AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.5

Question 1.

Use suitable identities to find the following products.

i) (x + 5) (x + 2)

Solution:

(x + 5) (x + 2)

= x^{2} + (5 + 2)x + 5 x 2

[ ∵ (x + a) (x + b) = x^{2} + (a + b) x + ab]

= x^{2} + 7x + 10

ii) (x – 5) (x – 5)

Solution:

(x – 5) (x – 5)

= (x – 5)^{2} = x^{2} – 2(x) (5) + 5^{2}

[ ∵(x – y)^{2} = x^{2} – 2xy + y^{2}]

= x^{2} – 10x + 25

iii) (3x + 2) (3x – 2)

Solution:

(3x + 2) (3x – 2) = (3x)^{2} – (2)^{2}

[∵ (x + y) (x – y) =x^{2 }– y^{2}]

= 9x^{2} – 4

iv) \(\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{2}-\frac{1}{x^{2}}\right)\)

Solution:

v) (1 + x) (1 + x)

Solution:

(1 + x) (1 + x)

= (1 + x)^{2} = 1^{2} + 2 (1) (x) + x^{2}

[∵(x + y)^{2} = x^{2} + 2xy + y^{2}]

= 1 + 2x + x^{2}

Question 2.

Evaluate the following products with¬out actual multiplication.

i) 101 x 99

Solution:

101 x 99

= (100 + 1) (100 – 1)

= 100^{2} – 1^{2}

= 10000 – 1

= 9999

ii) 999 x 999

Solution:

999 x 999

= 999^{2}

= (1000 – 1)^{2}

= 1000^{2} – 2 x (1000) x 1 + 1^{2}

= 1000000-2000 + 1

= 998001

iii) \(50 \frac{1}{2} \times 49 \frac{1}{2}\)

Solution:

iv) 501 x 501

Solution:

501 x 501

= (500 + 1) (500 + 1)

= (500 + 1)^{2}

= 500^{2} + 2 x (500) x 1 + 1^{2}

= 250000 + 1000 + 1 = 251001

v) 30.5 x 29.5 = (30 + 0.5) (30 – 0.5)

= 30^{2} – (0.5)^{2}

= 900 – 0.25

= 899.75

Question 3.

Factorise the following using appro-priate identities.

i) 16x^{2} + 24xy + 9y^{2}

Solution:

16x^{2} + 24xy + 9y2

= (4x)^{2} + 2 (4x) (3y) + (3y)^{2}

= (4x + 3y)^{2} = (4x + 3y) (4x + 3y)

[ ∵ (x + y)^{2} = x^{2} + 2xy + y^{2}]

ii) 4y^{2} – 4y + 1

Solution:

4y^{2} – 4y + 1

= (2y)^{2} – 2 (2y) (1) + (1)^{2}

[ ∵ (x -y)^{2} = x^{2} – 2xy + y^{2}]

= (2y -1)^{2} = (2y – 1) (2y-1)

iii) \(4 x^{2}-\frac{y^{2}}{25}\)

Solution:

iv) 18a^{2} – 50

Solution:

18a^{2} – 50 = 2 (9a^{2} – 25)

= 2[(3a)^{2} – (5)^{2}]

[ ∵ x^{2} – y^{2} = (x + y) (x – y)]

= 2 (3a + 5) (3a – 5)

v) x^{2} + 5x + 6

Solution:

x^{2} + 5x + 6 = x^{2} + (3 + 2) x + 3 x 2

[ ∵ (x + a) (x + b) = x^{2} + (a + b) x + a . b]

= (x + 3) (x + 2)

vi) 3p^{2} – 24p + 36

Solution:

3p^{2} – 24p + 36

= 3[p^{2} – 8p + 12]

= 3[p^{2} + (- 6 – 2)p + (- 6) (- 2)]

[ ∵ (x + a) (x + b) = x^{2} + (a + b) x + ab]

= 3 (p – 6) (p – 2)

Question 4.

Expand each of the following, using suitable identities.

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

(x + 2y + 4z)^{2} = (x)^{2} + (2y)^{2} + (4z)^{2} + 2(x) (2y) + 2 (2y) (4z) + 2 (4z) (x)

[ ∵ (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx]

= x^{2} + 4y^{2} + 16z^{2} + 4xy + 16yz + 8zx

ii) (2a – 3b)^{3}

Solution:

(2a – 3b)^{3} = (2a)^{3} – 3 (2a)^{2} (3b) + 3 (2a) (3b)^{2} – (3b)^{3}

[ ∵ (a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3}]

= 8a^{3} – 3(4a^{2}) (3b) + 3 (2a) (9b^{2}) – 27b^{3}

= 8a^{3} – 36a^{2}b + 54ab^{2}-27b^{3}

(or)

∵ (a – b)^{3} = a^{3} – b^{3}– 3ab (a – b)]

= (2a)^{3} – (3b)^{3} – 3(2a) (3b) (2a – 3b)

= 8a^{3} – 27b^{3} – 18ab (2a – 3b)

iii) (- 2a + 5b – 3c)^{2}

Solution:

(- 2a + 5b – 3c)^{2}

= (- 2a)^{2} + (5b)^{2} + (- 3c)^{2} + 2 (- 2a) (5b) + 2 (5b) (- 3c) + 2 (- 3c) (- 2a)

= 4a^{2} + 25b^{2} + 9c^{2} – 20ab – 30bc + 12ca

[ ∵ (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy +2yz +2za]

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

Solution:

v) (p + 1)^{3}

Solution:

(p + 1)^{3}

= (P)^{3} + 3 (p)^{2} (1) + 3 (p) (1)^{2} + (1)^{3}

[ ∵ (x + y)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3}]

= p^{3} + 3p^{2} + 3p + 1

vi) \(\left(x-\frac{2}{3} y\right)^{3}\)

Solution:

Question 5.

Factorise

i) 25x^{2} + 16y^{2} + 4z^{2} – 40xy + 16yz – 20xz

Solution:

25x^{2} + 16y^{2} + 4z^{2} – 40xy + 16yz – 20xz

= (5x)^{2} + (- 4y)^{2} + (- 2z)^{2} + 2(5x) (- 4y) + 2 (- 4y) (- 2z) + 2 (- 2z) (5x)

= (5x – 4y – 2z)^{2} = (- 5x + 4y +, 2z)^{2}

ii) 9a^{2} + 4b^{2} + 16c^{2} + 12ab – 16bc – 24ca

Solution:

9a^{2} + 4b^{2} + 16c^{2} + 12ab – 16bc -24ca

= (3a)^{2} + (2b)^{2} + (- 4c)^{2}+ 2 (3a) (2b) + 2 (2b) (- 4c) + 2(- 4c) (3a)

= (3a + 2b – 4c)^{2}

Question 6.

If a + b + c = 9 and ab + be + ca = 26, find a^{2} + b^{2} + c^{2}.

Solution:

Given that a + b + c = 9

Squaring on both sides,

(a + b + c)^{2} = 9^{2}

⇒ a^{2}+ b^{2} + c^{2}+ 2 (ab + be + ca) = 81 ⇒ a^{2} + b^{2} + c^{2} = 81 – 2 (ab + be + ca)

(by problem)

= 81 – 2 x 26

= 81 – 52 = 29

Question 7.

Evaluate the following by using suit¬able identities. m EachgM)

i) (99)^{3}

Solution:

(99)^{2} = (100 – 1)^{3}

= 100^{3} – 3 (100)^{2} (1) + 3 (100) (1)^{2} – 1^{3}

[ ∵ (x – y)^{3} = x^{3} – 3x^{2}y + 3xy^{2} + y^{3}]

= 1000000 – 30000 + 300 – 1

= 970299

ii) (102)^{3}

Solution:

(102)^{3} = (100 + 2)^{3}

= 100^{3} + 3 (100)^{2} (2) + 3 (100) (2)^{2} + 2^{3}

[ ∵ (x + y)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3}]

= 1000000 + 60000 + 1200 + 8

= 1061208

iii) (998)^{3}

Solution:

(998)^{3} =(1000 – 2)^{3}

[ ∵ (x – y)^{3} = x^{3} – 3x^{2}y + 3xy^{2} – y^{3}] = 1000^{3}– 3(1000)^{2}(2) + 3(1000)(2)^{2}– 2^{3}

= 1000000000 – 6000000 + 12000 – 8

= 994011992

iv) (1001)^{3}

Solution:

(1001)^{3} = (1000 + 1)^{3} .

[ ∵ (x + y)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3}] = 1000^{3} + 3(1000)^{2}(1) + 3(1000) (1)^{2} + 1^{3}

= 1000000000 + 3000000 + 3000 + 1

= 1003003001

Question 8.

Factorise each of the following.

i) 8a^{3} + b^{3} + 12a^{2} b + 6ab^{2}

Solution:

8a^{3} + b^{3} + 12a^{2} b + 6ab^{2}

= (2a)^{3} + (b)^{3} + 3 (2a)^{2} (b) + 3 (2a) (b)^{2}

= (2a + b)^{3}

ii) 8a^{3} – b^{3} – 12a^{2} b + 6ab^{2}

Solution:

8a^{3} – b^{3} – 12a^{2} b + 6ab^{2}

= (2a)^{3} – (b)^{3} – 3 (2a)^{2} (b) + 3 (2a) (b)^{2}

= (2a – b)^{3}

iii) 1 – 64a^{3} -12a + 48a^{2}

Solution:

1 – 64a^{3} – 12a + 48a^{2}

= (1)^{3} – (4a)^{3} – 3(1)^{2} (4a) + 3(1) (4a)^{2}

= (1 – 4a)^{3}

iv) \(8 p^{3}-\frac{12}{5} p^{2}+\frac{6}{25} p-\frac{1}{125}\)

Solution:

Question 9.

Verify i) x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2});

ii) x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

Using some non-zero positive integers and check by actual multiplication. Can you

call these as identities ?

i) x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})

Solution:

Given x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})

L.H.S = x^{3} + y^{3}

R.H.S = (x + y) (x^{2} – xy + y^{2})

= x (x^{2} – xy + y^{2}) + y (x^{2} – xy + y^{2})

= x^{3} -x^{2}y + xy^{2} + x^{2}y – xy^{2} + y^{3}

= x^{3} + y^{3}

= L.H.S

∴ L.H.S = R.H.S

Take x = 3, y = 2

L.H.S = 3^{3} + 2^{3} = 27 + 8 = 35

R.H.S = (3 + 2) (3^{2} – 3 x 2 + 2^{2})

= 5 x (9 – 6 + 4)

= 5 x 7 = 35

∴ L.H.S = R.H.S

ii) x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

Solution:

Given that x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

L.H.S = x^{3} – y^{3}

R.H.S = (x – y) (x^{2} + xy + y^{2})

= x (x^{2} + xy + y^{2}) – y (x^{2} + xy + y^{2})

= x^{3} + x^{2}y + xy^{2} – x^{2}y – xy^{2} – y^{3}

= x^{3} – y^{3}= L.H.S

L.H.S = 3^{3} – 2^{3} = 27 – 8 = 19

R.H.S = (3 – 2) (3^{2} + 3 x 2 + 2^{2})

= 1 x (9 + 6 + 4)

= 1 x 19 = 19

∴ L.H.S = R.H.S

We can call the above two expressions as identities

Question 10.

Factorise by using the above results (identities).

i) 27a^{3} + 64b^{3}

Solution:

27a^{3}+ 64b^{3} = (3a)^{3} + (4b)^{3}

= (3a + 4b) {(3a)^{2} – (3a) (4b) + (4b)^{2}}

= (3a + 4b) (9a^{2} – 12ab + 16b^{2})

ii) 343y^{3} – 1000

Solution:

343y^{3} – 1000 = (7y)^{3} – (10)^{3}

= (7y – 10) [(7y)^{2} + (7y) (10) + (10)^{2}]

= (7y – 10) (49y^{2} + 70y + 100)

Question 11.

Factorise 27x^{3} + y^{3} + z^{3} – 9xyz using identity.

Solution:

Given 27x^{3} + y^{3} + z^{3} – 9xyz

= (3x)^{3} + (y)^{3} + (z)^{3} – 3 (3x) (y) (z)

= (3x + y + z)

[(3x)^{2} + y^{2} + z^{2} – (3x) (y) – (y) (z) – (z) (3x)]

[ ∵ (x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2}– xy – yz – zx)

= (3x + y + z) (9x^{2} + y^{2} + z^{2} – 3xy – yz – 3xz)

Question 12.

Verify that x^{3}+ y^{3} + z^{3} – 3xyz = 1/2 (x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2} ]

(OR)

Verify that

p^{3} + q^{3} + r^{3} – 3pqr = 1/2 (p + q + r)

[(p – q)^{2} + (q – r)^{2} + (r – p)^{2}]

Solution:

Given x^{3}+ y^{3} + z^{3} – 3xyz = 1/2 (x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2} ]

R-H.S = 1/2 (x + y + z) [(x – y)^{2} + (y – z)^{2}+ (z – x)^{2}]

= 1/2 (x + y + z) [x^{2} + y^{2} – 2xy + y^{2} + z^{2} – 2yz + z^{2} + x^{2} – 2xz]

= 1/2 (x + y + z) [2x^{2} + 2y^{2} + 2z^{2} – 2xy – 2yz – 2zx]

= 1/2 (x + y + z) (2) [x^{2} + y^{2} + z^{2} – xy – yz – zx]

= (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)

= L.H.S

Hence proved.

Question 13.

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

Solution:

Given x + y + z = 0

To prove x^{3} + y^{3} + z^{3} = 3xyz

We have an identity

(x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)

= x^{3} + y^{3} + z^{3} – 3xyz

Substituting x + y + z = 0in the above equation, we get

0 x (x^{2} + y^{2} + z^{2} -xy-yz-zx)

= x^{3} + y^{3} + z^{3} – 3xyz

⇒ x^{3} + y^{3} + z^{3} – 3xyz = 0

⇒ x^{3} + y^{3} + z^{3} = 3xyz

Question 14.

Without actual calculating the cubes, find the value of each of the following.

i) (- 10)^{3} + 7^{3} + 3^{3}

Solution:

Given (-10)^{3} + 7^{3} + 3^{3}

Sum of the bases = -10 + 7 + 3 = = 0

∴ (- 10)^{3} + 7^{3} + 3^{3}

= 3 (- 10) x (7) x 3

= -630

[ ∵ x + y + z = 0 then x^{3} + y^{3} + z^{3} = 3xyz]

ii) (28)^{3} + (- 15)^{3} + (- 13)^{3}

Solution:

Given (28)^{3} + (- 15)^{3}+ (- 13)^{3}

Sum of the bases = 28 + (- 15) + (- 13) = 0

∴ (28)^{3} + (- 15)^{3} + (- 13)^{3}

= 3 x 28 x (- 15) x (- 13)

= 16380

iii) \(\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}\) read it as \(\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}+\left(\frac{-5}{6}\right)^{3}\)

Solution:

iv) (0.2)^{3} – (0.3)^{3} + (0.1)^{3}

Solution:

Given that (0.2)^{3} – (0.3)^{3} + (0.1)^{3}

= (0.2)^{3} + (- 0.3)^{3} + (0.1)^{3}

Sum of the bases = 0.2 – 0.3 + 0.1 = 0

∴ (0.2)^{3} + (-0.3)^{3} + (0.1)^{3}

= 3 x (0.2) (- 0.3) (0.1)

= -0.018

Question 15.

Give possible expressions for the length and breadth of the rectangle whose area is given by

i) 4a^{2} + 4a – 3

Given that area = 4a^{2} + 4a – 3

= 4a^{2} + 6a – 2a – 3

= 2a (2a + 3) – 1 (2a + 3)

= (2a – 1) (2a + 3)

∴ Length = (2a + 3); breadth = (2a – 1).

ii) 25a^{2 }– 35a + 12

Solution:

Given that area = 25a^{2} – 35a +12

= 25a^{2} – 20a – 15a + 12

= 5a (5a – 4) – 3 (5a – 4)

= (5a – 4) (5a – 3)

∴ (5a – 4) (5a – 3) are the length and breadth.

Question 16.

What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below ?

i) 3x^{3} – 12x

Solution:

Volume = 3x^{3 }– 12x

= 3x (x^{2} – 4)

= 3x (x + 2) (x – 2) are the dimensions.

ii) 12y^{2} + 8y – 20

Solution:

Given that volume = 12y^{2} + 8y – 20

= 4 (3y^{2} + 2y – 5)

= 4 [3y^{2} + 5y – 3y – 5]

= 4 [y (3y + 5) – 1 (3y + 5)]

= 4 (3y + 5) (y – 1)

Hence 4, (3y + 5) and (y – 1) are the dimensions.

Question 17.

Show that if 2 (a^{2} + b^{2} ) = (a + b)^{2} then a = b

Solution:

Given that 2 (a^{2} + b^{2} ) = (a + b)^{2}

To prove a = b

As 2 (a^{2} + b^{2} ) = (a + b)^{2}

We have

2a^{2} + 2b^{2} = a^{2} + 2ab + b^{2}

2a^{2} – a^{2} + 2b^{2} – b^{2} = 2ab

a^{2} + b^{2} = 2ab

This is possible only when a = b

∴ a = b