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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Edited By Ramraj Saini | Updated on Oct 06, 2023 08:15 AM IST

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Polynomials Class 9 Questions And Answers are provided here. 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 NCERT syllabus Class 9 chapter, you will learn the operations of two or more polynomials. NCERT solutions which are prepared by subjects expert at Careers360 keeping in mind of latest CBSE syllabus 2023, are there to help you while solving the problems related to this NCERT book Class 9 Maths chapter. Polynomials Class 9, introduces a lot of important concepts that will be helpful for those students who are targeting exams like JEE, CAT, SSC, etc.

This chapter talk about Polynomials in One Variable, Zeroes of a Polynomial, Remainder Theorem, Factorisation of Polynomials, and Algebraic Identities. 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. Here students will get NCERT solutions for Class 9 also.

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Polynomials Class 9 Questions And Answers PDF Free Download

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Polynomials Class 9 Solutions - Important Formulae

The general form of a polynomial is: p(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0

where a0, a1, a2, …. an are constants, and an ≠ 0.

Every one-variable linear polynomial will contain a unique zero, which is a real number that is a zero of the zero polynomial, and a non-zero constant polynomial that does not have any zeros.

>> Remainder Theorem: If p(x) has a degree greater than or equal to 1, and you divide p(x) by the linear polynomial (x - a), the remainder will be p(a).

>> Factor Theorem: The linear polynomial (x - a) will be a factor of the polynomial p(x) whenever p(a) = 0. Similarly, if (x - a) is a factor of p(x), then p(a) = 0.

Free download NCERT Solutions for Class 9 Maths Chapter 2 Polynomials for CBSE Exam.

Polynomials Class 9 NCERT Solutions (Intext Questions and Exercise)

NCERT Polynomials class 9 solutions Exercise: 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

Polynomials class 9 NCERT solutions Exercise: 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, values 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 polynomial p(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

Class 9 maths chapter 2 NCERT solutions Exercise: 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

1639996048817
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

1639996081431 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

1639996102222
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

1639996129502
Therefore, the remainder is 1-3\pi + 3\pi^2-\pi^3

Q1 (v) Find the remainder when x^ 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

1639996167148
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

1639996211310
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

1639996248244
Since, remainder is not equal to 0

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

Class 9 polynomials NCERT solutions Exercise: 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 the 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

1639996302612 We know that Dividend = (Divisor × 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 factors of the given polynomial.

By long division method, we will get

1639996323635 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

1639996347806 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

1639996378575 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)

Class 9 maths chapter 2 question answer Exercise: 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)

More About NCERT Solutions For Class 9 Maths Chapter 2 Polynomials

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 NCERT textbook 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. NCERT solutions for Class 9 Maths chapter 2 Polynomials will cover the detailed solution to each and every question present in the practice exercises including optional exercises.

Interested students can practice class 9 maths ch 2 question answer using the following exercises.

NCERT Solutions for Class 9 Maths Chapter Wise

Chapter No.
Chapter Name
Chapter 1
Chapter 2
Polynomials
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15

Key Features of NCERT Solutions For chapter 2 maths class 9

  • Comprehensive coverage of topics related to Polynomials in one variable, Zeros of a Polynomial, Real Numbers and their Decimal Expansions, and more.
  • Maths chapter 2 class 9 solutions are designed in a clear and concise language to help students understand the concepts with ease.
  • The step-by-step approach of the ch 2 maths class 9 solutions assists students in learning and solving mathematical problems in a structured manner.
  • Inclusion of a wide range of solved examples and exercises to aid students in practicing and assessing their understanding of the concepts.

The class 9 chapter 2 maths solutions are prepared by experts at Careers360 who have extensive knowledge and experience in Mathematics.

NCERT Solutions for Class 9 Subject Wise

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 practising 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.

NCERT Books and NCERT Syllabus

Keep Working Hard and Happy Learning!

Frequently Asked Questions (FAQs)

1. What are the important topics in NCERT Class 9 syllabus chapter Polynomials ?

The NCERT class 9 maths chapter 2 includes topics such as definition of a polynomial, zeroes, coefficient, degrees, and terms of a polynomial, different types of a polynomial, remainder and factor theorems, and the factorization of polynomials. students should practice these NCERT solutions to get indepth understanding of these concepts which ultimately lead to score well in the exam.

2. What is the number of exercises included in NCERT Solutions for class 9 maths polynomials?

Maths chapter 2  includes five exercises covering topics such as Polynomials in one variable, Zeros of a Polynomial, Real Numbers and their Decimal Expansions, Representing Real Numbers on the Number Line, Operations on Real Numbers, and Laws of Exponents for Real Numbers. Practicing these exercises of NCERT maths class 9 chapter 2  is crucial for achieving a better understanding of the concepts and scoring well in Mathematics. To help students gain confidence, Careers360 experts have designed these solutions to provide comprehensive explanations of the concepts covered in this chapter.

3. What are the advantages of choosing NCERT Solutions for Class 9 Maths Chapter 2?

NCERT Solutions for Class 9 Maths Chapter 2 use straightforward language to explain the concepts, making it accessible even for students who struggle with Mathematics. These solutions are meticulously crafted by a team of experts at Careers360 with the objective of helping students prepare for their CBSE exams effectively.

4. Is it challenging to grasp the concepts in NCERT Solutions for polynomials class 9 solutions?

Regular practice with NCERT Solutions for Class 9 Maths Chapter 2 can enable students to excel in their CBSE exams. These solutions are created by a team of skilled Maths experts at Careers360, and by solving all the questions and comparing their answers with the solutions, students can aim for high scores in their exams.

5. Where can I find the complete solutions of NCERT for Class 9 Maths ?

Here, students can get detailed NCERT solutions for Class 9 Maths which includes solutions to all the exercise of each chapters. 

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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