RBSE Class 9 Chapter 3: Polynomials Important Questions and Solutions

RBSE Class 9 Maths Chapter 3 Important questions and solutions are available at BYJU’S. The important questions and solutions for chapter 3 of class 9 are prepared by expert faculty at BYJU’S. All these important questions are provided by considering the new pattern of RBSE so that students can get thorough knowledge for their exams.

Chapter 3 of the RBSE Class 9 Maths helps the students in learning different terms related to polynomials. Also, they can learn different methods to find the zeroes of the given polynomial. Besides, students can get knowledge about how to use algebraic identities for factorisation. However, the important questions provided by BYJU’S here cover all these concepts with detailed explanations.

RBSE Maths Chapter 3: Exercise 3.1 Textbook Important Questions and Solutions

Question 1: How many terms are there in the trinomial?

  1. 1
  2. 3
  3. 2
  4. 4

Solution:

Correct answer: (B)

As per the definition, a polynomial with three terms is called a trinomial. In polynomials, a trinomial always has 3 terms. For example x2 + 2x + 2, y2 + 5y – 4, x3 – 6x + 11 and so on.

Question 2: A polynomial with degree 2 is called

  1. Quadratic polynomial
  2. Monomial
  3. Binomial
  4. Cubic polynomial

Solution:

Correct answer: (A)

The highest power of a polynomial is called the degree of the polynomial and a polynomial with degree 2 is called a quadratic polynomial. For example 5x2 + 4x – 3, 2t2 – 11 and so on.

Question 3: Write the number of terms of the below given polynomials and write the name of the polynomial.

(i) x2 – 5√3

(ii) 3y2 -13x + 5

(iii) x + 4/x

(iv) 2x3 + 7x – x2 – 14

Solution:

(i) x2 – 5√3 has two terms and it is called binomial.

(ii) 3y2 has only one term and is called monomial.

(iii) x + 4/x is not a polynomial, since the variable is in the denominator.

(iv) 2x3 + 7x – x2 has three terms and is called trinomial.

Question 4: Write the name of the polynomials given below based on the degree.

(i) 3√x – 4x

(ii) t2 + t + 5

(iii) 9x3 – 2x2 + x – 6

(iv) 3y

Solution:

(i) 3√x – 4x is not a polynomial, since the degree is not a whole number.

(ii) t2 + t + 5 is called quadratic polynomial since its highest degree is 2.

(iii) 9x3 – 2x2 + x – 6 is called cubic polynomial since its highest degree is 3.

(iv) 3y is called linear polynomial since its highest degree is 1.

Question 5: Write the coefficients of x in the following expressions.

(i) 12x2 + 5x – 3

(ii) 11 – x3

(iii) √7x – 3

(iv) x2 + πx + 4

Solution:

(i) 12x2 + 5x – 3

Coefficient of x = 5

(ii) 11 – x3

Here, there is no x term. Hence, its coefficient is equal to 0.

(iii) √7x – 3

Coefficient of x = √7

(iv) x2 + πx + 4

Coefficient of x = π

Question 6: Write an example of a binomial of degree 25.

Solution:

We know that binomial has two terms.

Therefore, a binomial with degree 25 is of the form ax25 + b or ax25 + bxn, where n is less than 25.

Question 7: Write an example of a monomial of degree 110.

Solution:

Monomial has only one term.

Therefore, a monomial with degree 110 is of the form ax110, where a is any real number.

Question 8: Write an example of a trinomial of degree 5.

Solution:

Trinomial has three terms.

Therefore, an example for a trinomial with degree 5 is: x5 + 2x3 – 4

Question 9: Write the degree of the polynomials given below.

(i) √y + 5

(ii) 3x2 – 2x + 15

(iii) 5x – 3

(iv) y3 + 4x2 – 2x + 7

Solution:

(i) √y + 5 is not a polynomial since the degree of the expression is not a whole number.

(ii) The highest power of 3x2 – 2x + 15 is 2. Hence, the degree of the polynomial is 2.

(iii) The highest power of 5x – 3 is 1. Therefore, the degree of the polynomial is 1.

(iv) The highest power of y3 + 4x2 – 2x + 7 is 3 so the degree of the polynomial is 3

Question 10: Identify the polynomials from the following expressions.

(i) 7

(ii) y – 3/y

(iii) 8x3 – x2 + 5x + 1

(iv) t2 + 2√3

Solution:

(i) 7 is a constant polynomial.

(ii) y – 3/y is not a polynomial since the variable is there in the denominator.

(iii) 8x3 – x2 + 5x + 1 is also a polynomial.

(iv) t2 + 2√3 is a polynomial.

RBSE Maths Chapter 3: Exercise 3.2 Textbook Important Questions and Solutions

Question 11: Find the value of the polynomial p(x) = 3x3 – 2x2 – 2x + 4 at x = 2.

Solution:

Given polynomial is:

p(x) = 3x3 – 2x2 – 2x + 4

To find the value of given polynomial at x = 2, substitute x = 2 in p(x),

p(2) = 3(2)3 – 2(2)2 – 2(2) + 4

= 3(8) – 2(4) – 4 + 4

= 24 – 8

= 16

Therefore, the value of p(x) at x = 2 is 16

Question 12: Find the value of polynomial p(x) = 2x3 – 13x2 + 12x + 17 at x = -1.

Solution:

Given polynomial is:

p(x) = 2x3 – 13x2 + 12x + 17

To find the value of given polynomial at x = -1, substitute that value in p(x),

p(-1) = 2(-1)3 – 13(-1)2 + 12(-1) + 17

= 2(-1) – 13 – 12 + 17

= -2 – 13 – 12 + 17

= -10

Therefore, the value of the given polynomial p(x) at x = -1 is -10.

Question 13: For the polynomial p(x) = (x + 3)(x – 2), find p(0), p(1) and p(-2).

Solution:

Given,

p(x) = (x + 3)(x – 2)

To find the value of given polynomial at x = a, substitute x = a in p(x),

p(0) = (0 + 3)(0 – 2)

= 3(-2)

= -6

p(1) = (1 + 3)(1 – 2)

= 4(-1)

= -4

p(-2) = (-2 + 3)(-2 – 2)

= 1(-4)

= -4

Question 14: Find the value of the polynomial p(y) = -y3 at -3, 2, and -1.

Solution:

Given,

p(y) = -y3

p(y) at y = -3:

p(-3) = -(-3)3

= -(-27)

= 27

p(y) at y = 2:

p(2) = -(2)3

= -8

p(y) at y = -1,

p(-1) = -(-1)3

= -(-1)

= 1

Question 15: Check whether 2 and -2 are the zeroes of the polynomial p(x) = x2 – 4.

Solution:

Given polynomial is:

p(x) = x2 – 4

Substituting x = 2 in p(x),

p(2) = (2)2 – 4

= 4 – 4 = 0

Substituting x = -2 in p(x),

p(-2) = (-2)2 – 4

= 4 – 4

= 0

Therefore, 2 and -2 are the zeroes of the given polynomial.

Question 16: Check whether y = -½ is a zero of the polynomial p(y) = 2y + 1.

Solution:

Given,

p(y) = 2y + 1

Substitute y = -½ in p(y),

p(-½) = 2(-½) + 1

= -1 + 1

= 0

Therefore, y = -½ is the zero of the given polynomial.

Question 17: Find the zero of the polynomial p(x) = 5x – 4.

Solution:

Given,

p(x) = 5x – 4

To find the zero of any polynomial, equate the given polynomial to 0.

Now, 5x – 4 = 0

5x = 4

x = ⅘

(or)

Comparing the given linear polynomial with p(x) = ax + b

a = 5, b = 4

Zero: x = -b/a

= -(-4)/5

= ⅘

Question 18: Find the zeroes of the polynomials given below.

(i) p(x) = 5x, p(x) = 9x + 7

Solution:

(i) p(x) = 5x

Comparing with the standard form,

a = 5, b = 0

x = -b/a = -0/5 = 0

Therefore, zero of the polynomial is 0.

(ii) p(x) = 9x + 7

Comparing with the standard form,

a = 9, b = 7

x = -b/a

= -7/9

Therefore, zero of the given polynomial is -7/9.

Question 19: Find the value of the polynomial p(x) = 2x3 – 3x2 + 4x – 5 at x = 1/2.

Solution:

Given polynomial is p(x) = 2x3 – 3x2 + 4x – 5

p(x) at x = ½

p(½) = 2(½)3 – 3(½)2 + 4(½) – 5

= 2(⅛) – 3(¼) + 2 – 5

= ¼ – ¾ – 3

= (1 – 3 – 12)/4

= -14/4

= -7/2

Question 20: Find the zero of the polynomial p(x) = px + q, where p and q are real numbers.

 

Solution:

Given,

p(x) = px + q

Comparing with the standard form,

a = p, b = q

x = -b/a

= -q/p

RBSE Maths Chapter 3: Exercise 3.3 Textbook Important Questions and Solutions

Question 21: Find the remainder, if the polynomial p(x) = 3x4 + x3 – x2 + 6x + 1 is divided by x + 1.

Solution:

Given,

p(x) = 3x4 + x3 – x2 + 6x + 1

Zero of x + 1 is -1.

Substituting x = -1 in p(x),

p(-1) = 3(-1)4 + (-1)3 – (-1)2 + 6(-1) + 1

= 3 – 1 – 1 – 6 + 1

= 4 – 8

= -4

Therefore, the remainder is -4.

Question 22: Find the remainder, when the polynomial 6x3 + x2 + 3x – 15 is divided by 3x – 2.

Solution:

Let the given polynomial be p(x) = 6x3 + x2 + 3x – 15

Zero of 3x – 2 is ⅔.

Substituting x = ⅔ in p(x),

p(⅔) = 6(⅔)3 + (⅔)2 + 3(⅔) – 15

= 6(8/27) + 4/9 + 2 – 15

= 16/9 + 4/ 9 – 13

= (16 + 4 – 117)/9

= -97/9

Question 23: Divide the polynomial p(x) = 7x2 + 5x – 3 by x – 1 and write the quotient and remainder.

Solution:

Given,

p(x) = 7x2 + 5x – 3

Let g(x) = x – 1

RBSE class 9 chapter 3 imp que 23 solution

Therefore, quotient = q(x) = 7x + 12 and the remainder = 9

Question 24: Find the remainder, when the polynomial 2x3 + 2ax2 – 5x + a is divided by x + a.

Solution:

Given polynomial is p(x) = 2x3 + 2ax2 – 5x + a

Zero of x + a is -a.

Substituting x = -a in p(x),

p(-a) = 2(-a)3 + 2a(-a)2 – 5(-a) + a

= -2a3 + 2a3 + 5a + a

= 6a

Question 25: Check whether x + 2 is a factor of x3 + 3x2 + 3x + 1 or not.

Solution:

Let the given polynomial be p(x) = x3 + 3x2 + 3x + 1

If x – a is a factor of p(x), then p(a) = 0.

To check whether x + 2 is a factor of the given polynomial, substitute x = -2 in p(x).

p(-2) = (-2)3 + 3(-2)2 + 3(-2) + 1

= -8 + 3(4) – 6 + 1

= -8 + 12 – 6 + 1

= – 1

Therefore, x + 2 is not a factor of the given polynomial.

Question 26: Find the value of b, if polynomials x3 + x2 – 4x + b and 2x3 + bx2 + 3x – 3 result in the same remainder when divided by x – 2.

Solution:

Let f(x) = x3 + x2 – 4x + b

Zero of x – 2 is 2.

Substituting x = 2 in f(x),

f(2) = (2)3 + (2)2 – 4(2) + b

= 8 + 4 – 8 + b

= 4 + b

The remainder of f(x) is 4 + b

Let g(x) = 2x3 + bx2 + 3x – 3

Substituting x = 2 in g(x),

g(2) = 2(2)3 + b(2)2 + 3(2) – 3

= 2(8) + 4b + 6 – 3

= 16 + 4b + 6 – 3

= 19 + 4b

Given that, the remainder is the same for both the polynomials.

4 + b = 19 + 4b

4 – 19 = 4b – b

-15 = 3b

-15/3 = b

-5 = b

Therefore, the value of b is -5.

RBSE Maths Chapter 3: Exercise 3.4 Textbook Important Questions and Solutions

Question 27: Determine whether x – 2 is a factor of x3 – x2 -(2 + √5)x + 2√5 or not.

Solution:

Let the given polynomial be p(x) = x3 – x2 -(2 + √5)x + 2√5

If x – a is a factor of p(x), then p(a) = 0.

To check whether x – 2 is a factor of the given polynomial, substitute x = 2 in p(x).

p(2) = (2)3 – (2)2 – (2 + √5)(2) + 2√5

= 8 – 4 – (4 + 2√5) + 2√5

= 4 – 4 – 2√5 + 2√5

= 0

Hence, x – 2 is the factor of the given polynomial.

Question 28: Determine whether x – 1 is a factor of the polynomial p(x) = x4 – 3x3 + 3x2 – x + 2 or not.

Solution:

Given,

p(x) = x4 – 3x3 + 3x2 – x + 2

If x – a is a factor of p(x), then p(a) = 0.

To check whether x – 1 is a factor of the given polynomial, substitute x = 1 in p(x).

p(1) = (1)4 – 3(1)3 + 3(1)2 – (1) + 2

= 1 – 3 + 3 – 1 + 2

= 2

Here, p(1) is not equal to 0.

Hence, x – 1 is not the factor of the given polynomial.

Question 29: Using the factor theorem, check whether g(x) = x + 1 is a factor of p(x) = 3x3 + x2 – 3x – 1 or not.

Solution:

Given,

p(x) = 3x3 + x2 – 3x – 1

g(x) = x + 1

Let g(x) = 0

x + 1 = 0

x = -1

Thus, zero of g(x) = x + 1 is -1.

Substituting x = -1 in p(x),

p(-1) = 3(-1)3 + (-1)2 – 3(-1) – 1

= 3(-1) + 1 + 3 – 1

= -3 + 3

= 0

Therefore, g(x) is the factor of p(x).

Question 30: Using the factor theorem, verify that g(x) = 2x + 1 is a factor of f(x) = 2x3 + x2 – 2x + 1 or not.

Solution:

Given,

f(x) = 2x3 + x2 – 2x + 1

g(x) = 2x + 1

Let g(x) = 0

2x + 1 = 0

2x = -1

x = -½

Thus, zero of g(x) = 2x + 1 is -½.

Substituting x = -½ in f(x),

f(-½) = 2(-½)3 + (-½)2 – 2(-½) + 1

= 2(-⅛) + ¼ + 1 + 1

= -¼ + ¼ + 2

= 2 ≠ 0

Hence, g(x) is not a factor of f(x).

Question 31: Find the value of k, when x – 5 is a factor of the polynomial p(x) = x3 – 3x2 + kx – 10.

Solution:

Given polynomial is:

p(x) = x3 – 3x2 + kx – 10

If x – a is a factor of p(x), then p(a) = 0.

Given that x – 5 is the factor of p(x).

That means, p(5) = 0

(5)3 – 3(5)2 + k(5) – 10 = 0

125 – 3(25) + 5k – 10 = 0

125 – 75 + 5k – 10 = 0

5k + 40 = 0

5k = -40

k = -40/5

k = -8

Therefore, the value of k is -8.

Question 32: If x – 1 is a factor of the polynomial 2x2 + kx + √2, then find the value of k.

Solution:

Let the given polynomial be p(x) = 2x2 + kx + √2

If x – a is a factor of p(x), then p(a) = 0.

Given that x – 1 is the factor of p(x).

That means, p(1) = 0

2(1)2 + k(1) + √2 = 0

2 + k + √2 = 0

k = -2 – √2

Or

k = -(2 + √2)

Hence, the value of k is -2 – √2 or -(2 + √2).

Question 33: Find the value of k and m if x – 1 and x + 1 are the factors of the polynomial x4 + kx3 – 3x2 + 2x + m.

Solution:

Let the given polynomial be p(x) = x4 + kx3 – 3x2 + 2x + m

If x – a is a factor of p(x), then p(a) = 0.

Given that x – 1 and x + 1 are the factors of p(x).

p(1) = 0

(1)4 + k(1)3 – 3(1)2 + 2(1) + m = 0

1 + k – 3 + 2 + m = 0

m + k = 0 ….(i)

Now, p(-1) = 0

(-1)4 + k(-1)3 – 3(-1)2 + 2(-1) + m = 0

1 – k – 3 – 2 + m = 0

m – k – 4 = 0

m – k = 4 ….(ii)

Adding (i) and (ii),

m + k + m – k = 0 + 4

2m = 4

m = 4/2

m = 2

Substituting m = 2 in (i),

2 + k = 0

k = -2

Therefore, m = 2 and k = -2.

Question 34: Factorise the polynomial 4x2 – x – 3.

Solution:

Given polynomial is: 4x2 – x – 3

Product of 4 and -3 is 4 × (-3) = -12

Factors of 12 are 1, 2, 3, 4, 6, 12

Sum of the factors should be -1.

So, -4 and 3 can be the factors of -12 here.

Now splitting the middle term of the given polynomial based on these factors.

4x2 – x – 3 = 4x2 – 4x + 3x – 3

= 4x(x – 1) + 3(x – 1)

= (4x + 3)(x – 1)

Hence, (4x + 3) and (x – 1) are the factors of 4x2 – x – 3.

Question 35: Factorise: 6x2 + 5x – 6

Solution:

Given polynomial is: 6x2 + 5x – 6

Product of 6 and -6 is 6 × (-6) = -36.

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.

Sum of the factors should be 5.

9 and -4 can be the factors of -36 here.

Now splitting the middle term of the given polynomial as shown below.

6x2 + 5x – 6 = 6x2 + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3)

= (3x – 2)(2x + 3)

Hence, (3x – 2) and (2x + 3) are the factors of 6x2 + 5x – 6.

Question 36: Find the zeroes of the polynomial x3 + 2x2 – x – 2.

Solution:

Let the given polynomial p(x) = x3 + 2x2 – x – 2

The product of the coefficient of the term with highest power and the constant term is: 1 × (-2) = -2

Possible factors of 2 in this case are ±1 and ±2.

Hence, the possible zeros can be ±1 and ±2.

Let us substitute x = 1 in p(x)

p(1) = (1)3 + 2(1)2 – (1) – 2

= 1 + 2 – 1 – 2

= 0

Now substitute x = -1 in p(x)

p(-1) = (-1)3 + 2(-1)2 – (-1) – 2

= -1 + 2 + 1 – 2

= 0

Substituting x = 2 in p(x),

p(2) = (2)3 + 2(2)2 – (2) – 2

= 8 + 8 – 2 – 2

= 12

Substituting x = -2 in p(x)

p(-2) = (-2)3 + 2(-2)2 – (-2) – 2

= -8 + 8 + 2 – 2

= 0

p(x) is equal to zero at x = 1, x = -1 and x = -2.

Therefore, 1, -1 and -2 are the zeroes of the given polynomial.

Question 37: Find the zeroes of the polynomial x4 – 2x3 – 7x2 + 8x + 12.

Solution:

Let the given polynomial be p(x) = x4 – 2x3 – 7x2 + 8x + 12

The product of the coefficient of the term with highest power and the constant term is: 1 × (12) = 12

Possible factors of 12 in this case are ±1, ±2, ±3, ±4, ±6 and ±12.

Hence, the possible zeros can be ±1, ±2, ±3, ±4, ±6 and ±12.

By considering the signs, here positive terms have more value. So we are taking negative numbers at first.

Substituting x = -1,

p(-1) = (-1)4 – 2(-1)3 – 7(-1)2 + 8(-1) + 12

= 1 – 2(-1) – 7 – 8 + 12

= 1 + 2 – 15 + 12

= 0

p(-2) = (-2)4 – 2(-2)3 – 7(-2)2 + 8(-2) + 12

= 16 – 2(-8) – 7(4) – 16 + 12

= 16 + 16 – 28 – 16 + 12

= 0

p(2) = (2)4 – 2(2)3 – 7(2)2 + 8(2) + 12

= 16 – 2(8) – 7(4) + 16 + 12

= 16 – 16 – 28 + 16 + 12

= 0

p(3) = (3)4 – 2(3)3 – 7(3)2 + 8(3) + 12

= 81 – 2(27) – 7(9) + 24 + 12

= 81 – 54 – 63 + 24 + 12

= 0

p(x) is equal to zero at x = -1, x = -2, x = 2 and x = 3.

Therefore, -1, -2, 2 and 3 are the zeroes of the given polynomial.

Question 38: Find the zeroes of the polynomial x3 – 3x2 – 9x – 5.

Solution:

Let the given polynomial be p(x) = x3 – 3x2 – 9x – 5

The product of the coefficient of the term with highest power and the constant term is: 1 × (-5) = -5

Possible factors of 5 in this case are ±1 and ±5.

Hence, the possible zeroes can be ±1 and ±5.

Substituting x = 1 in p(x),

p(1) = (1)3 – 3(1)2 – 9(1) – 5

= 1 – 3 – 9 – 5

= -16

p(-1) = (-1)3 – 3(-1)2 – 9(-1) – 5

= -1 – 3 + 9 – 5

= 0

p(5) = (5)3 – 3(5)2 – 9(5) – 5

= 125 – 3(25) – 45 – 5

= 125 – 75 – 50

= 0

p(-5) = (-5)3 – 3(-5)2 – 9(-5) – 5

= -125 – 3(25) + 45 – 5

= -125 – 75 + 10

= -190

p(x) is equal to 0 at x = -1 and x = 5.

Therefore, -1 and 5 are the zeroes of the given polynomial.

RBSE Maths Chapter 3: Exercise 3.5 Textbook Important Questions and Solutions

Question 39: Use the suitable identity to find the product of (x + 3) and (x + 7).

Solution:

(x + 3)(x + 7)

Using the identity (x + a)(x + b) = x2 + (a + b)x + ab

(x + 3)(x + 7) = x2 + (3 + 7)x + 3(7)

= x2 + 10x + 21

Therefore, (x + 3)(x + 7) = x2 + 10x + 21

Question 40: Use the suitable identity to find the product: (x2 + 3/5) (x2 – 3/5)

Solution:

(x2 + ⅗)(x2 – ⅗)

Using the identity (a + b)(a – b) = a2 – b2

(x2 + ⅗)(x2 – ⅗) = (x2)2 – (⅗)2

= x4 – 9/25

Therefore,

(x2 + ⅗)(x2 – ⅗) = x4 – 9/25

Question 41: Using the algebraic identities, find the product of the following.

(i) 94 × 97 (ii) 103 × 97

Solution:

(i) 94 × 97 = (100 – 6)(100 – 3)

= [100 + (-6)][100 + (-3)]

Using the identity (x + a)(x + b) = x2 + (a + b)x + ab

= (100)2 + [(-6) + (-3)](100) + (-6)(-3)

= 10000 -9(100) + 18

= 10000 – 900 + 18

= 9118

(ii) 103 × 97 = (100 + 3)(100 – 3)

Using the identity (a + b)(a – b) = a2 – b2

= (100)2 – (3)3

= 10000 – 9

= 9991

Question 42: Use the suitable identities, factorise the following.

(i) x2 – 6xy + 9y2 (ii) x2 + 4x + 4

Solution:

(i) x2 – 6xy + 9y2

= x2 – 2(x)(3y) + (3y)2

Using the identity (a – b)2 = a2 – 2ab + b2

= (x – 3y)2

= (x – 3y)(x – 3y)

(ii) x2 + 4x + 4

= x2 + 2(x)(2) + (2)2

Using the identity (a + b)2 = a2 + 2ab + b2

= (x + 2)2

= (x + 2)(x + 2)

Question 43: Expand (2 + x – 2y)2 with the help of suitable identity.

Solution:

(2 + x – 2y)2 = [2 + x + (-2y)]2

Using the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

= (2)2 + x2 + (-2y)2 + 2(2)(x) + 2(x)(-2y) + 2(-2y)(2)

= 4 + x2 + 4y2 + 4x – 4xy – 8y

Question 44: Expand (x/y + y/z + z/x)2 with the help of suitable identity.

Solution:

(x/y + y/z + z/x)2

Using the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

= (x/y)2 + (y/z)2 + (z/x)2 + 2(x/y)(y/z) + 2(y/z)(z/x) + 2(z/x)(x/y)

= (x2/y2) + (y2/z2) + (z2/x2) + (2x)/z + (2y)/x + (2z)/y

Question 45: Factorise: x2 + 2y2 + 8z2 + 2√2 xy – 8yz – 4√2 xz

Solution:

x2 + 2y2 + 8z2 + 2√2 xy – 8yz – 4√2 xz

= x2 + (√2y)2 + (-2√2z)2 + 2(x)(√2y) + 2(√2y)(-2√2z) + 2(-2√2z)(x)

Using the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

= (x + √2y – 2√2z)2

= (x + √2y – 2√2z)(x + √2y – 2√2z)

Question 46: Expand: (1 – 2x)3

Solution:

(1 – 2x)3

Using the identity (a – b)3 = a3 + b3 – 3ab(a – b)

= (1)3 – (2x)3 – 3(1)(2x)(1 – 2x)

= 1 – 8x3 – 6x(1 – 2x)

= 1 – 8x3 – 6x + 12x2

Question 47: Expand: (x + 2y/3)3

Solution:

(x + 2y/3)3

Using the identity (a + b)3 = a3 + b3 + 3ab(a + b)

= x3 + (2y/3)3 + 3(x)(2y/3)(x + 2y/3)

= x3 + (8y3/27) + 2xy(x + 2y/3)

= x3 + (8y3/27) + 2x2y + (4xy2/3)

Question 48: Evaluate the following using suitable identity:

(103)3

Solution:

(103)3 = (100 + 3)3

Using the identity (a + b)3 = a3 + b3 + 3ab(a + b)

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

= 1000000 + 27 + 900(100 + 3)

= 1000000 + 27 + 90000 + 2700

= 1092727

Question 48: Evaluate the following using suitable identity:

(999)3

Solution:

(999)3 = (1000 – 1)3

Using the identity (a – b)3 = a3 + b3 – 3ab(a – b)

= (1000)3 – (1)3 – 3(1000)(1)(1000 – 1)

= 1000000000 – 1 – 3000(1000 – 1)

= 1000000000 – 1 – 3000000 + 3000

= 997002999

Question 49: Factorise: 125x3 – 64y3 – 300x2y + 240xy2

Solution:

125x3 – 64y3 – 300x2y + 240xy2

= (5x)3 – (4y)3 – 3(5x)2(4y) + 3(5x)(4y)2

= (5x)3 – (4y)3 – 3(5x)(4y)(5x – 4y)

Using the identity (a – b)3 = a3 + b3 – 3ab(a – b)

= (5x – 4y)3

= (5x – 4y)(5x – 4y)(5x – 4y)

Question 50: Verify that 27a3 + b3 + c3 – 9abc = (3a + b + c)[9a2 + b2 + c2 – 3ab – bc – 3ac]

Solution:

LHS = 27a3 + b3 + c3 – 9abc

= (3a)3 + b3 + c3 – 3(3a)bc

Using the identity x3 + y3 + z3 -3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

= (3a + b + c)[(3a)2 + b2 + c2 – (3a)(b) – bc – c(3a)]

= (3a + b + c)[9a2 + b2 + c2 – 3ab – bc – 3ac]

Question 51: Use the suitable identity compute: (30)3 + (20)3 + (-50)3

Solution:

(30)3 + (20)3 + (-50)3

Let x = 30, y = 20, z = -50

x + y + z = 30 + 20 – 50 = 0

Use the identity: if x + y + z = 0, then x3 + y3 + z3 = 3xyz

(30)3 + (20)3 + (-50)3 = 3(30)(20)(-50)

= -90000

RBSE Maths Chapter 3: Additional Important Questions and Solutions

Question 1: Expression 12 + 3x + 5x2 write the coefficient of x2.

Solution:

Given polynomial is 12 + 3x + 5x2

Coefficient of x2 = 5

Question 2: Using the suitable identities, factorise the following:

(i) 27a3 + 64b3

(ii) x2 + 6xy + 9y2

(iii) 9x2 – 4y2 + 16z2 – 12xy – 16yz + 24xy

Solution:

(i) 27a3 + 64b3

= (3a)3 + (4b)3

Using the identity a3 + b3 = (a + b)(a2 – ab + b2)

= (3a + 4b)[(3a)2 – (3a)(4b) + (4b)2]

= (3a + 4b)(9a2 – 12ab + 16b2)

(ii) x2 + 6xy + 9y2

= x2 + 2(x)(3y) + (3y)2

Using the identity (a + b)2 = a2 + 2ab + b2

= (x + 3y)2

= (x + 3y)(x + 3y)

(iii) 9x2 – 4y2 + 16z2 – 12xy – 16yz + 24xy

= (3x)2 + (-2y)2 + (4z)2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x)

Using the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

= (30 – 2y + 4z)2

Question 3: Write down the coefficient of x2 in 7 – 11x + x3

Solution:

Given polynomial is 7 – 11x + x3

This can be written as 7 – 11x + x3 + 0x2

Therefore, the coefficient of x2 = 0

Question 4: Find the zero(s) of given polynomial p(x) = 3x + 2

Solution:

Given polynomial is:

p(x) = 3x + 2

Comparing with the standard form p(x) = ax + b

a = 3, b = 2

x = -b/a

= -⅔

Therefore, zero of the given polynomial is -⅔.

Question 5: Factorise 125x3 – 8y3

Solution:

125x3 – 8y3

= (5x)3 – (2y)3

Using the formula a3 – b3 = (a – b)(a2 + ab + b2)

= (5x – 2y)[(5x)2 + (5x)(2y) + (2y)2]

= (5x – 2y)(25x2 + 10xy + 4y2)

Question 6: Factorise 12x2 – 7x + 1

Solution:

Given polynomial is 12x2 – 7x + 1

Product of 12 and 1 = 12

Factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12

Sum of the factors should be -7.

So we are considering the negative numbers.

Let us take -4 and -3 as factors for 12.

12x2 – 7x + 1 = 12x2 – 4x – 3x + 1

= 4x(3x – 1) – 1(3x – 1)

= (4x – 1)(3x – 1)

Therefore, (4x – 1) and (3x – 1) are the factors of 12x2 – 7x + 1.

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