NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

 

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials - A polynomial is an algebraic expression which consists of variables and coefficient with operations such as additions, subtraction, multiplication, and non-negative exponents. In this particular chapter, you will learn the operations of two or more polynomials. NCERT solutions for class 9 maths chapter 2 Polynomials are there to help you while solving the problems related to this particular chapter. NCERT class 9 Polynomials, introduces a lot of important concepts that will be helpful for those students who are targeting exams like JEE, CAT, SSC, etc. It is an important topic in maths that comes under the algebra unit which holds 20 marks in the CBSE class 9 maths final exams. In this particular chapter, you will study the definition of a polynomial, zeroes, coefficient, degrees, and terms of a polynomial, type of a polynomial. You will also study the remainder and factor theorems and the factorization of polynomials. In Polynomials, there are a total of 5 exercises that comprise of a total of 33 questions. Solutions of NCERT class 9 maths chapter 2 Polynomials will cover the detailed solution to each and every question present in the practice exercises including optional exercises. CBSE NCERT solutions for class 9 maths chapter 2 Polynomials can also be used while doing homework. It can be a good tool for the class 9 students as it is designed in such a manner so that a student can fetch the maximum marks available for the particular question. NCERT solutions are also available for other classes and subjects which can be downloaded by clicking on the link given.

NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.1

Q1 (i) Is the following expression polynomial in one variable? State reasons for your answer. 4x^2 - 3x + 7

Answer:

YES 
Given polynomial 4x^2 - 3x + 7 has only one variable which is x

Q1 (ii) Is the following expression polynomial in one variable? State reasons for your answer. y^2 + \sqrt2

Answer:

YES
Given polynomial has only one variable which is y

Q1 (iii) Is the following expression polynomial in one variable? State reasons for your answer. 3\sqrt t + t\sqrt2

Answer:

NO
Because we can observe that the exponent of variable t in term 3\sqrt t is  \frac{1}{2}  which is not a whole number.
Therefore this expression is not a polynomial.

Q1 (iv) Is the following expression polynomial in one variable? State reasons for your answer. y + \frac{2}{y}

Answer:

NO
Because we can observe that the exponent of variable y in term \frac{2}{y} is -1 which is not a whole number. Therefore this expression is not a polynomial.

Q1 (v) Is the following expression polynomial in one variable? State reasons for your answer. x^{10} + y^3 + t^{50}

Answer:

NO
Because in the given polynomial x^{10} + y^3 + t^{50}  there are 3 variables which are x, y, t. That's why this is polynomial in three variable not in one variable.

Q2 (i) Write the coefficients of x^2 in the following: 2 + x^2 +x

Answer:

Coefficient of x^2 in polynomial  2 + x^2 +x  is  1.

Q2 (ii) Write the coefficients of x^2 in the following: 2 - x^2 + x^3

Answer:

Coefficient of x^2 in  polynomial  2 - x^2 + x^3  is  -1.

Q2 (iii) Write the coefficients of x^2 in the following: \frac{\pi}{2}x^2 + x

Answer:

Coefficient of  x^2  in polynomial  \frac{\pi}{2}x^2 + x  is  \frac{\pi}{2}

Q2 (iv) Write the coefficients of x^2 in the following: \sqrt2 x - 1

Answer:

Coefficient of  x^2  in polynomial  \sqrt2 x - 1   is  0

Q3 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.
In binomial, there are two terms 
Therefore,  binomial of degree 35 is
Eg:-  x^{35}+1
In monomial, there is only one term in it.
Therefore, monomial of degree 100 can be written as y^{100}

Q4 (i) Write the degree the following polynomial: 5x^3 + 4x^2 + 7x

Answer:

Degree of a polynomial is the highest power of the variable in the polynomial.
Therefore, the degree of polynomial  5x^3 + 4x^2 + 7x  is  3.

Q4 (ii) Write the degree the following polynomial: 4 - y^2

Answer:

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

Therefore, the degree of polynomial  4 - y^2  is  2.

Q4 (iii) Write the degree the following polynomial: 5t - \sqrt7

Answer:

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

Therefore, the degree of polynomial  5t - \sqrt7  is  1

Q4 (iv) Write the degree the following polynomial: 3

Answer:

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

In this case, only a constant value 3 is there and the degree of a constant polynomial is always 0.

Q5 (i) Classify the following as linear, quadratic and cubic polynomial: x^2+x

Answer:

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

Given polynomial is  x^2+x  with degree 2

Therefore, it is a quadratic polynomial.

Q5 (ii) Classify the following as linear, quadratic and cubic polynomial: x - x^3

Answer:

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

Given polynomial is  x - x^3  with degree 3

Therefore, it is a cubic polynomial

Q5 (iii) Classify the following as linear, quadratic and cubic polynomial: y + y^2 + 4

Answer:

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

Given polynomial is  y + y^2 + 4  with degree 2

Therefore, it is quadratic polynomial.

Q5 (iv) Classify the following as linear, quadratic and cubic polynomial: 1 +x

Answer:

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

Given polynomial is  1 +x  with degree 1

Therefore, it is linear polynomial

Q5 (v) Classify the following as linear, quadratic and cubic polynomial: 3t

Answer:

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

Given polynomial is  3t  with degree 1

Therefore, it is linear polynomial

Q5 (vi) Classify the following as linear, quadratic and cubic polynomial: r^2

Answer:

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

Given polynomial is  r^2  with degree 2

Therefore, it is quadratic polynomial

Q5 (vii) Classify the following as linear, quadratic and cubic polynomial: 7x^3

Answer:

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

Given polynomial is  7x^3  with degree 3

Therefore, it is a cubic polynomial

NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.2

Q1 (i) Find the value of the polynomial 5x - 4x^2 +3 at x = 0

Answer:

Given polynomial is  5x - 4x^2 +3 

Now, at x = 0  value is 

\Rightarrow 5(0)-4(0)^2+3 = 0 - 0 + 3 = 3

Therefore, value of polynomial 5x - 4x^2 +3 at x = 0 is  3

Q1 (ii) Find the value of the polynomial 5x - 4x^2 +3 at x = -1

Answer:

Given polynomial is  5x - 4x^2 +3 

Now, at x = -1  value is 

\Rightarrow 5(-1)-4(-1)^2+3 = -5 - 4 + 3 = -6

Therefore, value of polynomial 5x - 4x^2 +3 at x = -1 is  -6

Q1 (iii) Find the value of the polynomial 5x - 4x^2 +3 at x = 2

Answer:

Given polynomial is  5x - 4x^2 +3 

Now, at x = 2  value is 

\Rightarrow 5(2)-4(2)^2+3 = 10 - 16 + 3 = -3

Therefore, value of polynomial 5x - 4x^2 +3 at x = 2 is  -3

Q2 (i) Find p(0), p(1) and p(2) for each of the following polynomials: p(y)= y^2 - y + 1

Answer:

Given polynomial is

p(y)= y^2 - y + 1

Now, 

p(0)= (0)^2 - 0 + 1= 1

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

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

Therefore, valures of p(0), p(1) and p(2) are 1 , 1 and 3 respectively.

Q2 (ii) Find p(0), p(1) and p(2) for each of the following polynomials: p(t) = 2 + t + 2t^2 - t^3

Answer:

Given polynomial is

p(t) = 2 + t + 2t^2 - t^3

Now, 

p(0) = 2 + 0 + 2(0)^2 - (0)^3 = 2

p(1) = 2 + 1 + 2(1)^2 - (1)^3 = 4

p(2) = 2 + 2 + 2(2)^2 - (2)^3 = 4

Therefore, values of p(0), p(1) and p(2) are 2 , 4 and 4 respectively

Q2 (iii) Find p(0), p(1) and p(2) for each of the following polynomials: p(x) = x^3

Answer:

Given polynomial is

p(x) = x^3

Now, 

p(0) = (0)^3 =0

p(1) = (1)^3=1

p(2) = (2)^3=8

Therefore, values of p(0), p(1) and p(2) are 0 , 1 and 8 respectively

Q2 (iv) Find p(0), p(1) and p(2) for each of the following polynomials: p(x)= (x-1)(x+ 1)

Answer:

Given polynomial is

p(x)= (x-1)(x+ 1) = x^2-1

Now, 

p(0) = (0)^2-1 = -1

p(1) = (1)^2-1 = 0

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

Therefore, values of p(0), p(1) and p(2) are -1 , 0 and 3 respectively

Q3 (i) Verify whether the following are zeroes of the polynomial, indicated against it. p(x) = 3x + 1, x = -\frac{1}{3}

Answer:

Given polynomial is   p(x) = 3x + 1

Now, at  x = -\frac{1}{3}   it's value is 

p\left ( -\frac{1}{3} \right )=3\times \left ( -\frac{1}{3} \right )+1 = -1+1=0

Therefore, yes x = -\frac{1}{3}  is a zero of  polynomial  p(x) = 3x + 1

Q3 (ii) Verify whether the following are zeroes of the polynomial, indicated against it. p(x) = 5x - \pi, x = \frac{4}{5}

Answer:

Given polynomial is   p(x) = 5x - \pi

Now, at  x =\frac{4}{5}   it's value is 

p\left ( \frac{4}{5} \right )=5\times \left ( \frac{4}{5} \right ) -\pi = 4-\pi \neq 0
Therefore, no  x =\frac{4}{5}   is not a zero of  polynomial  p(x) = 5x - \pi

Q3 (iii) Verify whether the following are zeroes of the polynomial, indicated against it. p(x) = x^2 -1, x = 1,-1

Answer:

Given polynomial is   p(x) = x^2-1

Now, at  x = 1   it's value is 

p(1) = (1)^2-1 = 1 -1 =0

And at x = -1

p(-1) = (-1)^2-1 = 1 -1 =0
Therefore, yes  x = 1 , -1   are zeros of  polynomial  p(x) = x^2-1

Q3 (iv) Verify whether the following are zeroes of the polynomial, indicated against it. p(x) = (x + 1)(x-2), x = -1,2

Answer:

Given polynomial is   p(x) = (x+1)(x-2)

Now, at  x = 2   it's value is 

p(2) = (2+1)(2-2) = 0

And at x = -1

p(-1) = (-1+1)(-1-2) = 0

Therefore, yes  x = 2 , -1   are zeros of  polynomial  p(x) = (x+1)(x-2)

Q3 (v) Verify whether the following are zeroes of the polynomial, indicated against it. p(x) = x^2. x =0

Answer:

Given polynomial is   p(x) = x^2

Now, at  x = 0   it's value is 

p(0) = (0)^2=0

Therefore, yes  x = 0  is a zeros of  polynomial  p(x) = (x+1)(x-2)

Q3 (vi) Verify whether the following are zeroes of the polynomial, indicated against it. p(x) = lx + m, \ x =- \frac{m}{l}

Answer:

Given polynomial is   p(x) = lx+m

Now, at  x = -\frac{m}{l}   it's value is 

p\left ( -\frac{m}{l} \right )= l \times \left ( -\frac{m}{l} \right )+m = -m+m =0


Therefore, yes  x = -\frac{m}{l}   is a zeros of  polynomial  p(x) = lx+m

Q3 (vii) Verify whether the following are zeroes of the polynomial, indicated against it. p(x) = 3x^2 - 1, \ x = -\frac{1}{\sqrt3}, \frac{2}{\sqrt3}

Answer:

Given polynomial is   p(x) = 3x^2-1

Now, at  x = -\frac{1}{\sqrt3}   it's value is 

p\left ( -\frac{1}{\sqrt3} \right )= 3 \times \left ( -\frac{1}{\sqrt3} \right )^2-1 = 1-1 =0

And at  x = \frac{2}{\sqrt3}

p\left ( \frac{2}{\sqrt3} \right )= 3 \times \left ( \frac{2}{\sqrt3} \right )^2-1 = 4-1 =3\neq 0

Therefore,   x = -\frac{1}{\sqrt3}  is a zeros of polynomialp(x) = 3x^2-1

whereas x = \frac{2}{\sqrt3}  is not a zeros of  polynomial p(x) = 3x^2-1

Q3 (viii) Verify whether the following are zeroes of the polynomial, indicated against it. p(x) = 2x +1,\ x = \frac{1}{2}

Answer:

Given polynomial is   p(x) = 2x+1

Now, at  x = \frac{1}{2}   it's value is 

p\left ( \frac{1}{2} \right )= 2 \times \left ( \frac{1}{2} \right )+1 = 1+1=2 \neq 0

Therefore,   x = \frac{1}{2}  is not a zeros of  polynomial  p(x) = 2x+1

Q4 (i) Find the zero of the polynomial in each of the following cases: p(x)= x + 5

Answer:

Given polynomial is  p(x)= x + 5

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

Now,

p(x)=0

\Rightarrow x+5 = 0

\Rightarrow x=-5

Therefore, x = -5 is the zero of polynomial  p(x)= x + 5

Q4 (ii) Find the zero of the polynomial in each of the following cases: p(x) = x - 5

Answer:

Given polynomial is  p(x)= x - 5

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

Now,

p(x)=0

\Rightarrow x-5 = 0

\Rightarrow x=5

Therefore, x = 5 is a zero of polynomial  p(x)= x - 5

Q4 (iii) Find the zero of the polynomial in each of the following cases: p(x)= 2x + 5

Answer:

Given polynomial is  p(x)= 2x + 5

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

Now,

p(x)=0

\Rightarrow 2x+5 = 0

\Rightarrow x=-\frac{5}{2}

Therefore, x=-\frac{5}{2} is a zero of polynomial  p(x)= 2x + 5

Q4 (iv) Find the zero of the polynomial in each of the following cases: p(x) = 3x - 2

Answer:

Given polynomial is  p(x) = 3x - 2

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

Now,

p(x)=0

\Rightarrow 3x-2 = 0

\Rightarrow x=\frac{2}{3}
Therefore, x=\frac{2}{3}  is a zero of polynomial  p(x) = 3x - 2

Q4 (v) Find the zero of the polynomial in each of the following cases: p(x) = 3x

Answer:

Given polynomial is  p(x) = 3x

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

Now,

p(x)=0

\Rightarrow 3x = 0

\Rightarrow x=0

Therefore, x=0  is a zero of polynomial  p(x) = 3x

Q4 (vi) Find the zero of the polynomial in each of the following cases: p(x) = ax, \ a\neq 0

Answer:

Given polynomial is  p(x) = ax

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

Now,

p(x)=0

\Rightarrow ax = 0

\Rightarrow x=0

Therefore, x=0  is a zero of polynomial  p(x) = ax

Q4 (vii) Find the zero of the polynomial in each of the following cases: p(x) = cx + d, c\neq 0, c,d are real numbers

Answer:

Given polynomial is  p(x) = cx+d

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

Now,

p(x)=0

\Rightarrow cx+d = 0

\Rightarrow x=-\frac{d}{c}

Therefore,  x=-\frac{d}{c}  is a zero of polynomial  p(x) = cx+d

NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.3

Q1 (i) Find the remainder when x^3 + 3x^2 +3x + 1 is divided by x + 1

Answer:

When we divide  x^3 + 3x^2 +3x + 1  by  x + 1

By long division method, we will get 



Therefore, remainder is  0.

Q1 (ii) Find the remainder when x^3 + 3x^2 +3x + 1 is divided by x - \frac{1}{2}

Answer:

When we divide  x^3 + 3x^2 +3x + 1  by  x - \frac{1}{2}

By long division method, we will get 


Therefore, the remainder is  \frac{27}{8} 

Q1 (iii) Find the remainder when x^3 + 3x^2 +3x + 1 is divided by x

Answer:

When we divide  x^3 + 3x^2 +3x + 1  by  x

By long division method, we will get 



Therefore, remainder is  1.

Q1 (iv) Find the remainder when x^3 + 3x^2 +3x + 1 is divided by x + \pi

Answer:

When we divide  x^3 + 3x^2 +3x + 1  by  x + \pi

By long division method, we will get 



Therefore, the remainder is  1-3\pi + 3\pi^2-\pi^3 

Q1 (v) Find the remainder whenx^ 3 + 3x^ 2 + 3x + 1 is divided by 5+2x

Answer:

When we divide  x^3 + 3x^2 +3x + 1  by  5+2x

By long division method, we will get 



Therefore, the remainder is  -\frac{27}{8} 

 

Q2 Find the remainder when x^3 - ax^2 + 6x - a is divided by x - a.

Answer:

When we divide  x^3 - ax^2 + 6x - a  by  x - a

By long division method, we will get 



Therefore, remainder is  5a 

Q3 Check whether 7 + 3x is a factor of 3x^3 + 7x.

Answer:

When we divide  3x^3 + 7x  by  7 + 3x

We can also write  3x^3 + 7x  as  3x^3 +0x^2+ 7x

By long division method, we will get 



 Since, remainder is not equal to  0 

Therefore,  7 + 3x  is not a factore of  3x^3 + 7x

NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.4

Q1 (i) Determine which of the following polynomials has (x + 1) a factor : x^3 + x^2 +x + 1

Answer:

Zero of polynomial   (x + 1)  is -1.

If  (x + 1)  is a factor of polynomial    p(x)=x^3 + x^2 +x + 1  

Then, p(-1)  must be equal to zero 

Now,

\Rightarrow p(-1)=(-1)^3+(-1)^2-1+1

\Rightarrow p(-1)=-1+1-1+1 = 0

Therefore,  (x + 1)  is a factor of polynomial    p(x)=x^3 + x^2 +x + 1 

Q1 (ii) Determine which of the following polynomials has (x + 1) a factor : x^4 + x^3 + x^2 +x + 1

Answer:

Zero of polynomial   (x + 1)  is -1.

If  (x + 1)  is a factor of polynomial    p(x)=x^4 + x^3 + x^2 +x + 1  

Then, p(-1)  must be equal to zero 

Now,

\Rightarrow p(-1)=(-1)^4+(-1)^3+(-1)^2-1+1

\Rightarrow p(-1)=1-1+1-1+1 = 1\neq 0

Therefore,  (x + 1)  is not a factor of polynomial   p(x)=x^4 + x^3 + x^2 +x + 1

Q1 (iii) Determine which of the following polynomials has (x + 1) a factor : x^4 + 3x^3 + 3x^2 +x + 1

Answer:

Zero of polynomial   (x + 1)  is -1.

If  (x + 1)  is a factor of polynomial    p(x)=x^4 + 3x^3 + 3x^2 +x + 1  

Then, p(-1)  must be equal to zero 

Now,

\Rightarrow p(-1)=(-1)^4+3(-1)^3+3(-1)^2-1+1

\Rightarrow p(-1)=1-3+3-1+1 = 1\neq 0

Therefore,  (x + 1)  is not a factor of polynomial   p(x)=x^4 + 3x^3 + 3x^2 +x + 1

Q1 (iv) Determine which of the following polynomials has (x + 1) a factor : x^3 - x^2 -(2 + \sqrt2)x + \sqrt2

Answer:

Zero of polynomial   (x + 1)  is -1.

If  (x + 1)  is a factor of polynomial    p(x)=x^3 - x^2 -(2 + \sqrt2)x + \sqrt2  

Then, p(-1)  must be equal to zero 

Now,

\Rightarrow p(-1)=(-1)^3-(-1)^2-(2+\sqrt2)(-1)+\sqrt2

\Rightarrow p(-1)=-1-1+2+\sqrt2+\sqrt2 = 2\sqrt2\neq 0

Therefore,  (x + 1)  is not a factor of polynomial p(x)=x^3 - x^2 -(2 + \sqrt2)x + \sqrt2

Q2 (i) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:p(x) = 2x^3 + x^2 - 2x - 1,\ g(x) = x + 1

Answer:

Zero of polynomial  g(x)=x+1  is  -1

If  g(x)=x+1  is factor of polynomial  p(x) = 2x^3 + x^2 - 2x - 1 

Then,  p(-1)  must be equal to zero 

Now,

\Rightarrow p(-1)= 2(-1)^3+(-1)^2-2(-1)-1

\Rightarrow p(-1)= -2+1+2-1 = 0

Therefore,  g(x)=x+1   is factor of polynomial  p(x) = 2x^3 + x^2 - 2x - 1

Q2 (ii) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:p(x) = x^3 + 3x^2 + 3x + 1, \ g(x) = x + 2

Answer:

Zero of polynomial  g(x)=x+2  is  -2

If  g(x)=x+2  is factor of polynomial  p(x) = x^3 + 3x^2 + 3x + 1 

Then,  p(-2)  must be equal to zero 

Now,

\Rightarrow p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1

\Rightarrow p(-2)= -8+12-6+1 = -1\neq 0

Therefore,  g(x)=x+2   is not a factor of polynomial  p(x) = x^3 + 3x^2 + 3x + 1

Q2 (iii) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:p(x) = x^3 - 4x^2 + x + 6, \ g(x) = x - 3

Answer:

Zero of polynomial  g(x)=x-3  is  3

If  g(x)=x-3  is factor of polynomial  p(x) = x^3 - 4x^2 + x + 6 

Then,  p(3)  must be equal to zero 

Now,

\Rightarrow p(3) = (3)^3 - 4(3)^2 + 3 + 6

\Rightarrow p(3) = 27-36+3+6=0

Therefore,  g(x)=x-3   is a factor of polynomial  p(x) = x^3 - 4x^2 + x + 6

Q3 (i) Find the value of k, if x - 1 is a factor of p(x) in the following case:p(x) = x^2 + x + k

Answer:

Zero of polynomial   x - 1 is  1

If  x - 1  is factor of polynomial   p(x) = x^2 + x + k

Then,  p(1)  must be equal to zero 

Now,

\Rightarrow p(1) = (1)^2 + 1 + k

\Rightarrow p(1) =0

\Rightarrow 2+k = 0

\Rightarrow k = -2

Therefore,  value of k is -2

Q3 (ii) Find the value of k, if x - 1 is a factor of p(x) in the following case: p(x) = 2x^2 + kx + \sqrt2

Answer:

Zero of polynomial   x - 1 is  1

If  x - 1  is factor of polynomial   p(x) = 2x^2 + kx + \sqrt2

Then,  p(1)  must be equal to zero 

Now,

\Rightarrow p(1) = 2(1)^2 + k(1) + \sqrt2

\Rightarrow p(1) =0

\Rightarrow 2+k +\sqrt2= 0

\Rightarrow k = -(2+\sqrt2)

Therefore,  value of k is -(2+\sqrt2)

Q3 (iii) Find the value of k, if x - 1 is a factor of p(x) in the following case: p(x) = kx^2-\sqrt2 x + 1

Answer:

Zero of polynomial   x - 1 is  1

If  x - 1  is factor of polynomial   p(x) = kx^2-\sqrt2 x + 1

Then,  p(1)  must be equal to zero 

Now,

\Rightarrow p(1) = k(1)^2 -\sqrt2(1) + 1

\Rightarrow p(1) =0

\Rightarrow k -\sqrt2 +1= 0

\Rightarrow k = -1+\sqrt2

Therefore,  value of k is -1+\sqrt2

Q3 (iv) the value of k, if x - 1 is a factor of p(x) in the following case: p(x) = kx^2 -3 x + k

Answer:

Zero of polynomial   x - 1 is  1

If  x - 1  is factor of polynomial   p(x) = kx^2 -3 x + k

Then,  p(1)  must be equal to zero 

Now,

\Rightarrow p(1) = k(1)^2 -3(1) + k

\Rightarrow p(1) =0

\Rightarrow k -3+k= 0

\Rightarrow k = \frac{3}{2}
Therefore,  value of k is  \frac{3}{2}

Q4 (i) Factorise : 12x^2 - 7x + 1

Answer:

Given polynomial is   12x^2 - 7x + 1

We need to factorise the middle term into two terms such that their product is equal to 12 \times 1 = 12 and their sum is equal to -7

We can solve it as 

\Rightarrow 12x^2 - 7x + 1

\Rightarrow 12x^2-3x-4x+1                                                       (\because -3 \times -4 = 12 \ \ and \ \ -3+(-4) = -7)

\Rightarrow 3x(4x-1)-1(4x-1)

\Rightarrow (3x-1)(4x-1)                             

Q4 (ii) Factorise : 2x^2 + 7x + 3

Answer:

Given polynomial is   2x^2 + 7x + 3

We need to factorise the middle term into two terms such that their product is equal to 2 \times 3 = 6 and their sum is equal to 7

We can solve it as 

\Rightarrow 12x^2 - 7x + 1

\Rightarrow 2x^2+6x+x+3                                                       (\because 6 \times 1 = 6 \ \ and \ \ 6+1 = 7)

\Rightarrow 2x(x+3)+1(x+3)

\Rightarrow (2x+1)(x+3)                             

Q4 (iii) Factorise : 6x^2 +5x - 6

Answer:

Given polynomial is   6x^2 +5x - 6

We need to factorise the middle term into two terms such that their product is equal to 6 \times -6 =-3 6 and their sum is equal to 5

We can solve it as 

\Rightarrow 6x^2 +5x - 6

\Rightarrow 6x^2 +9x -4x - 6                                       (\because 9 \times -4 = -36 \ \ and \ \ 9+(-4) = 5)

\Rightarrow 3x(2x+3)-2(2x+3)

\Rightarrow (2x+3)(3x-2)                             

Q4 (iv) Factorise : 3x^2 - x - 4

Answer:

Given polynomial is   3x^2 - x - 4

We need to factorise the middle term into two terms such that their product is equal to 3 \times -4 =-12 and their sum is equal to -1

We can solve it as 

\Rightarrow 3x^2 - x - 4

\Rightarrow 3x^2 -4 x+3x - 4                                      (\because 3 \times -4 = -12 \ \ and \ \ 3+(-4) = -1)

\Rightarrow x(3x-4)+1(3x-4)

\Rightarrow (x+1)(3x-4)                             

Q5 (i) Factorise : x^3 - 2x^2 - x +2

Answer:

Given polynomial is  x^3 - 2x^2 - x +2

Now, by hit and trial method we observed that  (x+1)  is one of the factors of the given polynomial.

By long division method, we will get


We know that  Dividend = (Divisor \times Quotient) + Remainder

 x^3 - 2x^2 - x +2 = (x+1)(x^2-3x+2)+0

                                       = (x+1)(x^2-2x-x+2)
             
                                       = (x+1)(x-2)(x-1)

Therefore, on factorization of  x^3 - 2x^2 - x +2 we will get  (x+1)(x-2)(x-1)

Q5 (ii) Factorise : x^3 - 3x^2 -9x -5

Answer:

Given polynomial is  x^3 - 3x^2 -9x -5

Now, by hit and trial method we observed that  (x+1)  is one of the factore of given polynomial.

By long division method, we will get


We know that  Dividend = (Divisor \times Quotient) + Remainder

 x^3 - 3x^2 -9x -5=(x+1)(x^2-4x-5)

                                       = (x+1)(x^2-5x+x-5)
             
                                       = (x+1)(x-5)(x+1)

Therefore, on factorization of  x^3 - 3x^2 -9x -5 we will get  (x+1)(x-5)(x+1)

Q5 (iii) Factorise : x^3 + 13x^2 + 32x + 20

Answer:

Given polynomial is  x^3 + 13x^2 + 32x + 20

Now, by hit and trial method we observed that  (x+1)  is one of the factore of given polynomial.

By long division method, we will get


We know that  Dividend = (Divisor \times Quotient) + Remainder

x^3 + 13x^2 + 32x + 20=(x+1)(x^2+12x+20)

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

Therefore, on factorization of  x^3 + 13x^2 + 32x + 20 we will get  (x+1)(x+10)(x+2)

Q5 (iv) Factorise : 2y^3 + y^2 - 2y - 1

Answer:

Given polynomial is  2y^3 + y^2 - 2y - 1

Now, by hit and trial method we observed that  (y-1)  is one of the factors of the given polynomial.

By long division method, we will get


We know that  Dividend = (Divisor \times Quotient) + Remainder

2y^3 + y^2 - 2y - 1= (y-1)(2y^2+3y+1)

                                       = (y-1)(2y^2+2y+y+1)
             
                                       = (y-1)(2y+1)(y+1)

Therefore, on factorization of  2y^3 + y^2 - 2y - 1 we will get  (y-1)(2y+1)(y+1)

NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.5

Q1 (i) Use suitable identities to find the following product: (x + 4) ( x + 10)

Answer:

We will use identity 

(x+a)(x+b)=x^2+(a+b)x+ab

Put   a = 4 \ \ and \ \ b = 10

(x+4)(x+10)= x^2+(10+4)x+10\times 4
     
                                  = x^2+14x+40

Therefore, (x + 4) ( x + 10)  is equal to  x^2+14x+40

Q1 (ii) Use suitable identities to find the following product: (x+8)(x-10)

Answer:

We will use identity 

(x+a)(x+b)=x^2+(a+b)x+ab

Put   a = 8 \ \ and \ \ b = -10

(x+8)(x-10)= x^2+(-10+8)x+8\times (-10)
     
                                  = x^2-2x-80

Therefore, (x+8)(x-10)  is equal to  x^2-2x-80

Q1 (iii) Use suitable identities to find the following product: (3x+4)(3x - 5)

Answer:

We can write  (3x+4)(3x - 5)  as 

(3x+4)(3x - 5)= 9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )

We will use identity 

(x+a)(x+b)=x^2+(a+b)x+ab

Put   a = \frac{4}{3} \ \ and \ \ b = -\frac{5}{3}

9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )= 9\left ( x^2+\left ( \frac{4}{3}-\frac{5}{3} \right )x+\frac{4}{3} \times \left ( -\frac{5}{3} \right ) \right )
     
                                              =9x^2-3x-20

Therefore, (3x+4)(3x - 5)  is equal to  9x^2-3x-20

Q1 (iv) Use suitable identities to find the following product: (y^2 + \frac{3}{2})(y^2 - \frac{3}{2})

Answer:

We will use identity 

(x+a)(x-a)=x^2-a^2

Put   x=y^2 \ \ and \ \ a = \frac{3}{2}

(y^2 + \frac{3}{2})(y^2 - \frac{3}{2}) = \left ( y^2 \right )^2-\left(\frac{3}{2} \right )^2
     
                                     = y^4-\frac{9}{4}

Therefore, (y^2 + \frac{3}{2})(y^2 - \frac{3}{2})  is equal to  y^4-\frac{9}{4}

Q1 (v) Use suitable identities to find the following product: (3 - 2x) (3 + 2x)

Answer:

We can write (3 - 2x) (3 + 2x)  as

(3 - 2x) (3 + 2x)=-4\left ( x-\frac{3}{2} \right )\left(x+\frac{3}{2} \right )

We will use identity 

(x+a)(x-a)=x^2-a^2

Put   a = \frac{3}{2}

-4(x + \frac{3}{2})(x- \frac{3}{2}) =-4\left ( \left ( x \right )^2-\left(\frac{3}{2} \right )^2 \right )
     
                                        =9-4x^2

Therefore, (3 - 2x) (3 + 2x)  is equal to  9-4x^2

Q2 (i) Evaluate the following product without multiplying directly: 103 \times 107

Answer:

We can rewrite  103 \times 107  as 

\Rightarrow 103 \times 107= (100+3)\times (100+7)

We will use identity

(x+a)(x+b)=x^2+(a+b)x+ab

Put  x =100 , a=3 \ \ and \ \ b = 7

(100+3)\times (100+7)= (100)^2+(3+7)100+3\times 7

                                               =10000+1000+21= 11021

Therefore, value of 103 \times 107  is  11021

Q2 (ii) Evaluate the following product without multiplying directly: 95 \times 96

Answer:

We can rewrite  95 \times 96  as 

\Rightarrow 95 \times 96= (100-5)\times (100-4)

We will use identity

(x+a)(x+b)=x^2+(a+b)x+ab

Put  x =100 , a=-5 \ \ and \ \ b = -4

(100-5)\times (100-4)= (100)^2+(-5-4)100+(-5)\times (-4)

                                               =10000-900+20= 9120

Therefore, value of  95 \times 96  is  9120

Q2 (iii) Evaluate the following product without multiplying directly: 104 \times 96

Answer:

We can rewrite  104 \times 96  as 

\Rightarrow 104 \times 96= (100+4)\times (100-4)

We will use identity

(x+a)(x-a)=x^2-a^2

Put  x =100 \ \ and \ \ a=4

(100+4)\times (100-4)= (100)^2-(4)^2

                                               =10000-16= 9984

Therefore, value of  104 \times 96  is  9984

Q3 (i) Factorise the following using appropriate identities: 9x^2 + 6xy + y^2

Answer:

We can rewrite 9x^2 + 6xy + y^2 as 

\Rightarrow 9x^2 + 6xy + y^2 = (3x)^2+2\times 3x\times y +(y)^2

Using identity  \Rightarrow (a+b)^2 = (a)^2+2\times a\times b +(b)^2

Here, a= 3x \ \ and \ \ b = y

Therefore,

9x^2+6xy+y^2 = (3x+y)^2 = (3x+y)(3x+y)

Q3 (ii) Factorise the following using appropriate identities: 4y^2 - 4y + 1

Answer:

We can rewrite 4y^2 - 4y + 1 as 

\Rightarrow 4y^2 - 4y + 1=(2y)^2-2\times2y\times 1+(1)^2

Using identity  \Rightarrow (a-b)^2 = (a)^2-2\times a\times b +(b)^2

Here, a= 2y \ \ and \ \ b = 1

Therefore,

4y^2 - 4y + 1=(2y-1)^2=(2y-1)(2y-1)

Q3 (iii) Factorise the following using appropriate identities: x^2 - \frac{y^2}{100}

Answer:

We can rewrite   x^2 - \frac{y^2}{100}  as 

\Rightarrow x^2 - \frac{y^2}{100} = (x)^2-\left(\frac{y}{10} \right )^2

Using identity  \Rightarrow a^2-b^2 = (a-b)(a+b)

Here, a= x \ \ and \ \ b = \frac{y}{10}

Therefore,

x^2 - \frac{y^2}{100} = \left ( x-\frac{y}{10} \right )\left ( x+\frac{y}{10} \right )

Q4 (i) Expand each of the following, using suitable identities: (x + 2y+4z)^2

Answer:

Given is  (x + 2y+4z)^2

We will Use identity  

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

Here, a = x , b = 2y \ \ and \ \ c = 4z

Therefore,  

(x + 2y+4z)^2 = (x)^2+(2y)^2+(4z)^2+2.x.2y+2.2y.4z+2.4z.x

                                = x^2+4y^2+16z^2+4xy+16yz+8zx

Q4 (ii) Expand each of the following, using suitable identities: (2x - y + z)^2

Answer:

Given is  (2x - y + z)^2

We will Use identity  

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

Here, a = 2x , b = -y \ \ and \ \ c = z

Therefore,  

(2x -y+z)^2 = (2x)^2+(-y)^2+(z)^2+2.2x.(-y)+2.(-y).z+2.z.2x

                                = 4x^2+y^2+z^2-4xy-2yz+4zx

Q4 (iii) Expand each of the following, using suitable identities: (-2x + 3y + 2z)^2

Answer:

Given is  (-2x + 3y + 2z)^2

We will Use identity  

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

Here, a =- 2x , b = 3y \ \ and \ \ c = 2z

Therefore,  

(-2x +3y+2z)^2 = (-2x)^2+(3y)^2+(2z)^2+2.(-2x).3y+2.3y.2z+2.z.(-2x)

                                  = 4x^2+9y^2+4z^2-12xy+12yz-8zx

Q4 (iv) Expand each of the following, using suitable identities: (3a - 7b - c)^2

Answer:

Given is  (3a - 7b - c)^2

We will Use identity  

(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx

Here, x =3a , y = -7b \ \ and \ \ z = -c

Therefore,  

(3a - 7b - c)^2=(3a)^2+(-7b)^2+(-c)^2+2.3a.(-7b)+2.(-7b).(-c)+2.(-c).3a

                           = 9a^2+49b^2+c^2-42ab+14bc-6ca

Q4 (v) Expand each of the following, using suitable identities: (-2x + 5y -3z)^2

Answer:

Given is  (-2x + 5y -3z)^2

We will Use identity  

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

Here, a =- 2x , b = 5y \ \ and \ \ c = -3z

Therefore,  

(-2x +5y-3z)^2= (-2x)^2+(5y)^2+(-3z)^2+2.(-2x).5y+2.5y.(-3z)+2.(-3z).(-2x)

                                    = 4x^2+25y^2+9z^2-20xy-30yz+12zx

Q4 (vi) Expand each of the following, using suitable identities: \left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2

Answer:

Given is  \left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2

We will Use identity  

(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx

Here, x =\frac{a}{4} , y = -\frac{b}{2} \ \ and \ \ z = 1

Therefore,  

\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2=\left(\frac{a}{4} \right )^2+\left(-\frac{b}{2} \right )^2+(1)^2+2.\left(\frac{a}{4} \right ). \left(-\frac{b}{2} \right )+2. \left(-\frac{b}{2} \right ).1+2.1.\left(\frac{a}{4} \right )

                               = \frac{a^2}{16}+\frac{b^2}{4}+1-\frac{ab}{4}-b+\frac{a}{2}

Q5 (i) Factorise: 4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz

Answer:

We can rewrite  4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz  as

\Rightarrow 4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz = (2x)^2+(3y)^2+(-4z)^2+2.2x.3y+2.3y.(-4z)+2.(-4z).2x

We will Use identity  

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

Here, a = 2x , b = 3y \ \ and \ \ c = -4z

Therefore,  

4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz = (2x+3y-4z)^2
 
                                                                                      = (2x+3y-4z)(2x+3y-4z)

Q5 (ii) Factorise: 2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz

Answer:

We can rewrite  2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz  as

\Rightarrow 2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz = (-\sqrt2x)^2+(y)^2+(2\sqrt2z)^2+2.(-\sqrt2).y+2.y.2\sqrt2z+2.(-\sqrt2x).2\sqrt2z

We will Use identity  

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

Here, a = -\sqrt2x , b = y \ \ and \ \ c = 2\sqrt2z

Therefore,  

2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz=(-\sqrt2x+y+2\sqrt2z)^2
 
                                                                                       =(-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z)

Q6 (i) Write the following cubes in expanded form: (2x + 1)^3

Answer:

Given is  (2x + 1)^3

We will use identity

(a+b)^3=a^3+b^3+3a^2b+3ab^2

Here,  a= 2x \ \ and \ \ b= 1

Therefore, 

(2x+1)^3=(2x)^3+(1)^3+3.(2x)^2.1+3.2x.(1)^2

                     = 8x^3+1+12x^2+6x

Q6 (ii) Write the following cube in expanded form: (2a-3b)^3

Answer:

Given is  (2a-3b)^3

We will use identity

(x-y)^3=x^3-y^3-3x^2y+3xy^2

Here,  x= 2a \ \ and \ \ y= 3b

Therefore, 

(2a-3b)^3=(2a)^3-(3b)^3-3.(2a)^2.3b+3.2a.(3b)^2

                       = 8a^3-9b^3-36a^2b+54ab^2

Q6 (iii) Write the following cube in expanded form: \left[\frac{3}{2}x + 1\right ]^3

Answer:

Given is  \left[\frac{3}{2}x + 1\right ]^3

We will use identity

(a+b)^3=a^3+b^3+3a^2b+3ab^2

Here,  a= \frac{3x}{2} \ \ and \ \ b= 1

Therefore, 

\left(\frac{3x}{2}+1 \right )^3= \left(\frac{3x}{2} \right )^3+(1)^3+3.\left(\frac{3x}{2} \right )^2.1+3.\frac{3x}{2}.(1)^2

                        = \frac{27x^3}{8}+1+\frac{27x^2}{4}+\frac{9x}{2}

Q6 (iv) Write the following cube in expanded form: \left[x - \frac{2}{3} y\right ]^3

Answer:

Given is  \left[x - \frac{2}{3} y\right ]^3

We will use identity

(a-b)^3=a^3-b^3-3a^2b+3ab^2

Here,  a=x \ and \ \ b= \frac{2y}{3}

Therefore, 

\left[x - \frac{2}{3} y\right ]^3 = x^3-\left(\frac{2y}{3} \right )^3-3.x^2.\frac{2y}{3}+3.x.\left(\frac{2y}{3} \right )^2

                      = x^3-\frac{8y^3}{27}-2x^2y+\frac{4xy^2}{3}

Q7 (i) Evaluate the following using suitable identities: (99)^3

Answer:

We can rewrite   (99)^3  as

\Rightarrow (99)^3=(100-1)^3

We will use identity

(a-b)^3=a^3-b^3-3a^2b+3ab^2

Here,  a=100 \ and \ \ b= 1

Therefore, 

(100-1)^1=(100)^3-(1)^3-3.(100)^2.1+3.100.1^2

                      = 1000000-1-30000+300= 970299

Q7 (ii) Evaluate the following using suitable identities: (102)^3

Answer:

We can rewrite   (102)^3  as

\Rightarrow (102)^3=(100+2)^3

We will use identity

(a+b)^3=a^3+b^3+3a^2b+3ab^2

Here,  a=100 \ and \ \ b= 2

Therefore, 

(100+2)^1=(100)^3+(2)^3+3.(100)^2.2+3.100.2^2

                      = 1000000+8+60000+1200= 1061208

Q7 (iii) Evaluate the following using suitable identities: (998)^3

Answer:

We can rewrite   (998)^3  as

\Rightarrow (998)^3=(1000-2)^3

We will use identity

(a-b)^3=a^3-b^3-3a^2b+3ab^2

Here,  a=1000 \ and \ \ b= 2

Therefore, 

(1000-2)^1=(1000)^3-(2)^3-3.(0100)^2.2+3.1000.2^2

                        = 1000000000-8-6000000+12000= 994011992

Q8 (i) Factorise the following: 8a^3 + b^3 + 12a^2 b + 6ab^2

Answer:

We can rewrite   8a^3 + b^3 + 12a^2 b + 6ab^2  as

\Rightarrow 8a^3 + b^3 + 12a^2 b + 6ab^2 = (2a)^3+(b)^3+3.(2a)^2.b+3.2a.(b)^2

We will use identity

(x+y)^3=x^3+y^3+3x^2y+3xy^2

Here,  x=2a \ \ and \ \ y= b

Therefore, 

8a^3 + b^3 + 12a^2 b + 6ab^2 = (2a+b)^3

                                                   =(2a+b)(2a+b)(2a+b)

Q8 (ii) Factorise the following: 8a ^3 - b^3 - 12a^2 b + 6ab^2

Answer:

We can rewrite   8a ^3 - b^3 - 12a^2 b + 6ab^2  as

\Rightarrow 8a ^3 - b^3 - 12a^2 b + 6ab^2 = (2a)^3-(b)^3-3.(2a)^2.b+3.2a.(b)^2

We will use identity

(x-y)^3=x^3-y^3-3x^2y+3xy^2

Here,  x=2a \ \ and \ \ y= b

Therefore, 

8a ^3 - b^3 - 12a^2 b + 6ab^2 =(2a-b)^3

                                                   =(2a-b)(2a-b)(2a-b)

Q8 (iii) Factorise the following: 27 - 125a^ 3 - 135a + 225a^2

Answer:

We can rewrite   27 - 125a^ 3 - 135a + 225a^2  as

\Rightarrow 27 - 125a^ 3 - 135a + 225a^2^{} = (3)^3-(25a)^3-3.(3)^2.5a+3.3.(5a)^2

We will use identity

(x-y)^3=x^3-y^3-3x^2y+3xy^2

Here,  x=3 \ \ and \ \ y= 5a

Therefore, 

27 - 125a^ 3 - 135a + 225a^2 = (3-5a)^3

                                                          =(3-5a)(3-5a)(3-5a)

Q8 (iv) Factorise the following: 64a^3 - 27b^3 - 144a^2 b + 108ab^2

Answer:

We can rewrite   64a^3 - 27b^3 - 144a^2 b + 108ab^2  as

\Rightarrow 64a^3 - 27b^3 - 144a^2 b + 108ab^2 = (4a)^3-(3b)^3-3.(4a)^2.3b+3.4a.(3b)^2

We will use identity

(x-y)^3=x^3-y^3-3x^2y+3xy^2

Here,  x=4a \ \ and \ \ y= 3b

Therefore, 

64a^3 - 27b^3 - 144a^2 b + 108ab^2=(4a-3b)^2

                                                                  =(4a-3b)(4a-3b)(4a-3b)

Q8 (v) Factorise the following:27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p

Answer:

We can rewrite   27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p  as

\Rightarrow 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p = (3p)^3-\left(\frac{1}{6} \right )^3-3.(3p)^2.\frac{1}{6}+3.3p.\left(\frac{1}{6} \right )^2

We will use identity

(x-y)^3=x^3-y^3-3x^2y+3xy^2

Here,  x=3p \ \ and \ \ y= \frac{1}{6}

Therefore, 

27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p = \left ( 3p-\frac{1}{6} \right )^3

                                                   = \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right )

Q9 (i) Verify: x^3 + y^3 = (x +y)(x^2 - xy + y^2)

Answer:

We know that

(x+y)^3=x^3+y^3+3xy(x+y)

Now,

\Rightarrow x^3+y^3=(x+y)^3-3xy(x+y)

\Rightarrow x^3+y^3=(x+y)\left((x+y)^2-3xy \right )

\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2+2xy-3xy \right )                                                (\because (a+b)^2=a^2+b^2+2ab)                                           

\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2-xy \right )

Hence proved.

Q9 (ii) Verify: x^3 - y^3 = (x -y)(x^2 + xy + y^2)

Answer:

We know that

(x-y)^3=x^3-y^3-3xy(x-y)

Now,

\Rightarrow x^3-y^3=(x-y)^3+3xy(x-y)

\Rightarrow x^3-y^3=(x-y)\left((x-y)^2+3xy \right )

\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2-2xy+3xy \right )                                                (\because (a-b)^2=a^2+b^2-2ab)                                           

\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2+xy \right )

Hence proved.

Q10 (i) Factorise the following: 27y^3 + 125z^3

Answer:

We know that 

a^3+b^3=(a+b)(a^2+b^2-ab)

Now, we can write  27y^3 + 125z^3  as 

\Rightarrow 27y^3 + 125z^3 = (3y)^3+(5z)^3

Here,  a = 3y \ \ and \ \ b = 5z

Therefore,

27y^3+125z^3= (3y+5z)\left((3y)^2+(5z)^2-3y.5z \right )

27y^3+125z^3= (3y+5z)\left(9y^2+25z^2-15yz \right )

Q10 (ii) Factorise the following: 64m^3 - 343n^3

Answer:

We know that 

a^3-b^3=(a-b)(a^2+b^2+ab)

Now, we can write  64m^3 - 343n^3  as 

\Rightarrow 64m^3 - 343n^3 = (4m)^3-(7n)^3

Here,  a = 4m \ \ and \ \ b = 7n

Therefore,

64m^3-343n^3= (4m-7n)\left((4m)^2+(7n)^2+4m.7n \right )

64m^3-343n^3= (4m-7n)\left(16m^2+49n^2+28mn \right )

Q11 Factorise:     27x^3 + y^3 + z^3 - 9xyz

Answer:

Given is  27x^3 + y^3 + z^3 - 9xyz

Now, we know that 

a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

Now, we can write  27x^3 + y^3 + z^3 - 9xyz  as 

\Rightarrow 27x^3 + y^3 + z^3 - 9xyz =(3x)^3+(y)^3+(z)^3-3.3x.y.z

Here, a= 3x , b = y \ \ and \ \ c = z

Therefore,

27x^3 + y^3 + z^3 - 9xyz =(3x+y+z)\left((3x)^2+(y)^2+(z)^2-3x.y-y.z-z.3x \right )

                                                 =(3x+y+z)\left(9x^2+y^2+z^2-3xy-yz-3zx \right )

Q12 Verify that    x^3 + y^3 + z^3 -3xyz = \frac{1}{2} ( x + y + z)\left[(x-y)^2 + (y-z)^2 + (z-x)^2 \right ]

Answer:

We know that

x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)

Now, multiply and divide the R.H.S. by 2

x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)

                                            = \frac{1}{2}(x+y+z)(x^2+y^2-2xy+x^2+z^2-2zx+y^2+z^2-2yz)

                                            = \frac{1}{2}(x+y+z)\left((x-y)^2+(y-z)^2 +(z-x)^2\right )                                        \left(\because a^2+b^2-2ab=(a-b)^2 \right )

Hence proved.

Q13 If x + y + z = 0, show that x^3 + y^3 + z^3 = 3xyz.

Answer:

We know that

x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)

Now, It is given that  x + y + z = 0

Therefore,

x^3+y^3+z^3-3xyz =0(x^2+y^2+z^2-xy-yz-zx)

x^3+y^3+z^3-3xyz =0

x^3+y^3+z^3=3xyz

Hence proved.

Q14 (i) Without actually calculating the cubes, find the value of each of the following: (-12)^3 + (7)^3 + (5)^3

Answer:

Given is   (-12)^3 + (7)^3 + (5)^3

We know that   

If   x+y+z = 0   then ,   x^3+y^3+z^3 = 3xyz

Here, x = -12 , y = 7 \ \ an d \ \ z = 5

\Rightarrow x+y+z = -12+7+5 = 0

Therefore,

(-12)^3 + (7)^3 + (5)^3 = 3 \times (-12)\times 7 \times 5 = -1260

Therefore, value of  (-12)^3 + (7)^3 + (5)^3  is  -1260

Q14 (ii) Without actually calculating the cubes, find the value of the following: (28)^3 + (-15)^3 + (-13)^3

Answer:

Given is   (28)^3 + (-15)^3 + (-13)^3

We know that   

If   x+y+z = 0   then ,   x^3+y^3+z^3 = 3xyz

Here, x = 28 , y = -15 \ \ an d \ \ z = -13

\Rightarrow x+y+z =28-15-13 = 0

Therefore,

(28)^3 + (-15)^3 + (-13)^3 = 3 \times (28)\times (-15) \times (-13) = 16380

Therefore, value of  (28)^3 + (-15)^3 + (-13)^3  is  16380

Q15 (i) Give possible expressions for the length and breadth of the following rectangle, in which its area is given:

   25a^2 - 35a + 12

Answer:

We know that 

Area of rectangle is =  length \times breadth

It is given that area  =  25a^2-35a+12

Now, by splitting middle term method

\Rightarrow 25a^2-35a+12 = 25a^2-20a-15a+12

                                          = 5a(5a-4)-3(5a-4)

                                          = (5a-3)(5a-4)
Therefore, two answers are possible 

case (i) :-   Length = (5a-4)  and  Breadth = (5a-3)

case (ii) :-   Length = (5a-3)  and  Breadth = (5a-4)

Q15 (ii) Give possible expressions for the length and breadth of the following rectangle, in which its area is given:

    35y^2 + 13y- 12
 

Answer:

We know that 

Area of rectangle is =  length \times breadth

It is given that area  =  35y^2 + 13y- 12

Now, by splitting the middle term method

\Rightarrow 35y^2 + 13y- 12 =35y^2+28y-15y-12

                                          = 7y(5y+4)-3(5y+4)

                                          = (7y-3)(5y+4)

Therefore, two answers are possible 

case (i) :-   Length = (5y+4)  and  Breadth = (7y-3)

case (ii) :-   Length = (7y-3)  and  Breadth = (5y+4)

Q16 (i) What are the possible expressions for the dimensions of the cuboid whose volumes is given below?

    Volume :    3x^2 - 12x

Answer:

We know that 

Volume of cuboid is =  length \times breadth \times height

It is given that volume  =  3x^2-12x

Now, 

\Rightarrow 3x^2-12x=3\times x\times (x-4)

Therefore,one of the possible answer is possible 

Length = 3  and  Breadth = x  and  Height =  (x-4)

Q16 (ii) What are the possible expressions for the dimensions of the cuboid whose volumes is given below?

    Volume :    12ky^2 + 8ky - 20k

Answer:

We know that 

Volume of cuboid is =  length \times breadth \times height

It is given that volume  =  12ky^2+8ky-20k

Now, 

\Rightarrow 12ky^2+8ky-20k = k(12y^2+8y-20)

                                               = k(12y^2+20y-12y-20)

                                               = k\left(4y(3y+5)-4(3y+5) \right )

                                               = k(3y+5)(4y-4)

                                               = 4k(3y+5)(y-1)

Therefore,one of the possible answer is possible 

Length = 4k  and  Breadth = (3y+5)  and  Height =  (y-1)

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

How to use NCERT solutions for class 9 maths chapter 2 Polynomials?

  • First of all, learn some basics and concepts regarding chapter polynomials.
  • While reading the basics, go through the examples so that you can understand the applications of the concepts.
  • Once you have done the above two points, then you can directly move to the practice exercises.
  • While practicing the exercises, if you stuck anywhere then you can take the help of the NCERT solutions for class 9 maths chapter 2 Polynomials.
  • After the completion of practice exercises, you can go through some previous year question papers.

Keep Working Hard and Happy Learning!

 

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