NCERT Solutions for Class 9 Maths Chapter 2- Polynomials

NCERT Solutions For Class 9 Maths Chapter 2 PDF Free Download

NCERT solutions class 9 maths chapter 2 – Polynomials are provided here. These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. These expert faculties solve and provide the NCERT maths solution so that it should help students to solve the problems comfortably. They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT textbook for class 9.

In CBSE class 9 Polynomials chapter, students are introduced to a lot of important topics which will be helpful for those who wish to pursue mathematics as a subject in further classes. Based on these NCERT solutions. These solutions help students to prepare for their upcoming Board Exams by covering the whole the syllabus which follows NCERT guidelines.

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NCERT Solutions for class 9 Maths Chapter 2 Polynomials Part 1
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Access Answers of Maths NCERT class 9 Chapter 2 – Polynomials

Class 9 Maths Chapter 2 Exercise 2.1

Q1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x2 – 3x + 7

Solution:

The equation 4x2 – 3x + 7 can be written as 4x2 – 3x1 + 7x0

Since x is the only variable in the given equation and the powers of x (i.e., 2, 1 and 0) are whole numbers, we can say that the expression 4x2 – 3x + 7 is a polynomial in one variable.

(ii) y2 + √2

Solution:

The equation y2 + can be written as y2 + y0

Since y is the only variable in the given equation and the powers of y (i.e., 2 and 0) are whole numbers, we can say that the expression y2 + is a polynomial in one variable.

(iii) 3√t + t√2

Solution:

The equation 3 + t can be written as 3t1/2 + √2t

Though, t is the only variable in the given equation, the powers of t (i.e.,) is not a whole number. Hence, we can say that the expression 3 + t is not a polynomial in one variable.

(iv) y + 2/y

Solution:

The equation y + can be written as y+2y-1

Though, y is the only variable in the given equation, the powers of y (i.e.,-1) is not a whole number. Hence, we can say that the expression y + is not a polynomial in one variable.

(v) x10 + y3 + t50

Solution:

Here, in the equation x10 + y3 + t50

Though, the powers, 10, 3, 50, are whole numbers, there are 3 variables used in the expression

x10 + y3 + t50. Hence, it is not a polynomial in one variable.

 

Q2. Write the coefficients of x2 in each of the following:
(i) 2 + x2 + x

Solution:

The equation 2 + x2 + x can be written as 2 + (1) x2 + x

We know that, coefficient is the number which multiplies the variable.

Here, the number that multiplies the variable x2 is 1

, the coefficients of x2 in 2 + x2 + x is 1.

(ii) 2 – x2 + x3

Solution:

The equation 2 – x2 + x3 can be written as 2 + (–1) x2 + x3

We know that, coefficient is the number (along with its sign,i.e., – or +) which multiplies the variable.

Here, the number that multiplies the variable x2 is -1

, the coefficients of x2 in 2 – x2 + x3 is -1.

(iii) Π/2 x2 +x

Solution:

The equation Π/2x2 +x can be written as (Π/2 ) x2 + x

We know that, coefficient is the number (along with its sign,i.e., – or +) which multiplies the variable.

Here, the number that multiplies the variable x2 is

, the coefficients of x2 in Π/2x2 +x is Π/2.

(iv)√2x-1

Solution:

The equationx√2x-1 can be written as 0x2 +√2x-1 [Since 0x2 is 0]

We know that, coefficient is the number (along with its sign,i.e., – or +) which multiplies the variable.

Here, the number that multiplies the variable x2 is 0

, the coefficients of x2 in√2x-1 is 0.

Q3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Solution:

Binomial of degree 35: A polynomial having two terms and the highest degree 35 is called a binomial of degree 35

Eg.,  3x35+5

Monomial of degree 100: A polynomial having one term and the highest degree 100 is called a monomial of degree 100

Eg.,  4x100

Q4. Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 + 7x

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, 5x3 + 4x2 + 7x= 5x3 + 4x2 + 7x1

The powers of the variable x are: 3, 2, 1

, the degree of 5x3 + 4x2 + 7x is 3 as 3 is the highest power of x in the equation.

(ii) 4 – y2

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, in 4 – y2,

The power of the variable y is: 2

, the degree of 4 – y2 is 2 as 2 is the highest power of y in the equation.

(iii) 5t – √7

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, in 5t – √7,

The power of the variable y is: 1

, the degree of 5t – √7 is 1 as 1 is the highest power of y in the equation.

(iv) 3

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, 3== 3x0

The power of the variable here is: 0

, the degree of 3 is 0.

Q5. Classify the following as linear, quadratic and cubic polynomials:

Solution:

We know that,

Linear polynomial: A polynomial of degree one is called a linear polynomial.

Quadratic polynomial: A polynomial of degree two is called a quadratic polynomial.

Cubic polynomial: A polynomial of degree three a cubic polynomial.

(i) x2 + x

Solution:

The highest power of x2 + x is 2

, the degree is 2

Hence, x2 + x is a quadratic polynomial

(ii) x – x3

Solution:

The highest power of x – x3 is 3

, the degree is 3

Hence, x – x3 is a cubic polynomial

(iii) y + y2 + 4

Solution:

The highest power of y + y2 + 4 is 2

, the degree is 2

Hence, y + y2 + 4 is a quadratic polynomial

(iv) 1 + x

Solution:

The highest power of 1 + x is 1

, the degree is 1

Hence, 1 + x is a linear polynomial

(v) 3t

Solution:

The highest power of 3t is 1

, the degree is 1

Hence, 3t is a linear polynomial

(vi) r2

Solution:

The highest power of r2 is 2

, the degree is 2

Hence, r2 is a quadratic polynomial

(vii) 7x3

Solution:

The highest power of 7x3 is 3

, the degree is 3

Hence, 7x3 is a cubic polynomial

 

Exercise 2.2 Page: 34

Q1. Find the value of the polynomial (x)=5x−4x2+3 
(i) x= 0
(ii) x = – 1
(iii) x = 2

Solution:

Let f(x)= 5x−4x2+3

(i) When x=0

f(0)=5(0)+4(0)2+3

=3

(ii) When x= -1

f(x)=5x−4x2+3

f(−1)=5(−1) −4(−1)2+3

=−5–4+3

=−6

(iii) When x=2

f(x)=5x−4x2+3

f(2)=5(2) −4(2)2+3

=10–16+3

=−3

Q2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y)=y2−y+1

Solution:

p(y)=y2–y+1

∴p(0)=(0)2−(0)+1=1

p(1)=(1)2–(1)+1=1

p(2)=(2)2–(2)+1=3

(ii) p(t)=2+t+2t2−t3

Solution:

p(t)= 2+t+2t2−t3

∴p(0)=2+0+2(0)2–(0)3=2

p(1)=2+1+2(1)2–(1)3=2+1+2–1=4

p(2)=2+2+2(2)2–(2)3=2+2+8–8=4

(iii) p(x)=x3

Solution:

p(x)=x3

∴p(0)=(0)3=0

p(1)=(1)3=1

p(2)=(2)3=8

(iv) p(x)=(x−1)(x+1)

Solution:

p(x)=(x–1)(x+1)

∴p(0)=(0–1)(0+1)=(−1)(1)=–1

p(1)=(1–1)(1+1)=0(2)=0

p(2)=(2–1)(2+1)=1(3)=3

Q3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x)=3x+1, x=−1/3

Solution:

For, x=−1/3 , p(x)=3x+1

∴p(−1/3)=3(−1/3)+1=−1+1=0

∴−1/3 is a zero of p(x).

(ii) p(x)=5x–π, x=4/5

Solution:

For, x=4/5 p(x)=5x–π

∴p(4/5)=5(4/5)–π=4−π

∴ 4/5is not a zero of p(x).

(iii) p(x)=x2−1, x=1, −1

Solution:

For, x=1, −1;

p(x)=x2−1

∴p(1)=12−1=1−1=0

p(−1)=(-1)2−1=1−1=0

∴1, −1 are zeros of p(x).

(iv) p(x)=(x+1)(x–2), x= −1, 2

Solution:

For, x=−1,2;

p(x)=(x+1)(x–2)

∴p(−1)=(−1+1)(−1–2)

=((0)(−3))=0

p(2)=(2+1)(2–2)=(3)(0)=0

∴−1,2 are zeros of p(x).

(v) p(x)=x2, x=0

Solution:

For, x=0 p(x)= x2

p(0)=02=0

∴0 is a  zero of p(x).

(vi) p(x)=lx+m, x=−m/t

Solution:

For, x=−m/t;  p(x)=lx+m

∴p(−m/t)=l(−m/t)+m=−m+m=0

∴−m/tis a zero of p(x).

(vii) p(x)=3x2−1,x=−1/√3,2/√3,

Solution:

For, x=−1/√3,2/√3,; p(x)=3x2−1

∴p(−1/√3)=3(−1/√3)2−1=3(1/√3)−1=1−1=0

∴p(2/√3)=3(2/√3)2−1=3(4/3)−1=4−1=3≠0

∴−−1/√3 is a zero of p(x) but 2/√3 is not a zero of p(x).

(viii) p(x)=2x+1,x=1/2

Solution:

For, x=1/2 p(x)=2x+1

∴p(1/2)=2(1/2)+1=1+1=2≠0

∴ 1/2 is not a zero of p(x).

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

Solution:

p(x)=x+5

⇒x+5=0

⇒x=−5

∴-5 is a zero polynomial of the polynomial p(x).

(ii) p(x) = x – 5

Solution:

p(x)=x−5

⇒x−5=0

⇒x=5

∴5 is a zero polynomial of the polynomial p(x).

(iii) p(x) = 2x + 5

Solution:

p(x)=2x+5

⇒2x+5=0

⇒2x=−5

⇒x=−5/2

∴x= −5/2  is a zero polynomial of the polynomial p(x).

(iv) p(x) = 3x – 2 

Solution:

p(x)=3x–2

⇒3x−2=0

⇒3x=2

⇒x=2/3

∴x=2/3 is a zero polynomial of the polynomial p(x).

(v) p(x) = 3x 

Solution:

p(x)=3x

⇒3x=0

⇒x=0

∴0 is a zero polynomial of the polynomial p(x).

(vi) p(x) = ax, a0

Solution:

p(x)=ax

⇒ax=0

⇒x=0

∴x=0 is a zero polynomial of the polynomial p(x).

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Solution:

p(x)= cx + d

⇒ cx + d =0

⇒x=-d/c

∴ x=-d/c is a zero polynomial of the polynomial p(x).

Class 9 Maths Chapter 2 Exercise 2.3 Page: 40

Q1. Find the remainder when x3+3x2+3x+1 is divided by
(i) x+1

Solution:

x+1=0

⇒x=−1

∴Remainder:

p(−1)=(−1)3+3(−1)2+3(−1)+1

=−1+3−3+1

=0

(ii) x−1/2

Solution:

x−1/2=0

⇒x= 1/2

∴Remainder:

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

=1/8+3/4+3/2+1

=27/8

(iii) x

Solution:

x=0

∴Remainder:

p(0)=(0)3+3(0)2+3(0)+1

=1

(iv) x+π

Solution:

x+π=0

⇒x=−π

∴Remainder:

p(0)=(−π)3+3(−π)2+3(−π)+1

=−π3+3π2−3π+1

(v) 5+2x

Solution:

5+2x=0

⇒2x=−5

⇒x=− 5/2

∴Remainder:

(−5/2)3+3(−5/2)2+3(−5/2)+1=−125/8+75/4−15/2+1

=−27/8

Q2. Find the remainder when x3−ax2+6x−a is divided by x-a.

Solution:

Let p(x)=x3−ax2+6x−a

x−a=0

∴x=a

Remainder:

p(a)= (a)3 −a(a2)+6(a)−a

=a3−a3+6a−a = 5a

Q3. Check whether 7+3x is a factor of 3x3+7x.

Solution:

7+3x=0

⇒3x=−7 only if 7+3x divides 3x3+7x leaving no remainder.

⇒x=-7/3

∴Remainder:

3(7/3)3+7(7/3)= −343/9+(-49/3)

= −343+(-49)3/9

= -343-147/9

= -490/9≠0

∴7+3x is not a factor of 3x3+7x

Exercise 2.4 Page: 43

Q1. Determine which of the following polynomials has (x + 1) a factor:

(i) x3+x2+x+1

Solution:

Let p(x)= x3+x2+x+1

The zero of x+1 is -1. [x+1=0 means x=-1]

p(−1)=(−1)3+(−1)2+(−1)+1

=−1+1−1+1

=0

∴By factor theorem, x+1 is a factor of x3+x2+x+1

(ii) x4 + x3 + x2 + x + 1

Solution:

Let p(x)= x4 + x3 + x2 + x + 1

The zero of x+1 is -1. . [x+1=0 means x=-1]

p(−1)=(−1)4+(−1)3+(−1)2+(−1)+1

=1−1+1−1+1

=1≠0

∴By factor theorem, x+1 is not a factor of x4 + x3 + x2 + x + 1

(iii) x4 + 3x3 + 3x2 + x + 1 

Solution:

Let p(x)= x4 + 3x3 + 3x2 + x + 1

The zero of x+1 is -1.

p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1

=1−3+3−1+1

=1≠0

∴By factor theorem, x+1 is not a factor of x4 + 3x3 + 3x2 + x + 1

(iv) x3 – x2 – (2 +√2 )x +√2

Solution:

Let p(x)= x3 – x2 – (2 + √2)x +√2

The zero of x+1 is -1.

p(−1)=(−1)3–(−1)2–(2+√2)(−1)+√2

=−1−1+2+√2+√2

= 2√2

∴By factor theorem, x+1 is not a factor of x3 – x2 – (2 +√2 )x +√2

Q2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x)=2x3+x2–2x–1, g(x) = x + 1

Solution:

p(x)= 2x3+x2–2x–1, g(x) = x + 1

g(x)=0

⇒x+1=0

⇒x=−1

∴Zero of g(x) is -1.

Now,

p(−1)=2(−1)3+(−1)2–2(−1)–1

=−2+1+2−1

=0

∴By factor theorem, g(x) is a factor of p(x).

(ii) p(x)=x3+3x2+3x+1, g(x) = x + 2

Solution:

p(x)=x3+3×2+3x+1, g(x) = x + 2

g(x)=0

⇒x+2=0

⇒x=−2

∴Zero of g(x) is -2.

Now,

p(−2)=(−2)3+3(−2)2+3(−2)+1

=−8+12−6+1

=−1≠0

∴By factor theorem, g(x) is not a factor of p(x).

(iii) p(x)=x3–4x2+x+6, g(x) = x – 3

Solution:

p(x)= x3–4x2+x+6, g(x) = x -3

g(x)=0

⇒x−3=0

⇒x=3

∴Zero of g(x) is 3.

Now,

p(3)=(3)3−4(3)2+(3)+6

=27−36+3+6

=0

∴By factor theorem, g(x) is a factor of p(x).

Q3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
(i) p(x)=x2+x+k

Solution:

If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

⇒(1)2+(1)+k=0

⇒1+1+k=0

⇒2+k=0

⇒k=−2

(ii) p(x)=2x2+kx+√2

Solution:

If x-1 is a factor of p(x), then p(1)=0

⇒2(1)2+k(1)+√2 =0

⇒2+k+√2=0

⇒k = −(2+√2)

(iii) p(x)=kx2–√2x+1

Solution:

If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

⇒k(1)2−√2(1)+1=0

⇒k = √2−1

(iv) p(x)=kx2–3x+k

Solution:

If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

⇒k(1)2–3(1)+k=0

⇒k−3+k=0

⇒2k−3=0

⇒k= \(\frac{3}{2}\)

Q4. Factorize:

(i) 12x2–7x+1

Solution:

Using the splitting the middle term method,

We have to find a number whose sum=-7 and product=112=12

We get -3 and -4 as the numbers [-3+-4=-7 and -3-4=12]

12x2–7x+1=12x2-4x-3x+1

=4x (3x-1)-1(3x-1)

= (4x-1)(3x-1)

(ii) 2x2+7x+3

Solution:

Using the splitting the middle term method,

We have to find a number whose sum=7 and product=2=6

We get 6 and 1 as the numbers [6+1=7 and 6=6]

2x2+7x+3 =2x2+6x+1x+3

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

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

(iii) 6x2+5x-6 

Solution:

Using the splitting the middle term method,

We have to find a number whose sum=5 and product=6= -36

We get -4 and 9 as the numbers [-4+9=5 and -4=-36]

6x2+5x-6=6x2+ 9x – 4x – 6

=3x (2x + 3) – 2 (2x + 3)

= (2x + 3) (3x – 2)

(iv) 3x2 – x – 4 

Solution:

Using the splitting the middle term method,

We have to find a number whose sum=-1 and product=3= -12

We get -4 and 3 as the numbers [-4+3=-1 and -4=-12]

3x2 – x – 4 =3x2–x–4

=3x2–4x+3x–4

=x(3x–4)+1(3x–4)

=(3x–4)(x+1)

Q5. Factorize:
(i) x3–2x2–x+2

Solution:

Let p(x)=x3–2x2–x+2

Factors of 2 are ±1 and ± 2

By trial method, we find that

p(1) = 0

So, (x+1) is factor of p(x)

Now,

p(x)= x3–2x2–x+2

p(−1)=(−1)3–2(−1)2–(−1)+2

=−1−2+1+2

=0

Therefore, (x+1) is the factor of  p(x)

ncert solutions for class 9 maths chapter 1 fig 1

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x2–3x+2) =(x+1)(x2–x–2x+2)

=(x+1)(x(x−1)−2(x−1))

=(x+1)(x−1)(x-2)

(ii) x3–3x2–9x–5

Solution:

Let p(x) = x3–3x2–9x–5

Factors of 5 are ±1 and ±5

By trial method, we find that

p(5) = 0

So, (x-5) is factor of p(x)

Now,

p(x) = x3–3x2–9x–5

p(5) = (5)3–3(5)2–9(5)–5

=125−75−45−5

=0

Therefore, (x-5) is the factor of  p(x)

ncert solutions for class 9 maths chapter 1 fig 2

Now, Dividend = Divisor × Quotient + Remainder

(x−5)(x2+2x+1) =(x−5)(x2+x+x+1)

=(x−5)(x(x+1)+1(x+1))

=(x−5)(x+1)(x+1)

(iii) x3+13x2+32x+20

Solution:

Let p(x) = x3+13x2+32x+20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial method, we find that

p(-1) = 0

So, (x+1) is factor of p(x)

Now,

p(x) = x3+13x2+32x+20

p(-1) = (−1)3+13(−1)2+32(−1)+20

=−1+13−32+20

=0

Therefore, (x+1) is the factor of  p(x)

ncert solutions for class 9 maths chapter 1 fig 3

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x2+12x+20) =(x+1)(x2+2x+10x+20)

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

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

(iv) 2y3+y2–2y–1

Solution:

Let p(y) = 2y3+y2–2y–1

Factors = 2×(−1)= -2 are ±1 and ±2

By trial method, we find that

p(1) = 0

So, (y-1) is factor of p(y)

Now,

p(y) = 2y3+y2–2y–1

p(1) = 2(1)3+(1)2–2(1)–1

=2+1−2

=0

Therefore, (y-1) is the factor of  p(y)

ncert solutions for class 9 maths chapter 1 fig 4

Now, Dividend = Divisor × Quotient + Remainder

(y−1)(2y2+3y+1) =(y−1)(2y2+2y+y+1)

=(y−1)(2y(y+1)+1(y+1))

=(y−1)(2y+1)(y+1)

Exercise 2.5 Page: 48

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

Solution:

Using the identity, (x + a) (x + b) = x 2 + (a + b)x + ab

[Here, a=4 and b=10]

We get,

(x+4)(x+10) =x2+(4+10)x+(4×10)

=x2+14x+40

(ii) (x + 8) (x – 10)     

Solution:

Using the identity, (x + a) (x + b) = x 2 + (a + b)x + ab

[Here, a=8 and b= −10]

We get,

(x+8)(x−10) =x2+(8+(−10))x+(8×(−10))

=x2+(8−10)x–80

=x2−2x−80

(iii) (3x + 4) (3x – 5)

Solution:

Using the identity, (x + a) (x + b) = x 2 + (a + b)x + ab

[Here, x=3x, a=4 and b= −5]

We get,

(3x+4)(3x−5) =(3x)2+4+(−5)3x+4×(−5)

=9x2+3x(4–5)–20

=9x2–3x–20

(iv) (y2+3/2)(y2–3/2 )

Solution:

Using the identity, (x + y) (x – y) = x 2 – y 2

[Here, x=y2 and y=3/2]

We get,

(y2+3/2)(y2–3/2) = (y2)2–(3/2)2

=y4–(9/4)

Q2. Evaluate the following products without multiplying directly:
(i) 103 × 107

Solution:

103×107=(100+3)×(100+7)

Using identity, [(x+a)(x+b)=x2+(a+b)x+ab

Here, x=100

a=3

b=7

We get, 103×107=(100+3)×(100+7)

=(100)2+(3+7)100+(3×7))

=10000+1000+21

=11021

(ii) 95 × 96  

Solution:

95×96=(100-5)×(100-4)

Using identity, [(x-a)(x-b)=x2+(a+b)x+ab

Here, x=100

a=-5

b=-4

We get, 95×96=(100-5)×(100-4)

=(100)2+100(-5+(-4))+(-5×-4)

=10000-900+20

=9120

(iii) 104 × 96

Solution:

104×96=(100+4)×(100–4)

Using identity, [(a+b)(a-b)= a2-b2]

Here, a=100

b=4

We get, 104×96=(100+4)×(100–4)

=(100)2–(4)2

=10000–16

=9984

Q3. Factorize the following using appropriate identities:
(i) 9x2+6xy+y2

Solution:

9x2+6xy+y2=(3x)2+(2×3x×y)+y2

Using identity, x2 + 2xy + y2= (x + y)2

Here, x=3x

y=y

9x2+6xy+y2=(3x)2+(2×3x×y)+y2

=(3x+y)2

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

(ii) 4y2−4y+1

Solution:

4y2−4y+1=(2y)2–(2×2y×1)+12

Using identity, x2 – 2xy + y2= (x – y)2

Here, x=2y

y=1

4y2−4y+1=(2y)2–(2×2y×1)+12

=(2y–1)2

=(2y–1)(2y–1)

(iii) x2–y2/100

Solution:

x2–y2/100 = x2–(y/10)2

Using identity, x2 – y2= (x – y) (x y)

Here,

x=x

y=y/10

x2 – y2/100= x2–(y/10)2

=(x–y/10)(x+y/10)

Q4. Expand each of the following, using suitable identities:
(i) (x+2y+4z)2
(ii) (2x−y+z)2
(iii) (−2x+3y+2z)2
(iv) (3a – 7b – c)2
(v) (–2x + 5y – 3z)2
(vi) (a–b+1)2

Solutions:

(i) (x+2y+4z)2

Solution:

Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Here, x=x

y=2y

z=4z

(x+2y+4z)2 =x2+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x)

=x2+4y2+16z2+4xy+16yz+8xz

(ii) (2x−y+z)2 

Solution:

Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Here, x=2x

y=−y

z=z

(2x−y+z)2 =(2x)2+(−y)2+z2+(2×2x×−y)+(2×−y×z)+(2×z×2x)

=4x2+y2+z2–4xy–2yz+4xz

(iii) (−2x+3y+2z)2

Solution:

Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Here, x= −2x

y=3y

z=2z

(−2x+3y+2z)2 =(−2x)2+(3y)2+(2z)2+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)

=4x2+9y2+4z2–12xy+12yz–8xz

(iv) (3a – 7b – c)2

Solution:

Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Here, x= 3a

y= – 7b

z= – c

(3a – 7b – c)2 =(3a)2+(– 7b)2+(– c)2+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)

=9a2 + 49b2 + c2– 42ab+14bc–6ca

(v) (–2x + 5y – 3z)2

Solution:

Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Here, x= –2x

y= 5y

z= – 3z

(–2x+5y–3z)2 =(–2x)2+(5y)2+(–3z)2+(2×–2x × 5y)+(2× 5y ×– 3z)+(2×–3z ×–2x)

=4x2 + 25y2 + 9z2– 20xy–30yz+12zx

(vi) (1/4a – 1/2b+1)2

Solution:

Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Here, x= 1/4a

y= –1/2b

z= 1

(1/4a –1/2b +1)2 =(1/4a)2+(–1/2b)2+(1)2+(2×1/4a × –1/2b)+(2× –1/2b ×1)+(2×1×1/4a)

=1/16a2+1/4b2+12–2/8ab– 2/2b +2/4a

= 1/16a2+1/4b2+1–1/4ab – b +1/2a

Q5. Factorize:

(i) 4x2+9y2+16z2+12xy–24yz–16xz
(ii) 2x2+y2+8z2–2xy+4yz–8xz

Solutions:

(i) 4x2+9y2+16z2+12xy–24yz–16xz

Solution:

Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

We can say that, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2

4x2+9y2+16z2+12xy–24yz–16xz =(2x)2+(3y)2+(−4z)2+(2×2x×3y)+(2×3y×−4z)+(2×−4z×2x)

=(2x+3y–4z)2

=(2x+3y–4z)(2x+3y–4z)

(ii) 2x2+y2+8z2–2√2xy+4√2yz–8xz

Solution:

Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

We can say that, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2

2x2+y2+8z2–2√2xy+4√2yz–8xz  =(-√2x)2+(y)2+(2√2z)2+(2×−√2x×y)+(2×y×2√2z)+(2×2√2√2z×−√2x)

=(−√2x+y+2√2z)2

=(−√2x+y+2√2z)(−√2x+y+2√2z)

Q6. Write the following cubes in expanded form:
(i) (2x+1)3
(ii) (2a−3b)3
(iii) (x+1)3
(iv) (x−y)3

Solutions:

(i) (2x+1)3

Solution:

Using identity, (x + y)3 = x3 + y3 + 3xy (x + y)

(2x+1)3=(2x)3+13+(3×2x×1)(2x+1)

=8x3+1+6x(2x+1)

=8x3+12x2+6x+1

(ii) (2a−3b)3

Solution:

Using identity, (x – y)3 = x3 – y3 – 3xy(x – y)

(2a−3b)3=(2a)3−(3b)3–(3×2a×3b)(2a–3b)

=8a3–27b3–18ab(2a–3b)

=8a3–27b3–36a2b+54ab2

(iii) (3/2x+1)3

Solution:

Using identity, (x + y)3 = x3 + y3 + 3xy (x + y)

(3/2x+1)3 =(3/2x)3+13+(3×3/2x×1)(3/2x+1)

=27/8x3+1+9/2x(3/2x+1)

=27/8x3+1+27/4x2+9/2x

=27/8x3+27/4x2+9/2x+1

(iv) (x−2/3y)3

Solution:

Using identity, (x – y)3 = x3 – y3 – 3xy(x – y)

(x−2/3y)3 =(x)3–(2/3y)3–(3×x×2/3y)(x–2/3y)

=(x)3–8/27y3–2xy(x– 2/3y)

=(x)3–8/27y3–2x2y+4/3xy2

Q7. Evaluate the following using suitable identities: 
(i) (99)3
(ii) (102)3
(iii) (998)3

Solutions:

(i) (99)3

Solution:

We can write 99 as 100–1

Using identity, (x – y)3 = x3 – y3 – 3xy(x – y)

(99)3 = (100–1)3

=(100)3–13–(3×100×1)(100–1)

= 1000000 – 1 – 300(100 – 1)

= 1000000 – 1 – 30000 + 300

= 970299

(ii) (102)3

Solution:

We can write 102 as 100+2

Using identity, (x + y)3 = x3 + y3 + 3xy (x + y)

(100+2)3 =(100)3+23+(3×100×2)(100+2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3

Solution:

We can write 99 as 1000–2

Using identity, (x – y)3 = x3 – y3 – 3xy(x – y)

(998)3 =(1000–2)3

=(1000)3–23–(3×1000×2)(1000–2)

= 1000000000 – 8 – 6000(1000 – 2)

= 1000000000 – 8- 6000000 + 12000

= 994011992

Q8. Factorise each of the following:
(i) 8a3+b3+12a2b+6ab2
(ii) 8a3–b3–12a2b+6ab2
(iii) 27 – 125a3 – 135a + 225a2   
(iv) 64a3–27b3–144a2b+108ab2
(v) 27p3 –1/216 −(9/2)p2+(1/4)p

Solutions:

(i) 8a3+b3+12a2b+6ab2

Solution:

The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2

8a3+b3+12a2b+6ab2 =(2a)3+b3+3(2a)2b+3(2a)(b)2

=(2a+b)3

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

Here, the identity, (x + y)3 = x3 + y3 + 3xy (x + y) is used.

 

(ii) 8a3–b3–12a2b+6ab2

Solution:

The expression, 8a3–b3−12a2b+6ab2 can be written as (2a)3–b3–3(2a)2b+3(2a)(b)2

8a3–b3−12a2b+6ab2 =(2a)3–b3–3(2a)2b+3(2a)(b)2

=(2a–b)3

=(2a–b)(2a–b)(2a–b)

Here, the identity, (x – y)3 = x3 – y3 – 3xy(x – y) is used.

 

(iii) 27 – 125a3 – 135a + 225a2 

Solution:

The expression, 27 – 125a3 – 135a + 225a2 can be written as 33–(5a)3–3(3)2(5a)+3(3)(5a)2

27–125a3–135a+225a2 = 33–(5a)3–3(3)2(5a)+3(3)(5a)2

=(3–5a)3

=(3–5a)(3–5a)(3–5a)

Here, the identity, (x – y)3 = x3 – y3 – 3xy(x – y) is used.

(iv) 64a3–27b3–144a2b+108ab2

Solution:

The expression, 64a3–27b3–144a2b+108ab2 can be written as (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2

64a3–27b3–144a2b+108ab2 = (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2

=(4a–3b)3

=(4a–3b)(4a–3b)(4a–3b)

Here, the identity, (x – y)3 = x3 – y3 – 3xy(x – y) is used.

(v) 27p3 – 1/216 − 9/2p2+ 1/4p

Solution:

The expression, 27p3 –1/216 − 9/2p2+ 1/4p can be written as (3p)3–(1/6)3–3(3p)2(1/6)+3(3p)(1/6)2

27p3 – 1/216 − 9/2p2+ 1/4p = (3p)3–(1/6)3–3(3p)2(1/6)+3(3p)(1/6)2

= (3p–(1/6))3

= (3p–(1/6))(3p–(1/6))(3p–(1/6))

 

Q9. Verify:
(i) x3+y3=(x+y)(x2–xy+y2)
(ii) x3–y3=(x–y)(x2+xy+y2)

Solutions:

(i) x3+y3=(x+y)(x2–xy+y2)

We know that, (x+y)3 =x3+y3+3xy(x+y)

⇒x3+y3=(x+y)3–3xy(x+y)

⇒x3+y3=(x+y)[(x+y)2–3xy]

Taking(x+y) common ⇒x3+y3=(x+y)[(x2+y2+2xy)–3xy]

⇒x3+y3=(x+y)(x2+y2–xy)

(ii) x3–y3=(x–y)(x2+xy+y2)

We know that,(x–y)3 =x3–y3–3xy(x–y)

⇒x3−y3=(x–y)3+3xy(x–y)

⇒x3−y3=(x–y)[(x–y)2+3xy]

Taking(x+y) common⇒x3−y3=(x–y)[(x2+y2–2xy)+3xy]

⇒x3+y3=(x–y)(x2+y2+xy)

Q10. Factorize each of the following:
(i) 27y3+125z3
(ii) 64m3–343n3

Solutions:

(i) 27y3+125z3

The expression, 27y3+125z3 can be written as (3y)3+(5z)3

27y3+125z3 =(3y)3+(5z)3

We know that, x3+y3=(x+y)(x2–xy+y2)

27y3+125z3 =(3y)3+(5z)3

=(3y+5z)[(3y)2–(3y)(5z)+(5z)2]

=(3y+5z)(9y2–15yz+25z2)

(ii) 64m3–343n3

The expression, 64m3–343n3 can be written as (4m)3–(7n)3

64m3–343n3 = (4m)3–(7n)3

We know that, x3–y3=(x–y)(x2+xy+y2)

64m3–343n3 =(4m)3–(7n)3

=(4m-7n)[(4m)2+(4m)(7n)+(7n)2]

=(4m-7n)(16m2+28mn+49n2)

 

Q11. Factorise : 27x3+y3+z3–9xyz 

Solution:

The expression 27x3+y3+z3–9xyz can be written as (3x)3+y3+z3–3(3x)(y)(z)

27x3+y3+z3–9xyz  =(3x)3+y3+z3–3(3x)(y)(z)

We know that, x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

27x3+y3+z3–9xyz  =(3x)3+y3+z3–3(3x)(y)(z)

=(3x+y+z)(3x)2+y2+z2–3xy–yz–3xz

=(3x+y+z)(9x2+y2+z2–3xy–yz–3xz)

Q12. Verify that:

x3+y3+z3–3xyz=(x+y+z)[(x–y)2+(y–z)2+(z–x)2]

Solution:

We know that,

x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2–xy–yz–xz)

⇒x3+y3+z3–3xyz =×(x+y+z)[2(x2+y2+z2–xy–yz–xz)]

= (x+y+z)(2x2+2y2+2z2–2xy–2yz–2xz)

= (x+y+z)[(x2+y2−2xy)+(y2+z2–2yz)+(x2+z2–2xz)]

= (x+y+z)[(x–y)2+(y–z)2+(z–x)2]

Q13. If x + y + z = 0, show that x3+y3+z3=3xyz.

Solution:

We know that,

x3+y3+z3=3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – xz)

Now, according to the question, let (x + y + z) = 0,

then, x3+y3+z3=3xyz =(0)(x2+y2+z2–xy–yz–xz)

⇒x3+y3+z3–3xyz =0

⇒ x3+y3+z3 =3xyz

Hence Proved

Q14. Without actually calculating the cubes, find the value of each of the following:
(i) (−12)3+(7)3+(5)3
(ii) (28)3+(−15)3+(−13)3

(i) (−12)3+(7)3+(5)3

Solution:

(−12)3+(7)3+(5)3

Let a= −12

b= 7

c= 5

We know that if x + y + z = 0, then x3+y3+z3=3xyz.

Here, −12+7+5=0

(−12)3+(7)3+(5)3 = 3xyz

=

=

(ii) (28)3+(−15)3+(−13)3

Solution:

(28)3+(−15)3+(−13)3

Let a= 28

b= −15

c= −13

We know that if x + y + z = 0, then x3+y3+z3=3xyz.

Here, x + y + z = 28 – 15 – 13 = 0

(28)3+(−15)3+(−13)3= 3xyz

= 0+3(28)(−15)(−13)

=16380

Q15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 
(i) Area : 25a2–35a+12
(ii) Area : 35y2+13y–12

Solution:

(i) Area : 25a2–35a+12

Using the splitting the middle term method,

We have to find a number whose sum= -35 and product=2512=300

We get -15 and -20 as the numbers [-15+-20=-35 and -3-4=300]

25a2–35a+12 =25a2–15a−20a+12

=5a(5a–3)–4(5a–3)

=(5a–4)(5a–3)

Possible expression for length  = 5a – 4

Possible expression for breadth  = 5a – 3

(ii) Area : 35y2+13y–12

Using the splitting the middle term method,

We have to find a number whose sum= 13 and product=3512=420

We get -15 and 28 as the numbers [-15+28=-35 and -15=420]

35y2+13y–12 =35y2–15y+28y–12

=5y(7y–3)+4(7y–3)

=(5y+4)(7y–3)

Possible expression for length  = (5y + 4)

Possible expression for breadth  = (7y – 3)

Q16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? 
(i) Volume : 3x2–12x
(ii) Volume : 12ky2+8ky–20k

Solution:

(i) Volume : 3x2–12x

3x2–12x can be written as 3x(x – 4) by taking 3x out of both the terms.

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x – 4)

(ii) Volume : 12ky2+8ky –20k

12ky2+8ky –20k can be written as 4k(3y2+2y–5) by taking 4k out of both the terms.

12ky2+8ky–20k =4k(3y2+2y–5)

[Here, 3y2+2y–5 can be written as 3y2+5y–3y–5 using splitting the middle term method.]

=4k(3y2+5y–3y–5)

=4k[y(3y+5)–1(3y+5)]

=4k(3y+5)(y–1)

Possible expression for length = 4k

Possible expression for breadth = (3y +5)

Possible expression for height = (y -1)

NCERT Solutions for class 9 Maths Chapter 2- Polynomials

As this is one of the important topics in maths, It comes under the unit – Algebra which has a weightage of 20 marks in class 9 maths board exams.

  • This chapter talks about
  • Polynomials in One Variable
  • Zeroes of a Polynomial
  • Remainder Theorem
  • Factorization of Polynomials
  • Algebraic Identities

List of Exercises in class 9 Maths Chapter 2 :

Exercise 2.1 Solutions 5 Questions

Exercise 2.2 Solutions 4 Questions

Exercise 2.3 Solutions 3 Questions

Exercise 2.4 Solutions 5 Questions

Exercise 2.5 Solutions 16 Questions

NCERT Solutions for class 9 Maths Chapter 2- Polynomials

NCERT solutions for class 9 maths chapter 2 – Polynomials is the second chapter of class 9 Maths. Polynomials are introduced and discussed in detail here. The chapter discusses the Polynomials and their applications. The introduction of the chapter includes whole numbers, integers, and rational numbers.

The chapter starts with the introduction of Polynomials in section 2.1 followed by two very important topics in section 2.2 and 2.3

  • Polynomials in one Variable – Discussion of Linear, Quadratic and Cubic Polynomial.
  • Zeroes of a Polynomial – A zero of a polynomial need not be zero and can have more than one zero.
  • Real Numbers and their Decimal Expansions – Here you study the decimal expansions of real numbers and see whether it can help in distinguishing between rational and irrationals.

Next, it discusses the following topics.

  • Representing Real Numbers on the Number Line – In this the solutions for 2 problems in Exercise 2.4.
  • Operations on Real Numbers – Here you explore some of the operations like addition, subtraction, multiplication, and division on the irrational numbers.
  • Laws of Exponents for Real Numbers – Use these laws of exponents to solve the questions.

Key advantages of NCERT Solutions for class 9 Maths Chapter 2- Polynomials

  • These NCERT Solutions for class 9 Maths helps you solve and revise the whole syllabus of class 9.
  • After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
  • It follows NCERT guidelines which help in preparing the students accordingly.
  • It contains all the important questions from the examination point of view.
  • It helps in scoring well in maths in board exams.

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