Class 9 Exam  >  Class 9 Notes  >  Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

Chapter 2 - Polynomials, Solved Examples, Class 9, Maths PDF Download

Ex 1. : Consider a polynomial f(x) = 3x2 - 4x + 2, find the value at x = 3.

Sol. replace x by 3 everywhere

So, the value of f(x) = 3x2 - 4x + 2, at x = 3 is

f(3) = 3 x 32 - 4 x 3 + 2 = 27 -12 + 2 = 17

Similarly the value of polynomiak f(x) = 3x2 - 4x + 2

at x = -2 is, 
Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

at x = 0 is,

Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

at x = 1/2 is,

Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

DO YOUR SELF

Find the value of polynomial 5x - 4x2 + 3 at :

(i) x = 0

(ii) x = -1

(iii) x = 2


(iv) ZEROES OF A POLYNOMIAL

A real number α is a zero of a polynomial p(x) if the value of the polynomial p(x) is zero at x = α. i.e. p(α) = 0

OR

The value of the variable x, for which the polynomial p(x) becomes zero is called zero of the polynomial.

Ex : consider, a polynomial p(x) = x2 - 5x + 6 ; replace x by 2 and 3.
p(2) = (2)2 - 5 × 2 + 6 = 4 - 10 + 6 = 0,

p(3) = (3)2 - 5 × 3 + 6 = 9 - 15 + 6 = 0

∴ 2 and 3 are the zeroes of the polynomial p(x).

REMARK 

1. The constant polynomial has no zero.

2. Every linear polynomial has one and only one zero or root.
consider a linear polynomial p(x) = ax + b, a ≠ 0 ⇒ p(x) = 0 ⇒ ax + b  = 0 ⇒ ax = -b ⇒ x = -b/a  is a zero of the polynomial.

3. A given polynomial can have more than one zero or root.

4. If the degree of a polynomial is n, the maximum number of zeroes it can have is also n.
Ex. : If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes, if the degree of
polynomial is 8, maximum number of zeroes it can have is 8.

5. A zero of a polynomial need not be 0

Ex . : Consider a polynomial f(x) = x2 - 4, then f(2) = (2)2 - 4 = 4 - 4 = 0 here, zero of the polynomial
f(x) = x2 - 4 is 2 which itself is not 0.

6. 0 may be zero of the polynomial.

Ex. Consider a polynomial f(x) = x2 - x, then f(0) = 02 - 0 = 0 here, 0 is the zero of the polynomial

f(x) = x2 - x

Ex 2. Find which of the following algebraic expression is a polynomial.
 (i) 
Chapter 2 - Polynomials, Solved Examples, Class 9, Maths 

Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

(iv) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths
 

Sol. (i) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths Is a polynomial in one variable.

(ii) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths = x1 + x-1 Is not a polynomial as the second term 1/x has degree (-1).

(iii) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths= y1/2 - 8 Is not a polynomial since, the power of the first term √y is 1/2, which is not a whole number.

(iv) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths
Since, the exponent of the second term is 1/3, which in not a whole number. Therefore, the given expression is not a polynomial.


Ex 3. Find the degree of the polynomial :
 (i) 5x2 - 6x3 + 8x7 + 6x2 

(ii) 2y12 + 3y10 - y15 + y + 3 

(iii) x 

(iv) 8

Sol. (i) Since the term with highest exponent (power) is 8x7 and its power is 7.
∴ The degree of given polynomial is 7.
(ii) The highest power of the variable is 15. ⇒ degree = 15
(iii) x = x⇒ degree is 1.
(iv) 8 = 8x0 ⇒ degree is 0.

Ex 4. Write the coefficient of x2 in each of the following :
 (i) 
Chapter 2 - Polynomials, Solved Examples, Class 9, Maths 

(ii) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

(iii) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

(iv) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

Sol. (i) In the polynomial Chapter 2 - Polynomials, Solved Examples, Class 9, Maths the coefficient of x2 is 1.
(ii) In the polynomial Chapter 2 - Polynomials, Solved Examples, Class 9, Maths the coefficient of x2 is - 1.
(iii) In the polynomial Chapter 2 - Polynomials, Solved Examples, Class 9, Maths the coefficient of xis 1/2.

(iv) In the polynomial Chapter 2 - Polynomials, Solved Examples, Class 9, Maths the coefficient of x2 = 0. [We may write Chapter 2 - Polynomials, Solved Examples, Class 9, Maths = ax2 +Chapter 2 - Polynomials, Solved Examples, Class 9, Maths. So the
coefficient of x2 is 0]

Ex 5. Classify the following as linear, quadratic and cubic polynomials :
 (i) x2 + x 

(ii) x - x3 

(iii) 1 + x 

(iv) 3t
 (v) r2 

(vi) 7x3

(vii) y + y2 + 5 

(viii) 3xyz

Sol. (i) The polynomial x2 + x is a quadratic polynomial as its degree is 2.

(ii) Degree of the polynomial x - x3 is 3. It is a cubic polynomial

(iii) Degree of the polynomial 1 + x is 1. It is a linear polynomial.

(iv) Degree of the polynomial 3t is 1. It is a linear polynomial.

(v) Degree of the polynomial r2 is 2. It is a quadratic polynomial.

(vi) Degree of the polynomial 7x3 is 3. It is a cubic polynomial.

(vii)Degree of the polynomial y + y2 + 5 is 2. It is a quadratic polynomial.

(viii) 3xyz is a polynomial in 3 variables x, y and z. Its degree is 1 + 1 + 1 = 3. It is a cubic polynomial.


Ex 6. Find q(0), q(1) and q(2) for each of the following polynomials :
 (i) q(x) = x2 + 3x 

(ii) q(y) = 2 + y + 2y2 - 5y3 

(iii) q(t) = t3

Sol.

q (x) = x2 + 3x
q(0) - (0)2 + 3 x 0 = 0
q(l) = (l)2 + 3 x 1 = 4

(i) q(2) = (2)2+ 3 x 2 = 4 + 6 = 10
q(y) = 2 + y + 2y2 - 5y3 
q(0) = 2 + 0 + 2 (0)2 - 5 (0)3 = 2

(ii) q(1) = 2 + 1 + 2 (1)2 - 5 (1)3 = 2 + 1 + 2 - 5 = 0

q(t) = t3 
q(0) =0
q(1) = (1)3 = 1

(iii) q(2) = (2)3 = 8.

E7. Check whether 0 and 3 are zeroes of the polynomial x2 - 3x.
Sol. 

let p (x) = x1 - 3x

Then p(0) = (0)" - 3*0 = 0

∴ 0 is a zero of the given polymcinial
Again,    p(3) = (3)2 -3 x 3 = 9 - 9 = 0
∴ 3 is alari a zero cf the given polynomial.
Hence 0 and 3 are both zeroes of the polynomial x2 - 3x

Ex 8. Show that 3 is a zero of the polynomial x3 - 8x2 + 8x + 21.

Sol . Let p(x) = x3 - 8x2 + 8x + 21.

Now p(3) = (3)3 - 8 (3)2 + 8.3 + 21 = 27 - 72 + 24 + 21 = 0

∴ 3 is zero of the polynomial - x3 + 8x2 + 8x + 21.

Ex 9. Which of the number 1, -1, and -3 are zeroes of the polynomial 2x4 + 9x3 + 11x2 + 4x - 6.
 Sol. 

Let f(x) = 2x4 + 9x3 + 11x2 + 4x - 6
f(1) = 2(1)4 + 9(1)3 + 11(1)2 + 4(1) - 6 = 2 + 9 + 11 + 4 - 6 = 20  0
∴ 1 is not a zero of the polynomial f(x)

Again f(-1) = 2(-1)4 + 9(-1)3 + 11(-1)2 + 4(-1) - 6 = 2 - 9 + 11 - 4 - 6 = - 6  0
∴ -1 is not a zero the polynomial f(x)

Also f(-3) = 2(-3)4 + 9(-3)3 + 11(-3)2 + 4(-3) - 6 = 162 - 243 + 99 - 12 - 6 = 0
∴ -3 is a zero of the polynomial f(x).
Thus 1 and -1 are not zeroes of f(x) whereas -3 is a zero of f(x).

Ex 10. Verify whether the indicated numbers are zeroes (roots) of the polynomial corresponding to them in the following cases:
 (i) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

(ii) p(x) = (x + 1) (x - 2), x = -1, 2

(iii) p(x) = x2, x = 0 

(iv)Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

(v) Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

Sol. 

(i) p(x) = 3x + 1

Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

∴ Chapter 2 - Polynomials, Solved Examples, Class 9, Maths is a zero of the polynomial.

(ii)  p(x) - (x + 1) (x- 2)
⇒ P(-1) = (-1 + 1) (-1- 2) = 0 x (-3) = 0 and, p (2) = (2 + 1) [2 - 2) = 3 x 0 = 0
∴' x = -1 and x = 2 are zeroes of the given polynomial

(iii) p(x) = x2   ⇒    p(0) = (0)2 = 0

∴  x = 0 is a zero of the given polynomial

(iv)  p (x) - ℓx + m Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

Chapter 2 - Polynomials, Solved Examples, Class 9, Maths  is a zero of the given polynomial.

(v) p(x) = 2x + 1      Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

∴ x = 1/2 is not a zero of the given polynomial.

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FAQs on Chapter 2 - Polynomials, Solved Examples, Class 9, Maths

1. What are polynomials?
Ans. Polynomials are algebraic expressions that consist of variables and coefficients, along with the operations of addition, subtraction, and multiplication. They are used to represent various mathematical functions, and can be of different degrees, based on the highest power of the variable present in the expression.
2. Can you give an example of a polynomial?
Ans. Yes, an example of a polynomial is: 3x^2 + 2x + 1. Here, x is the variable, and 3, 2, and 1 are the coefficients. The expression has a degree of 2, since the highest power of x is 2.
3. How do you add polynomials?
Ans. To add polynomials, you need to combine the like terms. Like terms are terms that have the same variable with the same power. For example, to add 3x^2 + 2x + 1 and 2x^2 + 3x - 1, you need to combine the terms that have x^2, x, and the constant term separately. So, the sum of the two polynomials would be 5x^2 + 5x.
4. What is the degree of a polynomial?
Ans. The degree of a polynomial is the highest power of the variable present in the expression. For example, the degree of the polynomial 3x^2 + 2x + 1 is 2, since the highest power of x is 2.
5. How do you factorize a polynomial?
Ans. To factorize a polynomial, you need to find the factors that can be multiplied to give the original polynomial. One method is to look for common factors among the terms, and then use distributive property to express the polynomial as a product of these factors. Another method is to use the quadratic formula for polynomials of degree 2, and use long division for higher degree polynomials.
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