NCERT Important Question & Solutions for Class 9 Maths Chapter 2 Polynomials | EduGrown

NCERT Important Question & Solutions for Class 9 Maths Chapter 2 Polynomials

You can find Chapter 2 Polynomials Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important Questions & Solutions For Exercise 2.1 Polynomials

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) 3 √t + t√2                                    (iii) y+ 2y

Solution:
(i) We have 4x² – 3x + 7 = 4x² – 3x + 7x⁰
It is a polynomial in one variable i.e., x
because each exponent of x is a whole number.

(ii) We have 3 √t + t√2 = 3 √t^1/2 + √2.t
It is not a polynomial, because one of the exponents of t is 12,
which is not a whole number.

(iii) We have y + y+2y = y + 2.y^-1
It is not a polynomial, because one of the exponents of y is -1,
which is not a whole number.

2.Write the coefficients of x² in each of the following
(i) 2 + x² + x
(ii) 2 – x² + x³

Solution:
(i) The given polynomial is 2 + x² + x.
The coefficient of x² is 1.
(ii) The given polynomial is 2 – x² + x³.
The coefficient of x² is -1.

3.Write the degree of each of the following polynomials.
(i) 5x³+4x² + 7x
(ii) 4 – y²
(iii) 5t – √7
(iv) 3
Solution:
(i) The given polynomial is 5x³ + 4x² + 7x.
The highest power of the variable x is 3.
So, the degree of the polynomial is 3.

(ii) The given polynomial is 4- y². The highest
power of the variable y is 2.
So, the degree of the polynomial is 2.

(iii) The given polynomial is 5t – √7 . The highest power of variable t is 1. So, the degree of the polynomial is 1.
(iv) Since, 3 = 3x° [∵ x°=1]
So, the degree of the polynomial is 0.

4.Classify the following as linear, quadratic, and cubic polynomials.    (i) x²+ x    (ii) r²     (iii) 7x³

Solution:
(i) The degree of x² + x is 2. So, it is a quadratic polynomial.

(ii) The degree of r² is 2. So, it is a quadratic polynomial.

(iii) The degree of 7x³ is 3. So, it is a cubic polynomial.

NCERT Important Questions & Solutions For Exercise 2.2 Polynomials

1.Find the zero of the polynomials in each of the following cases

(i) p(x) = x + 5                  (ii) p(x) = x – 5                        (iii) p(x) = ax

 

2.Find the remainder when x³ + 3x² + 3x + 1 is divided by
(i) x + 1
(ii) x – 12
(iii) x
(iv) x + π
(v) 5 + 2x

Solution:
Let p(x) = x³ + 3x² + 3x +1
(i) The zero of x + 1 is -1.
∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1
= -1 + 3- 3 + 1 = 0
Thus, the required remainder = 0

(ii) The zero of x−12 is 12

Thus, the required remainder = 278

(iii) The zero of x is 0.
∴ p(0) = (0)³ + 3(0)² + 3(0) + 1
= 0 + 0 + 0 + 1 = 1
Thus, the required remainder = 1.

(iv) The zero of x + π is -π.
p(-π) = (-π)³ + 3(- π)²2 + 3(- π) +1
= -π³ + 3π² + (-3π) + 1
= – π³ + 3π² – 3π +1
Thus, the required remainder is -π³ + 3π² – 3π+1.

(v) The zero of 5 + 2x is −52 .


Thus, the required remainder is −278 .

NCERT Important Questions & Solutions For Exercise 2.3 Polynomials

1. Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Solution:
We have, p(x) = x³ – ax² + 6x – a and zero of x – a is a.
∴ p(a) = (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a = 5a
Thus, the required remainder is 5a.

2. Check whether 7 + 3x is a factor of 3x³+7x.
Solution:
We have, p(x) = 3x³+7x. and zero of 7 + 3x is −73.

Since,( −4909) ≠ 0
i.e. the remainder is not 0.
∴ 3x³ + 7x is not divisib1e by 7 + 3x.
Thus, 7 + 3x is not a factor of 3x³ + 7x.

NCERT Important Questions & Solutions For Exercise 2.4 Polynomials

1. Determine which of the following polynomials has (x +1) a factor. (i) x³+x²+x +1 (ii) x⁴ + x³ + x² + x + 1

Solution:
The zero of x + 1 is -1.
(i) Let p (x) = x³ + x² + x + 1
∴ p (-1) = (-1)³ + (-1)² + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p (- 1) = 0
So, (x+ 1) is a factor of x³ + x² + x + 1.

(ii) Let p (x) = x⁴ + x³ + x² + x + 1
∴ P(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1)+1
= 1 – 1 + 1 – 1 + 1
⇒ P (-1) ≠ 1
So, (x + 1) is not a factor of x⁴ + x³ + x² + x+ 1.

2.Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases
(i) p (x)= 2x³ + x² – 2x – 1, g (x) = x + 1
(ii) p(x)= x³ + 3x² + 3x + 1, g (x) = x + 2

Solution:
(i) We have, p (x)= 2x³ + x² – 2x – 1 and g (x) = x + 1
∴ p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
= 2(-1) + 1 + 2 – 1
= -2 + 1 + 2 -1 = 0
⇒ p(-1) = 0, so g(x) is a factor of p(x).

(ii) We have, p(x) x³ + 3x² + 3x + 1 and g(x) = x + 2
∴ p(-2) = (-2)³ + 3(-2)²+ 3(-2) + 1
= -8 + 12 – 6 + 1
= -14 + 13
= -1
⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

3. Find the value of k, if x – 1 is a factor of p (x) in each of the following cases
(i) p (x) = x² + x + k
(ii) p (x) = 2x² + kx + √2

Solution:
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x² + x + k
Since, p(1) = (1)² +1 + k
⇒ p(1) = k + 2 = 0
⇒ k = -2.

(ii) Here, p (x) = 2x² + kx + √2
Since, p(1) = 2(1)² + k(1) + √2
= 2 + k + √2 =0
k = -2 – √2 = -(2 + √2)

4.Factorise
(i) 12x² – 7x +1
(ii) 2x² + 7x + 3

Solution:
(i) We have,
12x² – 7x + 1 = 12x² – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x² -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x² + 7x + 3 = 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

5. Factories

(i) x³ + 13x² + 32x + 20
(ii) 2y³ + y² – 2y – 1

(i) We have, x³ + 13x² + 32x + 20
= x³ + x² + 12x² + 12x + 20x + 20
= x²(x + 1) + 12x(x +1) + 20(x + 1)
= (x + 1)(x² + 12x + 20)
= (x + 1)(x² + 2x + 10x + 20)
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)(x + 2)(x + 10)
Thus, x³ + 13x² + 32x + 20
= (x + 1)(x + 2)(x + 10)

(ii) We have, 2y³ + y² – 2y – 1
= 2y³ – 2y² + 3y² – 3y + y – 1
= 2y²(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)(2y² + 3y + 1)
= (y – 1)(2y² + 2y + y + 1)
= (y – 1)[2y(y + 1) + 1(y + 1)]
= (y – 1)(y + 1)(2y + 1)
Thus, 2y³ + y² – 2y – 1
= (y – 1)(y + 1)(2y +1)

NCERT Important Questions & Solutions For Exercise 2.4 Polynomials

1. Use suitable identities to find the following products

(i) (y²+ 3/2) (y²– 3/2)

2.Evaluate the following products without multiplying directly
(i) 103 x 107

Solution:
(i)We have, 103 x 107 = (100 + 3) (100 + 7)
= ( 100)² + (3 + 7) (100)+ (3 x 7)
[Using (x + a)(x + b) = x² + (a + b)x + ab]
= 10000 + (10) x 100 + 21
= 10000 + 1000 + 21=11021

3.Expand each of the following, using suitable identity

(i) (- 2x + 5y – 3z)²
(ii) [ 1/4a –1/4b + 1] ²

(i)(- 2x + 5y- 3z)² = (- 2x)² + (5y)² + (- 3z)² + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

4. Factorise
(i) 4 x² + 9y² + 16z² + 12xy – 24yz – 16xz

Solution:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
= (2x)² + (3y)² + (- 4z)² + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x)
= (2x + 3y – 4z)² = (2x + 3y + 4z) (2x + 3y – 4z)

5.

Solution:
We have, (x + y)³ = x³ + y³ + 3xy(x + y) …(1)
and (x – y)³ = x³ – y³ – 3xy(x – y) …(2)

6.Factorise each of the following

7. Verify that
x³ +y³ +z³ – 3xyz = 1/2 (x + y+z)[(x-y)² + (y – z)² +(z – x)²]
Solution:
R.H.S
= 1/2(x + y + z)[(x – y)²+(y – z)²+(z – x)²]
= 1/2 (x + y + 2)[(x² + y² – 2xy) + (y² + z² – 2yz) + (z² + x² – 2zx)]
= 1/2 (x + y + 2)(x² + y² + y² + z² + z² + x² – 2xy – 2yz – 2zx)
= 1/2 (x + y + z)[2(x² + y² + z² – xy – yz – zx)]
= 2 x 1/2 x (x + y + z)(x² + y² + z² – xy – yz – zx)
= (x + y + z)(x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz = L.H.S.
Hence, verified.

8. If x + y + z = 0, show that x³ + y³ + z³ = 3 xyz.
Solution:
Since, x + y + z = 0
⇒ x + y = -z (x + y)³ = (-z)³
⇒ x³ + y³ + 3xy(x + y) = -z³
⇒ x³ + y³ + 3xy(-z) = -z³ [∵ x + y = -z]
⇒ x³ + y³ – 3xyz = -z³
⇒ x³ + y³ + z³ = 3xyz
Hence, if x + y + z = 0, then
x³ + y³ + z³ = 3xyz

9.Without actually calculating the cubes, find the value of each of the following
(i) (- 12)³ + (7)³ + (5)³

Solution:
(i) We have, (-12)³ + (7)³ + (5)³
Let x = -12, y = 7 and z = 5.
Then, x + y + z = -12 + 7 + 5 = 0
We know that if x + y + z = 0, then, x³ + y³ + z³ = 3xyz
∴ (-12)³ + (7)³ + (5)³ = 3[(-12)(7)(5)]
= 3[-420] = -1260

NCERT Quick revision Notes Chapter-2 Polynomials

NCERT Solutions Chapter-2 Polynomials

NCERT MCQ Chapter-2 Polynomials