic division for dividing polynomials

###
**Factoring Polynomials:**

Teach students various factoring techniques such as:

GCF (Greatest Common Factor) factoring

Factoring by grouping

Difference of squares

Factoring trinomials (e.g., using the AC method or trial and error)

Emphasize the importance of factoring in finding polynomial solutions.

#####
**Finding Solutions (Roots) of Polynomials:**

Introduce the concept of roots or zeros of polynomials.

Explain how to find roots graphically and algebraically.

**EXERCISE 2.1**

- The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

**Solution:**

** **

(i) From the graph, the graph of the polynomial is parallel to X – axis. It does not cut X – axis at any point

∴ the polynomial has no zeroes

(ii) From the graph, the polynomial graph cuts the X – axis at only one point.

∴ the polynomial has one zero

(iii) From the graph, the polynomial graph cuts the X – axis at three points.

∴ the polynomial has three zeroes

(iv) From the graph, the polynomial graph cuts the X – axis at two points.

∴ the polynomial has two zeroes.

(v) From the graph, the polynomial graph cuts the X – axis at four points.

∴ the polynomial has four zeroes

(vi) From the graph, the polynomial graph cuts the X – axis at three points.

∴ the polynomial has three zeroes.

**Teach methods such as:**

Factoring to find roots

Using the quadratic formula for quadratic polynomials

Synthetic Division and the Remainder Theorem

Rational root theorem and synthetic division for higher-degree polynomials

Provide plenty of practice problems for students to apply these methods.

#####
**Graphing Polynomials:**

Show how to graph polynomial functions using their roots and leading coefficients.

Discuss end behavior and how it relates to the degree and leading coefficient of the polynomial.

######
**Real-World Applications:**

Illustrate how polynomials are used in various real-world scenarios such as finance, physics, and engineering.

Provide examples and problems related to these applications to demonstrate the practical significance of polynomial solutions.

Chapter 2 polynomials

######
**Review and Practice:**

Regularly review previous topics and provide ample opportunities for students to practice solving problems involving polynomials.

Offer additional resources such as worksheets, online practice problems, or interactive activities to reinforce learning.

By following these steps and adjusting the pace and depth of instruction based on students’ comprehension and progress, you can effectively teach polynomial solutions to your students.

**EXERCISE 2.2**

**1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.**

(i) x^{2} – 2x – 8 (ii) 4s^{2} – 4s + 1 (iii) 6x^{2} – 3 – 7x (iv) 4u^{2} + 8u

(v) t^{2} – 15 (vi) 3x^{2} – x – 4

**Solution:**

**(i)** Given polynomial is x^{2} – 2x – 8

** ** Let P (x) = x^{2} – 2x – 8

For the zeroes of the polynomial P (x) = 0

⟹ x^{2} – 2x – 8 = 0

x^{2} – 4x + 2x – 8 = 0

x (x – 4) + 2(x – 4) = 0

(x – 4) (x + 2) = 0

x – 4 = 0 or x + 2 = 0

x = 4 or x = – 2

**(ii)** Given polynomial is 4s^{2} – 4s + 1

Let P (s) = 4s^{2} – 4s + 1

For the zeroes of the polynomial P (s) = 0

⟹ 4s^{2} – 4s + 1 = 0

s^{2} – 2s – 2s + 1 = 0

s (2s – 1) – 1(2s – 1) = 0

(2s – 1) (2s– 1) = 0

2s – 1 = 0 or 2s – 1= 0

s = or s =

let α = and β =

**(iii)** Given polynomial is 6x^{2} – 3 – 7x

Let P (x) = 6x^{2} – 7x – 3

For the zeroes of the polynomial P (x) = 0

⟹ 6x^{2} – 7x – 3 = 0

6x^{2} – 9x + 2x – 3 = 0

3x (2x – 3) + 1(2x – 3) = 0

(2x – 3) (3x + 1) = 0

2x – 3= 0 or 3x + 1 = 0

x = or x =

**(iv)** Given polynomial is 4u^{2} + 8u

Let P (u) = 4u^{2} + 8u

For the zeroes of the polynomial P (u) = 0

⟹ 4u^{2} + 8u = 0

4u (u + 2) = 0

4u = 0 or u + 2 = 0

u = 0 or u = – 2

let α = – 2 and β = 0

**(v)** Given polynomial is t^{2} – 15

Let P (t) = t^{2} – 15

For the zeroes of the polynomial P (t) = 0

⟹ t^{2} – 15 = 0

t^{2} = 15

t = ±

t = + or t = −

let α = – and β =

**(vi)** Given polynomial is 3x^{2} – x – 4

Let P (x) = 3x^{2} – x – 4

For the zeroes of the polynomial P (x) = 0

⟹ 3x^{2} – x – 4= 0

3x^{2} – 4x + 3x – 4 = 0

x (3x – 4) + 1(3x – 4) = 0

(3x – 4) (x + 1) = 0

3x – 4 = 0 or x + 1 = 0

x = or x = – 1

let α = – 1 and β =

Chapter 2 polynomials

**2. Find a quadratic polynomial each with the given numbers as the sum **

** and product of its zeroes respectively.**

**Solution:**

**(i) **Given α + β = and α β = – 1

Quadratic polynomial is k[x^{2} – (α + β) x + α β]

When k =4, the quadratic polynomial is 4x^{2} – x – 4

**(ii) **Given α + β = and α β =

Quadratic polynomial is k[x^{2} – (α + β) x + α β]

= k[x^{2} – ( ) x + ]

= k[x^{2} – x + ]

When k =3, the quadratic polynomial is 3x^{2} – 3 x + 1

**(iii) **Given α + β = 0 and α β =

Quadratic polynomial is k[x^{2} – (α + β) x + α β]

= k[x^{2} – (0) x +]

= k[x^{2} + ]

When k =1, the quadratic polynomial is x^{2} +

**(iv) **Given α + β = 1 and α β = 1

Quadratic polynomial is k[x^{2} – (α + β) x + α β]

= k[x^{2} – (1) x + 1]

= k[x^{2} – x + 1]

When k =1, the quadratic polynomial is x^{2} – x + 1

**(v) **Given α + β = – and α β =

Quadratic polynomial is k[x^{2} – (α + β) x + α β]

When k =4, the quadratic polynomial is 4x^{2} + x + 1

**(vi) **Given α + β = 4 and α β = 1

Quadratic polynomial is k[x^{2} – (α + β) x + α β]

= k[x^{2} – (4) x + 1]

= k[x^{2} –4x + 1]

When k =1, the quadratic polynomial is x^{2} –4x + 1