NCERT Solutions for Class 9 Maths Chapter 2 Polynomials ( New Syllabus)

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1 NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 ,Page 29

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials are provided here. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials which will help students to solve problems easily. 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 NCERT Solutions for Class 9 Maths Chapter 2 Polynomials , students are introduced to many important topics that will be helpful for those who wish to pursue Mathematics as a subject in higher studies. NCERT Solutions help students to prepare for their upcoming exams by covering the updated CBSE syllabus for 2023-24 and its guidelines. All the solutions are provided in new syllabus of NCERT Solutions for Class 9 Maths Chapter 2 Polynomials.

NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 ,Page 29

Question 1. 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+√2 can be written as y2+2y0
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+2 is a polynomial in one variable.
(iii) 3√t+t√2
Solution:
The equation 3√t+t√2 can be written as 3t1/2+√2t
Though t is the only variable in the given equation, the power of t (i.e., 1/2) is not a whole number. Hence, we can say that the expression 3√t+t√2 is not a polynomial in one variable.
(iv) y+2/y
Solution:
The equation y+2/y can be written as y+2y-1
Though is the only variable in the given equation, the power of y (i.e., -1) is not a whole number. Hence, we can say that the expression y+2/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.

Question -2 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 the coefficient is the number which multiplies the variable.

Here, the number that multiplies the variable x2 is 1

Hence, the coefficient of xin 2+x2+x is 1.

(ii) 2–x2+x3

Solution:

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

We know that the 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

Hence, the coefficient of xin 2–x2+xis -1.

(iii) (π/2)x2+x

Solution:

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

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

Here, the number that multiplies the variable x2 is π/2.

Hence, the coefficient of xin (π/2)x+x is π/2.

(iii)√2x-1

Solution:

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

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

Here, the number that multiplies the variable x2is 0

Hence, the coefficient of xin √2x-1 is 0.

Question 3. 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.

For example,  3x35+5

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

For example,  4x100

Question 4. 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 t 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 = 3×1 = 3× x0

The power of the variable here is: 0

Hence, the degree of 3 is 0.

5. 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 is called 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–xis 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 ris 2

The degree is 2

Hence, r2is a quadratic polynomial.

(vii) 7x3

Solution:

The highest power of 7xis 3

The degree is 3

Hence, 7x3 is a cubic polynomial.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

 EXERCISE 2.2 PAGE NO. 31

1. 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

2. 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)= 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)= 0

p(1) = (1)= 1

p(2) = (2)= 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

3. 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

∴ 4/5 is 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) = 0= 0

∴ 0 is a zero of p(x).

(vi) p(x) = lx+m, x = −m/l

Solution:

For, x = -m/; p(x) = lx+m

∴ p(-m/l)l(-m/l)+m = −m+m = 0

∴ -m/l is 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).

4. Find the zero of the polynomials 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, a≠0

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).

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials EXERCISE 2.3 PAGE -35

1. 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) x– 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 ≠ 0

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

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x) = 2x3+x2–2x–1, g(x) = x+1

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+3x2+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).

3. Find the value of k, if x–1 is a factor of p(x) in each of the following cases:

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

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) = kx22x+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= 3/2

4. Factorise:

(i) 12x2–7x+1

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -7 and product =1×12 = 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 = 2×3 = 6

We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 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×-6 = -36

We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -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×-4 = -12

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

3x2–x–4 = 3x2–4x+3x–4

= x(3x–4)+1(3x–4)

= (3x–4)(x+1)

5. Factorise:

(i) x3–2x2–x+2

Solution:

Let p(x) = x3–2x2–x+2

Factors of 2 are ±1 and ± 2

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 2 Polynomials

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:

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

(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 the 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 2 Polynomials

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 the 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 2 Polynomials

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)

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials EXERCISE 2.4 PAGE 40

1. 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) = x2–y 2

[Here, x = y2and y = 3/2]

We get,

(y2+3/2)(y2–3/2) = (y2)2–(3/2)2

= y4–9/4

2. 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

3. Factorise the following using appropriate identities:

(i) 9x2+6xy+y2

Solution:

9x2+6xy+y= (3x)2+(2×3x×y)+y2

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

Here, x = 3x

y = y

9x2+6xy+y= (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)+1

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

Here, x = x

y = y/10

x2–y2/100 = x2–(y/10)2

= (x–y/10)(x+y/10)

4. 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) ((1/4)a-(1/2)b +1)2

Solution:

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

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

Here, x = x

y = 2y

z = 4z

(x+2y+4z)= 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 

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

Here, x = 2x

y = −y

z = z

(2x−y+z)= (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)= (−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)= (3a)2+(– 7b)2+(– c)2+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)

= 9a2 + 49b+ 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)= (–2x)2+(5y)2+(–3z)2+(2×–2x × 5y)+(2× 5y×– 3z)+(2×–3z ×–2x)

= 4x2+25y+9z2– 20xy–30yz+12zx

(vi) ((1/4)a-(1/2)b+1)2

Solution:

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

Here, x = (1/4)a

y = (-1/2)b

z = 1

 

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

5. Factorise:

(i) 4x2+9y2+16z2+12xy–24yz–16xz

(ii) 2x2+y2+8z2–2√2xy+4√2yz–8xz

Solution:

(i) 4x2+9y2+16z2+12xy–24yz–16xz

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

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×−√2x)

= (−√2x+y+2√2z)2

= (−√2x+y+2√2z)(−√2x+y+2√2z)

6. Write the following cubes in expanded form:

(i) (2x+1)3

(ii) (2a−3b)3

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

(iv) (x−(2/3)y)3

Solution:

(i) (2x+1)3

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+1+6x(2x+1)

= 8x3+12x2+6x+1

(ii) (2a−3b)3

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

(2a−3b)= (2a)3−(3b)3–(3×2a×3b)(2a–3b)

= 8a3–27b3–18ab(2a–3b)

= 8a3–27b3–36a2b+54ab2

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

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

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

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

(iv)  (x−(2/3)y)3

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

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

7. 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)= (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)=(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)=(1000–2)3

=(1000)3–23–(3×1000×2)(1000–2)

= 1000000000–8–6000(1000– 2)

= 1000000000–8- 6000000+12000

= 994011992

8. 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+6ab= (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+6ab= (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+225a=
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+108ab2can 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)−(9/2) p2+(1/4)p

Solution:

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

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

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

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

Taking x = 3p and y = 1/6

= (3p–1/6)3

= (3p–1/6)(3p–1/6)(3p–1/6)

9. Verify:

(i) x3+y= (x+y)(x2–xy+y2)

(ii) x3–y= (x–y)(x2+xy+y2)

Solutions:

(i) x3+y= (x+y)(x2–xy+y2)

We know that, (x+y)3 = x3+y3+3xy(x+y)

⇒ x3+y= (x+y)3–3xy(x+y)

⇒ x3+y= (x+y)[(x+y)2–3xy]

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

⇒ x3+y= (x+y)(x2+y2–xy)

(ii) x3–y= (x–y)(x2+xy+y2

We know that, (x–y)3 = x3–y3–3xy(x–y)

⇒ x3−y= (x–y)3+3xy(x–y)

⇒ x3−y= (x–y)[(x–y)2+3xy]

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

10. Factorise each of the following:

(i) 27y3+125z3

(ii) 64m3–343n3

Solutions:

(i) 27y3+125z3

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

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

We know that, x3+y= (x+y)(x2–xy+y2)

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

= (3y+5z)[(3y)2–(3y)(5z)+(5z)2]

= (3y+5z)(9y2–15yz+25z2)

(ii) 64m3–343n3

The expression, 64m3–343n3can be written as (4m)3–(7n)3

64m3–343n= (4m)3–(7n)3

We know that, x3–y= (x–y)(x2+xy+y2)

64m3–343n= (4m)3–(7n)3

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

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

11. 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)

12. Verify that:

x3+y3+z3–3xyz = (1/2) (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 = (1/2)(x+y+z)[2(x2+y2+z2–xy–yz–xz)]

= (1/2)(x+y+z)(2x2+2y2+2z2–2xy–2yz–2xz)

= (1/2)(x+y+z)[(x2+y2−2xy)+(y2+z2–2yz)+(x2+z2–2xz)]

= (1/2)(x+y+z)[(x–y)2+(y–z)2+(z–x)2]

13. If  x+y+z = 0, show that x3+y3+z= 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+z-3xyz = (0)(x2+y2+z2–xy–yz–xz)

⇒ x3+y3+z3–3xyz = 0

⇒ x3+y3+z= 3xyz

Hence Proved

14. 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

Solution:

(i) (−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)= 3xyz

= 3×-12×7×5

= -1260

(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+z= 3xyz.

Here, x+y+z = 28–15–13 = 0

(28)3+(−15)3+(−13)= 3xyz

= 0+3(28)(−15)(−13)

= 16380

15. 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 =25×12 = 300

We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20 = 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 = 35×-12 = 420

We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=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)

16. 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)

 

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