NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

NCERT Solutions Class 9 Maths
Chapter – 2 (Polynomials)

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 2 Polynomials Exercise 2.5 has been provided here to help the students in solving the questions from this exercise.

Chapter 2: Polynomials

Exercise – 2.5

1. Use suitable identities to find the following products:
(i) (x + 4)(x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (y²+ $\mathbf{\frac{3}{2}}$ ) (y²– $\mathbf{\frac{3}{2}}$ )
(v) (3 – 2x) (3 + 2x)

Answer –

(i) (x + 4)(x + 10)
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)
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)
Using the identity, (x + a)(x + b) = x²+ (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+ $\mathbf{\frac{3}{2}}$ )(y²– $\mathbf{\frac{3}{2}}$ )
Using the identity, (x + y)(x – y) = x²– y²Here, x = y²and y = $\frac{3}{2}$
We get,
(y²+ $\frac{3}{2}$ )(y²– $\frac{3}{2}$ )

= (y²)²– ( $\frac{3}{2}$ )²

= y⁴– $\frac{9}{4}$

2. Evaluate the following products without multiplying directly:
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96

Answer –
(i) 103 × 107
103 × 107 = (100 + 3) × (100 + 7)
Using identity, [(x + a)(x + b) = x²+ (a + b)x + ab]
Here, x = 100, a = 3 and b = 7
We get,
103 × 107 = (100+3)×(100+7)
⇒ (100)²+ (3+7)100+(3×7)
⇒ 10000 + 1000 + 21
⇒ 11021

(ii) 95 × 96
95 × 96 = (100 – 5)×(100 – 4)
Using identity, [(x – a)(x – b) = x²– (a + b)x + ab]
Here, x = 100, a = -5 and b = -4
We get,
95 × 96 = (100 – 5)×(100 – 4)
⇒ (100)²+ 100(-5 + (-4)) + (-5×-4)
= 10000 – 900 + 20
= 9120

(iii) 104 × 96
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. Factorize the following using appropriate identities:
(i) 9x²+ 6xy + y²(ii) 4y²– 4y + 1
(iii) x² – $\mathbf{\frac{y^2}{100}}$

Answer –
(i) 9x²+ 6xy + y²= (3x)²+ (2 × 3x × y) + y²Using identity, x²+ 2xy + y²= (x + y)²Here, x = 3x, y = y
9x²+ 6xy + y²= (3x)²+ (2 × 3x × y) + y²= (3x + y)²= (3x + y) (3x + y)

(ii) 4y²−4y + 1
= (2y)²–(2 × 2y × 1) + 1
Using identity, x² – 2xy + y²= (x – y)²Here, x = 2y, y = 1
4y²−4y+1
= (2y)²– (2 × 2y × 1) + 1²= (2y – 1)²= (2y – 1) (2y – 1)

(iii) x² – $\mathbf{\frac{y^2}{100}}$

= x²– $\left ( \frac{y}{10} \right )^2$
Using identity, x²– y²= (x – y)(x + y)
Here, x = x, y = $\frac{y}{10}$
x²– $\frac{y^2}{100}$

= x²– $\left ( \frac{y}{10} \right )^2$

= (x – $\frac{y}{10}$ ) (x + $\frac{y}{10}$ )

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) $\mathbf{\left [ \frac{1}{4}a - \frac{1}{2}b +1\right ]}^2$

Answer –
(i) (x + 2y + 4z)²Using identity, (x + y + z)² = x²+ y²+ z²+ 2xy + 2yz + 2zx
Here, x = x, y = 2y and 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
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)²Using identity, (x + y + z)² = x²+ y²+ z²+ 2xy + 2yz + 2zx
Here, x = −2x, y = 3y and 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)²Using identity (x + y + z)² = x²+ y²+ z²+ 2xy + 2yz + 2zx
Here, x = 3a, y = – 7b and 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)²Using identity, (x + y + z)² = x²+ y²+ z²+ 2xy + 2yz + 2zx
Here, x = –2x, y = 5y and 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) $\mathbf{\left [ \frac{1}{4}a - \frac{1}{2}b +1\right ]}^2$

Using identity, (x + y + z)² = x²+ y²+ z²+ 2xy + 2yz + 2zx
Here, x = $\frac{1}{4}$ a, y = $-\frac{1}{2}$ b and z = 1
$\left [ \frac{1}{4}a - \frac{1}{2}b +1\right ]^2$

= $\left ( \frac{1}{4}a \right )^2 + \left ( \frac{1}{2}b \right )^2 + 1^2 + \left ( 2\times \frac{1}{4}a\times -\frac{1}{2}b\right ) + \left ( 2\times -\frac{1}{2}b\times 1\right) + \left ( 2\times 1\times \frac{1}{4}a\right )$

= $\frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{2}{8}ab -\frac{2}{2}b +\frac{2}{4}a$

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

5. Factorize:
(i) 4x²+ 9y²+ 16z²+ 12xy – 24yz – 16xz
(ii) 2x²+ y²+ 8z²– 2√2xy + 4√2yz – 8xz

Answer –
(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)²4x²+ 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)²2x²+ 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) $\mathbf{\left [ \frac{3}{2}x + 1\right ]^3}$ (iv) $\mathbf{\left [ x-\frac{2}{3}y \right ]^3}$

Answer –
(i) (2x + 1)³Using 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)³Using 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) $\mathbf{\left [ \frac{3}{2}x + 1\right ]^3}$

Using identity, (x + y)³ = x³+ y³+ 3xy(x + y)
$\left [ \frac{3}{2}x + 1\right ]^3$ = $\left ( \frac{3}{2}x \right )^3$ + 1³+ (3× $\frac{3}{2}x$ × 1)( $\frac{3}{2}x$ + 1)

= $\frac{27}{8}x^3 + 1 + \frac{9}{2}x \left ( \frac{3}{2}x +1 \right )$

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

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

(iv) $\mathbf{\left [ x-\frac{2}{3}y \right ]^3}$
Using identity, (x – y)³ = x³– y³– 3xy(x – y)

$\left [ x-\frac{2}{3}y \right ]^3$ = x³– $\left ( \frac{2}{3}y \right )^3$ – (3 × x × $\frac{2}{3}y$ ) (x – $\frac{2}{3}y$ )

= x³– $\frac{8}{27}y^3$ – 2xy (x – $\frac{2}{3}y$ )

= x³– $\frac{8}{27}y^3$ – 2x²y + $\frac{4}{3}xy^2$

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

Answer –

(i) (99)³We have, 99 = (100 -1)
∴ 99³ = (100 – 1)³= (100)³ – 1³ – 3(100)(1)(100 -1)
[Using (a – b)³ = a³ – b³ – 3ab (a – b)]
= 1000000 – 1 – 300(100 – 1)
= 1000000 -1 – 30000 + 300
= 1000300 – 30001
= 970299

(ii) (102)³We have, 102 =100 + 2
∴ 102³ = (100 + 2)³= (100)³ + (2)³ + 3(100)(2)(100 + 2)
[Using (a + b)³ = a³ + b³ + 3ab (a + b)]
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208

(iii) (998)³We have, 998 = 1000 – 2
∴ (998)³ = (1000-2)³
= (1000)³– (2)³ – 3(1000)(2)(1000 – 2)
[Using (a – b)³ = a³ – b³ – 3ab (a – b)]
= 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– $\mathbf{\frac{1}{216}}$ − $\mathbf{\frac{9}{2}}$ p²+ $\mathbf{\frac{1}{4}}$ p

Answer –

(i) 8a³+ b³+ 12a²b + 6ab²= (2a)³ + (b)³ + 6ab(2a + b)
= (2a)³ + (b)³ + 3(2a)(b)(2a + b)
= (2a + b)³[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (2a + b)(2a + b)(2a + b)

(ii) 8a³– b³– 12a²b + 6ab²= (2a)³ – (b)³ – 3(2a)(b)(2a – b)
= (2a – b)³[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (2a – b)(2a – b)(2a – b)

(iii) 27 – 125a³– 135a + 225a²
= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)
= (3 – 5a)³[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (3 – 5a)(3 – 5a)(3 – 5a)

(iv) 64a³– 27b³– 144a²b + 108ab²= (4a)³ – (3b)³ – 3(4a)(3b)(4a – 3b)
= (4a – 3b)³[Using a³ – b³ – 3 ab(a – b) = (a – b)³]
= (4a – 3b)(4a – 3b)(4a – 3b)

(v) 27p³– $\mathbf{\frac{1}{216}}$ − $\mathbf{\frac{9}{2}}$ p²+ $\mathbf{\frac{1}{4}}$ p

= (3p)³– $\left ( \frac{1}{6} \right )^3$ − $\left ( \frac{9}{2} \right )$ p²+ $\frac{1}{4}$ p

= (3p)³– $\left ( \frac{1}{6} \right )^3$ − 3(3p× $\frac{1}{6}$ )(3p – $\frac{1}{6}$ ) Using (x – y)³ = x³ – y³ – 3xy (x – y)
Taking x = 3p and y = $\frac{1}{6}$

= (3p – $\frac{1}{6}$ )³

= (3p – $\frac{1}{6}$ )(3p – $\frac{1}{6}$ )(3p – $\frac{1}{6}$ )

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

Answer –
(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)

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

Answer –
(i) 27y³+ 125z³We have, 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) 64m³–343n³We have, 64m³ – 343n³ = (4m)³ – (7n)³We 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)

11. Factorise : 27x³+ y³+ z³– 9xyz

Answer –
We have, 27x³ + y³ + z³ – 9xyz = (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
Using the identity, x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
We have, (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
= (3x + y + z)[(3x)³ + y³ + z³ – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x² + y² + z² – 3xy – yz – 3zx)

12. Verify that: x³+ y³+ z³– 3xyz = $\mathbf{\frac{1}{2}}$ (x + y + z)[(x – y)²+ (y – z)²+ (z – x)²]

Answer – We know that, x³+ y³+ z³− 3xyz = (x + y + z)(x²+ y²+ z²– xy – yz – xz)
⇒ x³+ y³+ z³– 3xyz = $\frac{1}{2}$ (x + y + z)[2(x²+ y²+ z²– xy – yz – xz)]
= $\frac{1}{2}$ (x + y + z)(2x²+ 2y²+ 2z²– 2xy – 2yz – 2xz)

= $\frac{1}{2}$ (x + y + z)[(x²+ y²− 2xy) + (y²+ z²– 2yz) + (x²+ z²– 2xz)]

= $\frac{1}{2}$ (x + y + z)[(x – y)²+ (y – z)²+ (z – x)²]

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

Answer – Since, x + y + z = 0
⇒ x + y = -z
⇒ (x + y)³ = (-z)³
⇒ x³ + y³ + 3xy(x + y) = -z³
⇒ x³ + y³ + 3xy(-z) = -z³ [∵ x + y = -z]
⇒ x³ + y³ – 3xyz = -z³
⇒ x³ + y³ + z³ = 3xyz
Hence, if x + y + z = 0, then
x³ + y³ + z³ = 3xyz

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

Answer –

(i) (−12)³+ (7)³+ (5)³Let x = −12, y = 7 and z= 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
= 3 × -12 × 7 × 5
= -1260

(ii) (28)³+ (−15)³+ (−13)³Let x = 28, y= −15 and z = −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

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

Answer –
Area of a rectangle = (Length) x (Breadth)
(i) Area : 25a²– 35a + 12
25a² – 35a + 12
= 25a² – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
= (5a – 4)(5a – 3)
Possible expression for length = 5a–4
Possible expression for breadth = 5a –3

(ii) Area : 35y²+ 13y – 12
35y²+ 13y -12
= 35y² + 28y – 15y – 12
= 7y(5y + 4) – 3(5y + 4)
= (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

Answer –
Volume of a cuboid = (Length) x (Breadth) x (Height)

(i) Volume : 3x²–12x
We have, 3x² – 12x = 3(x² – 4x)
= 3 × x × (x – 4)
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x – 4)

(ii) Volume : 12ky²+ 8ky – 20k
We have, 12ky² + 8ky – 20k
= 4[3ky² + 2ky – 5k]
= 4[k(3y² + 2y – 5)]
= 4 × k × (3y² + 2y – 5)
= 4k[3y² – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) × (y – 1)]
= 4k × (3y + 5) × (y – 1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y – 1)

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5