Class 9 Exam  >  Class 9 Notes  >  NCERT Solutions Chapter 2 - Polynomials (I), Class 9, Maths

NCERT Solutions Chapter 2 - Polynomials (I), Class 9, Maths PDF Download

1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i) 4x2 - 3x +  7

(ii) y2  + √2

(iii) 3√t +  t√2

(iv) y + 2/y

(v) x10 +  y3 +  t50

Answer

(i) 4x2 - 3x +  7
There is only one variable x with whole number power so this polynomial in one variable.

 

(ii)  y2 + √2

There is only one variable y with whole number power so this polynomial in one variable.

 

(iii) 3√2  + t√2 

There is only one variable t but in 3√t power of t is 1/2 which is not a whole number so 3√t +  t√2 is not a polynomial.

 

(iv) y +  2/y

There is only one variable but 2/y = 2y-1 so the power is not a whole number so y + 2/y is not a polynomial. 

 

(v) x10  + y3 +  t50

There are three variable x, y and t and there powers are whole number so this polynomial in three variable.

 

2. Write the coefficients of x2 in each of the following:
 (i) 2 + xx
 (ii) 2 - x2 +  x3

(iii) Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

(iv) √2x - 1

Answer

(i) coefficients of x2 = 1
(ii) coefficients of x2 = -1
(iii) coefficients of x2 = π/2
(iv) coefficients of x2 = 0



3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer

3x35 + 7 and 4x100


4. Write the degree of each of the following polynomials:

(i) 5x3 + 4x2 + 7x 

(ii) 4 – y2 

(iii) 5t – √7

(iv) 3

Answer

(i) 5x3 has highest power in the given polynomial which power is 3. Therefore, degree of polynomial is 3.

(ii) – y2  has highest power in the given polynomial which power is 2. Therefore, degree of polynomial is 2.

(iii) 5t has highest power in the given polynomial which power is 1. Therefore, degree of polynomial is 1.

(iv) There is no variable in the given polynomial. Therefore, degree of polynomial is 0.

5. Classify the following as linear, quadratic and cubic polynomial:

(i) xx
► Quadratic Polynomial

 

(ii) xx3
► Cubic Polynomial

 

(iii) y  + y2 + 4
► Quadratic Polynomial
 

(iv) 1 +  x
► Linear Polynomial
 

(v) 3t
►Linear Polynomial
 

(vi) r2

► Quadratic Polynomial

(vii) 7x3
► Cubic Polynomial



Exercise 2.2

 1. Find the value of the polynomial at 5x + 4x2 +  3 at

 (i) x = 0 

(ii) x = - 1 

(iii) x = 2

 

Answer

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A



2. Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y2 - y +1

(ii) p(t) = 2 +t + 2t2 - t3
 (iii) p(x) = x3 

(iv) p(x) = (x - 1) (x + 1)

 

Answer

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A
 


3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x  1, x = -1/3

(ii)  p(x) = 5x - π, x = 4/5

(iii) p(x) = x2 - 1, x = 1, -1

(iv) p(x) = (x + 1) (x - 2), x = -1, 2
 (v) p(x) = x2 , x = 0

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

(viii) p(x) = 2x +  1, x = 1/2

Answer

(i) If x = -1/3 is a zero of polynomial p(x) = 3x + 1 then p(-1/3) should be 0.
At, p(-1/3) = 3(-1/3)  1 = -1  1 = 0
Therefore, x = -1/3 is a zero of polynomial p(x) = 3x + 1.


(ii) If x = 4/5 is a zero of polynomial p(x) = 5x - π then p(4/5) should be 0.
At, p(4/5) = 5(4/5) - π = 4 - π
Therefore, x = 4/5 is not a zero of given polynomial p(x) = 5x - π.
 

(iii) If x = 1 and x = -1 are zeroes of polynomial p(x) = x2 - 1, then p(1) and p(-1) should be 0.
At, p(1) = (1)2 - 1 = 0 and
At, p(-1) = (-1)2 - 1 = 0
Hence, x = 1 and -1 are zeroes of the polynomial  p(x) = x2 - 1.

(iv) If x = -1 and x = 2 are zeroes of polynomial p(x) = (x + 1) (x - 2), then p( - 1) and (2) should be 0.
At, p(-1) = (-1 + 1) (-1 - 2) = 0 (-3) = 0, and
At, p(2) = (2 + 1) (2 - 2) = 3 (0) = 0
Therefore, x = -1 and x = 2 are zeroes of the polynomial p(x) = (x + 1) (x - 2).

 

(v) If x = 0 is a zero of polynomial p(x) = x2, then p(0) should be zero.
Here, p(0) = (0)2 = 0
Hence, x = 0 is a zero of the polynomial p(x) = x2.

NCERT Solutions Chapter 2 - Polynomials (I), Class 9, Maths

NCERT Solutions Chapter 2 - Polynomials (I), Class 9, Maths
NCERT Solutions Chapter 2 - Polynomials (I), Class 9, Maths
NCERT Solutions Chapter 2 - Polynomials (I), Class 9, Maths

(viii)  If x = 1/2 is a zero of polynomial p(x) = 2x +  1 then p(1/2) should be 0.
At, p(1/2) = 2(1/2) 1 = 1 + 1 = 2
Therefore, x = 1/2 is not a zero of given polynomial p(x) = 2x +  1.

4. Find the zero of the polynomial in each of the following cases:
 (i) p(x) = x +

(ii) p(x) = x - 5 

(iii) p(x) = 2x + 5
 (iv) p(x) = 3x - 2 

(v) p(x) = 3x 

(vi) p(x) = ax, a ≠ 0
 (vii) p(x) = cx  d, c ≠ 0, c, are real numbers.

Answer

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

 


Exercises 2.3

 1. Find the remainder when x3 + 3x2  + 3x + 1 is divided by
 (i) x + 1
 (ii) x - 1/2
 (iii) x
 (iv) x + π
 (v) 5 + 2x


Answer

(i) x + 1
By long division,

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

Therefore, the remainder is 0.

(ii) x - 1/2
By long division,

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

Therefore, the remainder is 27/8.

(iii) x

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

Therefore, the remainder is 1.

(iv) x + π

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

Therefore, the remainder is [1 - 3π 3π2 - π3].

(v) 5 + 2x

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

Therefore, the remainder is -27/8.

2. Find the remainder when x3 - ax2 + 6x - a is divided by x - a.

Answer

By Long Division,

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

Therefore, remainder obtained is 5when x3ax2 +  6xa is divided by x - a.



3. Check whether 7 + 3x is a factor of 3x3 + 7x.

Answer

We have to divide 3x3 + 7by 7 + 3x. If remainder comes out to be 0 then 7 3x will be a factor of 3x3 + 7x.
By Long Division,

Polynomials, NCERT,Solutions,9th,Class IX,CBSE, Mathematics, Question and Answer, Q and A

As remainder is not zero so 7 + 3x is not a factor of 3x3 + 7x.

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FAQs on NCERT Solutions Chapter 2 - Polynomials (I), Class 9, Maths

1. What are Polynomials?
Ans. Polynomials are expressions consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents. In simpler terms, a polynomial is a mathematical expression that consists of variables and coefficients, which are generally represented with the help of algebraic terms.
2. What are the different types of Polynomials?
Ans. Polynomials can be classified into different types based on their degrees. A polynomial can be: - Constant polynomial (degree 0) - Linear polynomial (degree 1) - Quadratic polynomial (degree 2) - Cubic polynomial (degree 3) - Quartic polynomial (degree 4) - Quintic polynomial (degree 5) - And so on.
3. What is the degree of a Polynomial?
Ans. The degree of a polynomial is the highest power of the variable present in the polynomial. For example, if a polynomial has the term x^5, then its degree is 5. If a polynomial has no variable, then it is a constant polynomial, and its degree is 0.
4. How to add and subtract Polynomials?
Ans. To add or subtract polynomials, we simply combine like terms. Like terms are terms that have the same variable and the same power. For example, to add the polynomials 3x^2 + 4x + 2 and 2x^2 + 3x - 1, we add the coefficients of like terms, i.e., 3x^2 + 2x^2 = 5x^2, 4x + 3x = 7x, and 2 - 1 = 1. Hence, the sum is 5x^2 + 7x + 1.
5. What is the Remainder Theorem?
Ans. The Remainder Theorem states that if a polynomial P(x) is divided by x - a, then the remainder is equal to P(a). In simpler terms, if we divide a polynomial by x - a, the remainder we get is the value of the polynomial when we substitute x = a. The remainder theorem is useful in finding the remainder value of a polynomial when it is divided by a linear factor.
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