## AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions and Answers.

### 8th Class Maths 12th Lesson Factorisation InText Questions and Answers

Do this

Question 1.

Express the given numbers in the form of product of primes. [Page No. 267]

(i) 48 (ii) 72 (iii) 96

(i) 48

Answer:

48 = 2 × 2 × 2 × 2 × 3

ii) 72

Answer:

72 = 2 × 2 × 2 × 3 × 3

iii) 96

Answer:

96 = 2 × 2 × 2 × 2 × 2 × 3

Question 2.

Find the factors of following: [Page No. 268]

(i) 8x^{2}yz (ii) 2xy (x + y) (iii) 3x + y^{3}z

Answer:

i) 8x^{2}yz = 2 × 2 × 2 × x × x × y × z

ii) 2xy (x + y) = 2 × x × y × (x + y)

iii) 3x + y^{3}z = (3 × x) + (y × y × y × z)

Question 3.

Factorise: [Page No. 270]

(i) 9a^{2} – 6a

(ii) 15a^{3}b – 35ab^{3}

(iii) 7lm – 21lmn

Answer:

(i) 9a^{2} – 6a = 3 × 3 × a × a – 2 × 3 × a

= 3 × a (3a – 2)

∴ 9a^{2} – 6a = 3a (3a – 2)

ii) 15a^{3}b – 35ab^{3}

= 3 × 5 × a × a × a × b – 7 × 5 × a × b × b × b

= 5 × a × b [3 × a × a – 7 × b × b]

= 5ab [3a^{2} – 7b^{2}]

iii) 7lm – 21lmn

= 7 × l × m – 7 × 3 × m × n × l

= 7 × l × m [1 – 3n]

= 7lm [1 – 3n]

Question 4.

Factorise:

i) 5xy + 5x + 4y + 4

ii) 3ab + 3a + 2b + 2 [Pg. No. 271]

Answer:

i) 5xy + 5x + 4y + 4

= (5xy + 5x) + (4y + 4)

= 5x(y + 1) + 4(y + 1)

= (y + 1) (5x + 4)

ii) 3ab + 3a + 2b + 2

= [3 × a × b + 3 × a] + [2 × b + 2]

= 3 × a [b + 1] + 2 [b + 1]

= (b + 1) (3a + 2)

Think, Discuss and Write

While solving some problems containing algebraic expressions in different operations, some students solved as given below. Gan you identify the errors made by them? Write correct answers. [Page No. 279]

Question 1.

Srilekha solved the given equation as shown below.

3x + 4x + x + 2x = 90

9x = 90 Therefore x = 10 What could say about the correctness of the solution?

Can you identify where Srilekha has gone wrong?

Answer:

Srilekha’s solution is wrong,

∵ 3x + 4x + x + 2x = 90

10x = 90

x = \(\frac{90}{10}\)

∴ x = 9

Question 2.

Abraham did the following. [Page No. 280]

For x = -4, 7x = 7 – 4 = -3.

Answer:

Abraham’s solution is wrong.

∴ If x = -4

⇒ 7x = 7(-4) = -28

Question 3.

John and Reshma have done the multiplication of an algebraic expression by the following methods: verify whose multiplication is correct. [Page No. 280]

Answer:

∴ John’s solutions are wrong and Reshma’s solutions are correct.

Question 4.

Harmeet does the division as (a + 5) ÷ 5 = a + 1 His friend Srikar done the same (a + 5) ÷ 5 = a/5 + 1 and his friend Rosy did it this way (a + 5) ÷ 5 = a Can you guess who has done it correctly? Justify! [Page No. 280]

Answer:

The solutions of Harmeet, Rosy are wrong.

(a + 5) ÷ 5 = \(\frac{a+5}{5}\)

= \(\frac{a}{5}\) + \(\frac{5}{5}\)

= \(\left(\frac{a}{5}+1\right)\)

∴ Srikar had done it correctly.