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Question 1:

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

(ii) $2x^{2}+y^{2}+_8z^{2}-2\sqrt{2}xy+4\sqrt{2}yz-8xz$

Solution 1:

It is known that,

(i)

(ii)

Comments

Tasneem

Question 2:

Write the following cubes in expanded form: (i) $\left ( 2x+_1 \right )^{3}$ (ii) $\left ( 2a-3b^ \right )^{3}$ (iii) $\left ( \frac{3}{2}x+1 \right )^{3}$ (iv) $\left ( x-\frac{2}{3}y \right )^{3}$

Solution 2:

It is known that,

(i)

(ii)

(iii)

(vi)

Comments

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Question 3:

Evaluate the following using suitable identities: (i) $\left ( 99 \right )^{3}$ (ii) $\left ( 102 \right )^{^{3}}$ (iii) $\left ( 998 \right )^{3}$

Solution 3:

(i) (99)³= (100 − 1)³

= (100)³ − (1)³ − 3(100) (1) (100 − 1)

= 970299

(ii) (102)³ = (100 + 2)³

= (100)³ + (2)³ + 3(100) (2) (100 + 2)

= 1061208.00

(iii) (998)³= (1000 − 2)³

= (1000)³ − (2)³ − 3(1000) (2) (1000 − 2)

= 994011992.00

Comments

Tasneem

Question 4:

Factorise each of the following: (i) $8a^{_{3}}+b^{3}+12a^{2}b+6ab^{2}$ (ii) $8a^{3}-b^{3}-12a^{2}b+6ab^{2}$ (iii) $27-125a^{3}-135a+225a^{2}$ (iv) $64a^{3}-27b^{3}-144a^{2}b+108ab^{2}$ (v) $27p^{3}-\frac{1}{216}-\frac{9}{2}p^{2}+\frac{1}{4}p$

Solution 4:

It is known that,

(i)

(ii)

(iii)

(iv)

(v)

Comments

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Question 5:

Verify: (i) $x^{3}+y^{3}=\left ( x+y \right )\left ( x^{2}-xy+y^{2} \right )$ (ii) $x^{3}-y^{3}=\left ( x-y \right )\left ( x^{2}+xy +y^{2}\right )$

Solution 5:

(i) It is known that,

(ii) It is known that,

Comments

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Question 6:

10. Factorise each of the following: (i) $27y^{3}+125z^{3}$ (ii) $64m^{3}-343n^{3}$ [Hint : See Question 9.]

Solution 6:

(i)

(ii)

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Question 7:

Factorise : (i) $27x^{3}+y^{3}+z^{3}-9xyz$

Solution 7:

It is known that,

Comments

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Question 8:

Verify that : $x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left ( x+y+z \right )\left \lfloor \left ( x-y \right )^{2}+\left ( y-z \right )^{2}+\left ( z-x \right )^{2} \right \rfloor$

Solution 8:

It is known that,

Comments

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Question 9:

If x + y + z = 0, show that $x^{3}+y^{3}+z^{3}=3xyz$

Solution 9:

It is known that,

Put x + y + z = 0,

Comments

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Question 10:

Without actually calculating the cubes, find the value of each of the following:
(i) $\left ( -12 \right )^{3}+\left ( 7 \ri^{3})+\left ( 5 \right )^{3}$
(ii) $\left ( 28 \right )^{3}+\left ( -15 \right )^{3}+\left ( -13 \right )^{3}$

Solution 10:

(i)

Let x = −12, y = 7, and z = 5

It can be observed that,

x + y + z = − 12 + 7 + 5 = 0

It is known that if x + y + z = 0, then

∴

= −1260

(ii)

Let x = 28, y = −15, and z = −13

It can be observed that,

x + y + z = 28 + (−15) + (−13) = 28 − 28 = 0

It is known that if x + y + z = 0, then

Comments

Tasneem

Question 11:

Give possible expressions for the length and breadth of each of the following
rectangles, in which their areas are given: (i) Area: $25a^{2}-35a+12$ (ii) Area: $35y^{2}+13y-12$

Solution 11:

Area = Length × Breadth

The expression given for the area of the rectangle has to be factorised. One of its factors will be its length and the other will be its breadth.

(i)

Therefore, possible length = 5a − 3

And, possible breadth = 5a − 4

(ii)

Therefore, possible length = 5y + 4

And, possible breadth = 7y − 3

Comments

Tasneem

Question 12:

What are the possible expressions for the dimensions of the cuboids whose volumes
are given below? (i) Volume : $3x^{2}-12x$ (ii) Volume : $12ky^{2}+8ky-20k$

Solution 12:

Volume of cuboid = Length × Breadth × Height

The expression given for the volume of the cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will be its height.

(i)

One of the possible solutions is as follows.

Length = 3, Breadth = x, Height = x − 4

(ii)

One of the possible solutions is as follows.

Length = 4k, Breadth = 3y + 5, Height = y − 1

Comments

Tasneem

Question 13:

Use suitable identities to find the following products: (i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5) (iv) $\left ( y^{2}+\frac{3}{2} \right )\left ( y^{2}-\frac{3}{2} \right )$ (v) (3 – 2x) (3 + 2x)

Solution 13:

(i) By using the identity ,

(ii) By using the identity ,

(iii)

By using the identity ,

(iv) By using the identity ,

(v) By using the identity ,

Comments

Tasneem

Question 14:

Evaluate the following products without multiplying directly: (i) 103x107 (ii) 95x96 (iii) 104x96

Solution 14:

(i) 103 × 107 = (100 + 3) (100 + 7)

= (100)² + (3 + 7) 100 + (3) (7)

[By using the identity, where

x = 100, a = 3, and b = 7]

= 10000 + 1000 + 21

= 11021

(ii) 95 × 96 = (100 − 5) (100 − 4)

= (100)² + (− 5 − 4) 100 + (− 5) (− 4)

[By using the identity, where

x = 100, a = −5, and b = −4]

= 10000 − 900 + 20

= 9120

(iii) 104 × 96 = (100 + 4) (100 − 4)

= (100)² − (4)²

= 10000 − 16

= 9984

Comments

Tasneem

Question 15:

Factorise the following using appropriate identities: (i) $9x^{2}+6xy+y^{2^{}}$ (ii) $4x^{2}-4y+1$ (iii) $x^{2}-\frac{y^{2}}{100}$

Solution 15:

(i)

(ii)

(iii)

Comments

Tasneem

Question 16:

Expand each of the following, using suitable identities:(i) $\left ( x+2y+4z \right )^{2}$ (ii) $\left ( 2x-y+z \right )^{2}$ (iii) $\left ( -3x+3y+2z \right )^{2}$ (iv) $\left ( 3a-7b-c \right )^{2}$ (v) $\left ( -2x+5y-3z \right )^{2}$ (vi) $\left \lfloor \frac{1}{4}a-\frac{1}{2}b+1 \right \rfloor^{_{2}}$

Solution 16:

It is known that,

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Comments

Tasneem

All Exercises - Chapter 2 - Polynomials for Class 9

Exercise 2.1

Exercise 2.2

Exercise 2.3

Exercise 2.4

Exercise 2.5

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Learn Exercise 2.5 with Free Lessons & Tips

All Exercises - Chapter 2 - Polynomials for Class 9

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