Mathematics NCERT Grade 9, Chapter 2: PolynomialsPolynomials are a particular type of algebraic expression. Students will also study the remainder theorem and factor theorem in this chapter. Some more algebraic identities will be discussed in this chapter. Their use in factorisation and in evaluating some given expressions will also be discussed. Different polynomials based on the number of terms are discussed in this chapter.
  • Polynomial with one term is Monomial
  • Polynomial with two terms is Binomial
  • Polynomial with three terms is Trinomial 
This section also includes the topic degree of a polynomial which refers to the highest power of the variable in a polynomial.
  • Polynomial of degree 1 is called linear polynomial.
  • Polynomial of degree 2 is called quadratic polynomial.
  • Polynomial of degree 3 is called cubic polynomial. 
 The chapter will move further to the topic zeroes of a polynomial. 3 solved examples are given to make the concept clear to students. 
  • A zero of a polynomial need not to be 0. 
  • 0 may be zero of a polynomial.
  • Every linear polynomial has one and only one zero.
  • A polynomial can have more than one zero. 
Exercise 2.2 is based on the same concept. 
The next section explains the topic remainder theorem. In this section, the solved example is given in which each step of the solution is explained in detail.
  • Remainder Theorem: If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial x a, then the remainder is p(a). 
Section 2.5 covers the topic- Factorisation of Polynomials.
  • Factor theorem: x a is a factor of polynomial p(x) if p(a) = 0. 
Exercise 2.6 is based on the topic- algebraic identities. 8 identities are discussed in this section. These identities are used to solve problems of factorisation. The last exercise of the chapter is Exercise 2.5. 
In the end summary of the chapter is given for quick revision of the chapter Polynomials. 

Page No 32:

Question 1:

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

(i) (ii) (iii)

(iv) (v)

Answer:

(i)

Yes, this expression is a polynomial in one variable x.

(ii)

Yes, this expression is a polynomial in one variable y.

(iii)

No. It can be observed that the exponent of variable t in term is , which is not a whole number. Therefore, this expression is not a polynomial.

(iv)

No. It can be observed that the exponent of variable y in termis −1, which is not a whole number. Therefore, this expression is not a polynomial.

(v)

No. It can be observed that this expression is a polynomial in 3 variables x, y, and t. Therefore, it is not a polynomial in one variable.

Video Solution for number systems (Page: 32 , Q.No.: 1)

NCERT Solution for Class 9 math - number systems 32 , Question 1

Page No 32:

Question 2:

Write the coefficients of in each of the following:

(i) (ii)

(iii) (iv)

Answer:

(i)

In the above expression, the coefficient of is 1.

(ii)

In the above expression, the coefficient of is −1.

(iii)

In the above expression, the coefficient of is.

(iv)

In the above expression, the coefficient of is 0.

Video Solution for number systems (Page: 32 , Q.No.: 2)

NCERT Solution for Class 9 math - number systems 32 , Question 2

Page No 32:

Question 3:

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

Answer:

Degree of a polynomial is the highest power of the variable in the polynomial.

Binomial has two terms in it. Therefore, binomial of degree 35 can be written as .

Monomial has only one term in it. Therefore, monomial of degree 100 can be written as x100.

Video Solution for number systems (Page: 32 , Q.No.: 3)

NCERT Solution for Class 9 math - number systems 32 , Question 3

Page No 32:

Question 4:

Write the degree of each of the following polynomials:

(i) (ii)

(iii) (iv) 3

Answer:

Degree of a polynomial is the highest power of the variable in the polynomial.

(i)

This is a polynomial in variable x and the highest power of variable x is 3. Therefore, the degree of this polynomial is 3.

(ii)

This is a polynomial in variable y and the highest power of variable y is 2. Therefore, the degree of this polynomial is 2.

(iii)

This is a polynomial in variable t and the highest power of variable t is 1. Therefore, the degree of this polynomial is 1.

(iv) 3

This is a constant polynomial. Degree of a constant polynomial is always 0.

Video Solution for number systems (Page: 32 , Q.No.: 4)

NCERT Solution for Class 9 math - number systems 32 , Question 4

Page No 32:

Question 5:

Classify the following as linear, quadratic and cubic polynomial:

(i) (ii) (iii) (iv) (v)

(vi) (vii)

Answer:

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively.

(i) is a quadratic polynomial as its degree is 2.

(ii) is a cubic polynomial as its degree is 3.

(iii) is a quadratic polynomial as its degree is 2.

(iv) 1 + x is a linear polynomial as its degree is 1.

(v) is a linear polynomial as its degree is 1.

(vi) is a quadratic polynomial as its degree is 2.

(vii) is a cubic polynomial as its degree is 3.

Video Solution for number systems (Page: 32 , Q.No.: 5)

NCERT Solution for Class 9 math - number systems 32 , Question 5



Page No 34:

Question 1:

Find the value of the polynomial at

(i) x = 0 (ii) x = −1 (iii) x = 2

Answer:

(i)

(ii)

(iii)


Video Solution for number systems (Page: 34 , Q.No.: 1)

NCERT Solution for Class 9 math - number systems 34 , Question 1

Page No 34:

Question 2:

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

(i) p(y) = y2y + 1 (ii) p(t) = 2 + t + 2t2t3

(iii) p(x) = x3 (iv) p(x) = (x − 1) (x + 1)

Answer:

(i) p(y) = y2y + 1

p(0) = (0)2 − (0) + 1 = 1

p(1) = (1)2 − (1) + 1 = 1

p(2) = (2)2 − (2) + 1 = 3

(ii) p(t) = 2 + t + 2t2t3

p(0) = 2 + 0 + 2 (0)2 − (0)3 = 2

p(1) = 2 + (1) + 2(1)2 − (1)3

= 2 + 1 + 2 − 1 = 4

p(2) = 2 + 2 + 2(2)2 − (2)3

= 2 + 2 + 8 − 8 = 4

(iii) p(x) = x3

p(0) = (0)3 = 0

p(1) = (1)3 = 1

p(2) = (2)3 = 8

(iv) p(x) = (x − 1) (x + 1)

p(0) = (0 − 1) (0 + 1) = (− 1) (1) = − 1

p(1) = (1 − 1) (1 + 1) = 0 (2) = 0

p(2) = (2 − 1 ) (2 + 1) = 1(3) = 3

Video Solution for number systems (Page: 34 , Q.No.: 2)

NCERT Solution for Class 9 math - number systems 34 , Question 2



Page No 35:

Question 3:

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

(i) (ii)

(iii) p(x) = x2 − 1, x = 1, − 1 (iv) p(x) = (x + 1) (x − 2), x = − 1, 2

(v) p(x) = x2 , x = 0 (vi) p(x) = lx + m

(vii) (viii)

Answer:

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

Therefore, is a zero of the given polynomial.

(ii) If is a zero of polynomial p(x) = 5x − π , thenshould be 0.

Therefore, is not a zero of the given polynomial.

(iii) If x = 1 and x = −1 are zeroes of polynomial p(x) = x2 − 1, then p(1) and p(−1) should be 0.

Here, p(1) = (1)2 − 1 = 0, and

p(− 1) = (− 1)2 − 1 = 0

Hence, x = 1 and −1 are zeroes of the given polynomial.

(iv) If x = −1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x − 2), then p(−1) and p(2)should be 0.

Here, p(−1) = (− 1 + 1) (− 1 − 2) = 0 (−3) = 0, and

p(2) = (2 + 1) (2 − 2 ) = 3 (0) = 0

Therefore, x = −1 and x = 2 are zeroes of the given polynomial.

(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 given polynomial.

(vi) If is a zero of polynomial p(x) = lx + m, then should be 0.

Here,

Therefore, is a zero of the given polynomial.

(vii) If and are zeroes of polynomial p(x) = 3x2 − 1, then

Hence, is a zero of the given polynomial. However, is not a zero of the given polynomial.

(viii) If is a zero of polynomial p(x) = 2x + 1, then should be 0.

Therefore, is not a zero of the given polynomial.

Video Solution for number systems (Page: 35 , Q.No.: 3)

NCERT Solution for Class 9 math - number systems 35 , Question 3

Page No 35:

Question 4:

Find the zero of the polynomial in each of the following cases:

(i) p(x) = x + 5 (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:

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

(i) p(x) = x + 5

p(x) = 0

x + 5 = 0

x = − 5

Therefore, for x = −5, the value of the polynomial is 0 and hence, x = −5 is a zero of the given polynomial.

(ii) p(x) = x − 5

p(x) = 0

x − 5 = 0

x = 5

Therefore, for x = 5, the value of the polynomial is0 and hence, x = 5 is a zero of the given polynomial.

(iii) p(x) = 2x + 5

p(x) = 0

2x + 5 = 0

2x = − 5

Therefore, for, the value of the polynomial is 0 and hence, is a zero of the given polynomial.

(iv) p(x) = 3x − 2

p(x) = 0

3x − 2 = 0

Therefore, for, the value of the polynomial is 0 and hence, is a zero of the given polynomial.

(v) p(x) = 3x

p(x) = 0

3x = 0

x = 0

Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vi) p(x) = ax

p(x) = 0

ax = 0

x = 0

Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vii) p(x) = cx + d

p(x) = 0

cx+ d = 0

Therefore, for, the value of the polynomial is 0 and hence, is a zero of the given polynomial.

Video Solution for number systems (Page: 35 , Q.No.: 4)

NCERT Solution for Class 9 math - number systems 35 , Question 4



Page No 40:

Question 1:

Find the remainder when x3 + 3x2 + 3x + 1 is divided by

(i) x + 1 (ii) (iii) x

(iv) x + π (v) 5 + 2x

Answer:

(i) x + 1

By long division,

Therefore, the remainder is 0.

(ii)

By long division,

Therefore, the remainder is.

(iii) x

By long division,

Therefore, the remainder is 1.

(iv) x + π

By long division,

Therefore, the remainder is

(v) 5 + 2x

By long division,

Therefore, the remainder is

Video Solution for number systems (Page: 40 , Q.No.: 1)

NCERT Solution for Class 9 math - number systems 40 , Question 1

Page No 40:

Question 2:

Find the remainder when x3ax2 + 6xa is divided by x − a.

Answer:

By long division,

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

Video Solution for number systems (Page: 40 , Q.No.: 2)

NCERT Solution for Class 9 math - number systems 40 , Question 2

Page No 40:

Question 3:

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

Answer:

Let us divide (3x3 + 7x) by (7 + 3x). If the remainder obtained is 0, then 7 + 3x will be a factor of 3x3 + 7x.

By long division,

As the remainder is not zero, therefore, 7 + 3x is not a factor of 3x3 + 7x.

Video Solution for number systems (Page: 40 , Q.No.: 3)

NCERT Solution for Class 9 math - number systems 40 , Question 3



Page No 43:

Question 1:

Determine which of the following polynomials has (x + 1) a factor:

(i) x3 + x2 + x + 1 (ii) x4 + x3 + x2 + x + 1

(iii) x4 + 3x3 + 3x2 + x + 1 (iv)

Answer:

(i) If (x + 1) is a factor of p(x) = x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

p(x) = x3 + x2 + x + 1

p(−1) = (−1)3 + (−1)2 + (−1) + 1

= − 1 + 1 − 1 + 1 = 0

Hence, x + 1 is a factor of this polynomial.

(ii) If (x + 1) is a factor of p(x) = x4 + x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

p(x) = x4 + x3 + x2 + x + 1

p(−1) = (−1)4 + (−1)3 + (−1)2 + (−1) + 1

= 1 − 1 + 1 −1 + 1 = 1

As p(− 1) ≠ 0,

Therefore, x + 1 is not a factor of this polynomial.

(iii) If (x + 1) is a factor of polynomial p(x) = x4 + 3x3 + 3x2 + x + 1, then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.

p(−1) = (−1)4 + 3(−1)3 + 3(−1)2 + (−1) + 1

= 1 − 3 + 3 − 1 + 1 = 1

As p(−1) ≠ 0,

Therefore, x + 1 is not a factor of this polynomial.

(iv) If(x + 1) is a factor of polynomial p(x) = , then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.

As p(−1) ≠ 0,

Therefore, (x + 1) is not a factor of this polynomial.

Video Solution for number systems (Page: 43 , Q.No.: 1)

NCERT Solution for Class 9 math - number systems 43 , Question 1

Page No 43:

Question 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

(iii) p(x) = x3 − 4 x2 + x + 6, g(x) = x − 3

Answer:

(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then p(−1) must be zero.

p(x) = 2x3 + x2 − 2x − 1

p(−1) = 2(−1)3 + (−1)2 − 2(−1) − 1

= 2(−1) + 1 + 2 − 1 = 0

Hence, g(x) = x + 1 is a factor of the given polynomial.

(ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then p(−2) must

be 0.

p(x) = x3 +3x2 + 3x + 1

p(−2) = (−2)3 + 3(−2)2 + 3(−2) + 1

= − 8 + 12 − 6 + 1

= −1

As p(−2) ≠ 0,

Hence, g(x) = x + 2 is not a factor of the given polynomial.

(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then p(3) must

be 0.

p(x) = x3 − 4 x2 + x + 6

p(3) = (3)3 − 4(3)2 + 3 + 6

= 27 − 36 + 9 = 0

Hence, g(x) = x − 3 is a factor of the given polynomial.

Video Solution for number systems (Page: 43 , Q.No.: 2)

NCERT Solution for Class 9 math - number systems 43 , Question 2



Page No 44:

Question 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)

(iii) (iv) p(x) = kx2 − 3x + k

Answer:

If x − 1 is a factor of polynomial p(x), then p(1) must be 0.

(i) p(x) = x2 + x + k

p(1) = 0

⇒ (1)2 + 1 + k = 0

⇒ 2 + k = 0
 

k = −2

Therefore, the value of k is −2.

(ii)

p(1) = 0

(iii)

p(1) = 0

(iv) p(x) = kx2 − 3x + k

p(1) = 0

k(1)2 − 3(1) + k = 0

k − 3 + k = 0

⇒ 2k − 3 = 0

Therefore, the value of k is.

Video Solution for number systems (Page: 44 , Q.No.: 3)

NCERT Solution for Class 9 math - number systems 44 , Question 3

Page No 44:

Question 4:

Factorise:

(i) 12x2 − 7x + 1 (ii) 2x2 + 7x + 3

(iii) 6x2 + 5x − 6 (iv) 3x2x − 4

Answer:

(i) 12x2 − 7x + 1

We can find two numbers such that pq = 12 × 1 = 12 and p + q = −7. They are p = −4 and q = −3.

Here, 12x2 − 7x + 1 = 12x2 − 4x − 3x + 1

= 4x (3x − 1) − 1 (3x − 1)

= (3x − 1) (4x − 1)

(ii) 2x2 + 7x + 3

We can find two numbers such that pq = 2 × 3 = 6 and p + q = 7.

They are p = 6 and q = 1.

Here, 2x2 + 7x + 3 = 2x2 + 6x + x + 3

= 2x (x + 3) + 1 (x + 3)

= (x + 3) (2x+ 1)

(iii) 6x2 + 5x − 6

We can find two numbers such that pq = −36 and p + q = 5.

They are p = 9 and q = −4.

Here,

6x2 + 5x − 6 = 6x2 + 9x − 4x − 6

= 3x (2x + 3) − 2 (2x + 3)

= (2x + 3) (3x − 2)

(iv) 3x2x − 4

We can find two numbers such that pq = 3 × (− 4) = −12

and p + q = −1.

They are p = −4 and q = 3.

Here,

3x2x − 4 = 3x2 − 4x + 3x − 4

= x (3x − 4) + 1 (3x − 4)

= (3x − 4) (x + 1)


Video Solution for number systems (Page: 44 , Q.No.: 4)

NCERT Solution for Class 9 math - number systems 44 , Question 4

Page No 44:

Question 5:

Factorise:

(i) x3 − 2x2x + 2 (ii) x3 + 3x2 −9x − 5

(iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 − 2y − 1

Answer:

(i) Let p(x) = x3 − 2x2x + 2

All the factors of 2 have to be considered. These are ± 1, ± 2.

By trial method,

p(−1) = (−1)3 − 2(−1)2 − (−1) + 2

= −1 − 2 + 1 + 2 = 0

Therefore, (x +1 ) is factor of polynomial p(x).

Let us find the quotient on dividing x3 − 2x2x + 2 by x + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

x3 − 2x2x + 2 = (x + 1) (x2 − 3x + 2) + 0

= (x + 1) [x2 − 2xx + 2]

= (x + 1) [x (x − 2) − 1 (x − 2)]

= (x + 1) (x − 1) (x − 2)

= (x − 2) (x − 1) (x + 1)

(ii) Let p(x) = x3 − 3x2 − 9x − 5

All the factors of 5 have to be considered. These are ±1, ± 5.

By trial method,

p(−1) = (−1)3 − 3(−1)2 − 9(−1) − 5

= − 1 − 3 + 9 − 5 = 0

Therefore, x + 1 is a factor of this polynomial.

Let us find the quotient on dividing x3 + 3x2 − 9x − 5 by x + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

x3 − 3x2 − 9x − 5 = (x + 1) (x2 − 4x − 5) + 0

= (x + 1) (x2 − 5x + x − 5)

= (x + 1) [(x (x − 5) +1 (x − 5)]

= (x + 1) (x − 5) (x + 1)

= (x − 5) (x + 1) (x + 1)

(iii) Let p(x) = x3 + 13x2 + 32x + 20

All the factors of 20 have to be considered. Some of them are ±1,

± 2, ± 4, ± 5 ……

By trial method,

p(−1) = (−1)3 + 13(−1)2 + 32(−1) + 20

= − 1 +13 − 32 + 20

= 33 − 33 = 0

As p(−1) is zero, therefore, x + 1 is a factor of this polynomial p(x).

Let us find the quotient on dividing x3 + 13x2 + 32x + 20 by (x + 1).

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0

= (x + 1) (x2 + 10x + 2x + 20)

= (x + 1) [x (x + 10) + 2 (x + 10)]

= (x + 1) (x + 10) (x + 2)

= (x + 1) (x + 2) (x + 10)

(iv) Let p(y) = 2y3 + y2 − 2y − 1

By trial method,

p(1) = 2 ( 1)3 + (1)2 − 2( 1) − 1

= 2 + 1 − 2 − 1= 0

Therefore, y − 1 is a factor of this polynomial.

Let us find the quotient on dividing 2y3 + y2 − 2y − 1 by y ­− 1.

p(y) = 2y3 + y2 − 2y − 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)

Video Solution for number systems (Page: 44 , Q.No.: 5)

NCERT Solution for Class 9 math - number systems 44 , Question 5



Page No 48:

Question 1:

Use suitable identities to find the following products:

(i) (ii)

(iii) (iv)

(v)

Answer:

(i) By using the identity ,

(ii) By using the identity ,

(iii)

By using the identity ,

(iv) By using the identity ,

(v) By using the identity ,


Video Solution for number systems (Page: 48 , Q.No.: 1)

NCERT Solution for Class 9 math - number systems 48 , Question 1

Page No 48:

Question 2:

Evaluate the following products without multiplying directly:

(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96

Answer:

(i) 103 × 107 = (100 + 3) (100 + 7)

= (100)2 + (3 + 7) 100 + (3) (7)

[By using the identity, where

x = 100, a = 3, and b = 7]

= 10000 + 1000 + 21

= 11021

(ii) 95 × 96 = (100 − 5) (100 − 4)

= (100)2 + (− 5 − 4) 100 + (− 5) (− 4)

[By using the identity, where

x = 100, a = −5, and b = −4]

= 10000 − 900 + 20

= 9120

(iii) 104 × 96 = (100 + 4) (100 − 4)

= (100)2 − (4)2

= 10000 − 16

= 9984

Video Solution for number systems (Page: 48 , Q.No.: 2)

NCERT Solution for Class 9 math - number systems 48 , Question 2

Page No 48:

Question 3:

Factorise the following using appropriate identities:

(i) 9x2 + 6xy + y2

(ii)

(iii)

Answer:

(i)

(ii)

(iii)


Video Solution for number systems (Page: 48 , Q.No.: 3)

NCERT Solution for Class 9 math - number systems 48 , Question 3



Page No 49:

Question 4:

Expand each of the following, using suitable identities:

(i) (ii)

(iii) (iv)

(v) (vi)

Answer:

It is known that,

(i)

(ii)

(iii)

(iv)

(v)

(vi)



[[VIDEO:16294>]]

Page No 49:

Question 5:

Factorise:

(i)

(ii)

Answer:

It is known that,

(i)

(ii)


Video Solution for number systems (Page: 49 , Q.No.: 5)

NCERT Solution for Class 9 math - number systems 49 , Question 5

Page No 49:

Question 6:

Write the following cubes in expanded form:

(i) (ii)

(iii) (iv)

Answer:

It is known that,

(i)

(ii)

(iii)

(vi)


Video Solution for number systems (Page: 49 , Q.No.: 6)

NCERT Solution for Class 9 math - number systems 49 , Question 6

Page No 49:

Question 7:

Evaluate the following using suitable identities:

(i) (99)3 (ii) (102)3 (iii) (998)3

Answer:

It is known that,

(i) (99)3 = (100 − 1)3

= (100)3 − (1)3 − 3(100) (1) (100 − 1)

= 1000000 − 1 − 300(99)

= 1000000 − 1 − 29700

= 970299

(ii) (102)3 = (100 + 2)3

= (100)3 + (2)3 + 3(100) (2) (100 + 2)

= 1000000 + 8 + 600 (102)

= 1000000 + 8 + 61200

= 1061208

(iii) (998)3= (1000 − 2)3

= (1000)3 − (2)3 − 3(1000) (2) (1000 − 2)

= 1000000000 − 8 − 6000(998)

= 1000000000 − 8 − 5988000

= 1000000000 − 5988008

= 994011992

Video Solution for number systems (Page: 49 , Q.No.: 7)

NCERT Solution for Class 9 math - number systems 49 , Question 7

Page No 49:

Question 8:

Factorise each of the following:

(i) (ii)

(iii) (iv)

(v)

Answer:

It is known that,

(i)




 

(ii)




 

(iii)

(iv)

(v)

Video Solution for number systems (Page: 49 , Q.No.: 8)

NCERT Solution for Class 9 math - number systems 49 , Question 8

Page No 49:

Question 9:

Verify:

(i)

(ii)

Answer:

(i) It is known that,

(ii) It is known that,


Video Solution for number systems (Page: 49 , Q.No.: 9)

NCERT Solution for Class 9 math - number systems 49 , Question 9

Page No 49:

Question 10:

Factorise each of the following:

(i)

(ii)

[Hint: See question 9.]

Answer:

(i)

(ii)


Video Solution for number systems (Page: 49 , Q.No.: 10)

NCERT Solution for Class 9 math - number systems 49 , Question 10

Page No 49:

Question 11:

Factorise:

Answer:

It is known that,



Video Solution for number systems (Page: 49 , Q.No.: 11)

NCERT Solution for Class 9 math - number systems 49 , Question 11

Page No 49:

Question 12:

Verify that

Answer:

It is known that,


Video Solution for number systems (Page: 49 , Q.No.: 12)

NCERT Solution for Class 9 math - number systems 49 , Question 12

Page No 49:

Question 13:

If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Answer:

It is known that,

Put x + y + z = 0,


Video Solution for number systems (Page: 49 , Q.No.: 13)

NCERT Solution for Class 9 math - number systems 49 , Question 13

Page No 49:

Question 14:

Without actually calculating the cubes, find the value of each of the following:

(i)

(ii)

Answer:

(i)

Let x = −12, y = 7, and z = 5

It can be observed that,

x + y + z = − 12 + 7 + 5 = 0

It is known that if x + y + z = 0, then

= −1260

(ii)

Let x = 28, y = −15, and z = −13

It can be observed that,

x + y + z = 28 + (−15) + (−13) = 28 − 28 = 0

It is known that if x + y + z = 0, then


Video Solution for number systems (Page: 49 , Q.No.: 14)

NCERT Solution for Class 9 math - number systems 49 , Question 14

Page No 49:

Question 15:

Give possible expressions for the length and breadth of each of thefollowing rectangles, in which their areas are given:

Answer:

Area = Length × Breadth

The expression given for the area of the rectangle has to be factorised. One of its factors will be its length and the other will be its breadth.

(i)

Therefore, possible length = 5a − 3

And, possible breadth = 5a − 4

(ii)

Therefore, possible length = 5y + 4

And, possible breadth = 7y − 3

Video Solution for number systems (Page: 49 , Q.No.: 15)

NCERT Solution for Class 9 math - number systems 49 , Question 15



Page No 50:

Question 16:

What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

Answer:

Volume of cuboid = Length × Breadth × Height

The expression given for the volume of the cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will be its height.

(i)

One of the possible solutions is as follows.

Length = 3, Breadth = x, Height = x − 4

(ii)

One of the possible solutions is as follows.

Length = 4k, Breadth = 3y + 5, Height = y − 1



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