AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.4

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4

Question 1.
Find the errors and correct the following mathematical sentences
(i) 3(x – 9) = 3x – 9
(ii) x(3x+2) = 3x2 + 2
(iii) 2x+3x = 5x2
(iv) 2x + x + 3x = sx
(v) 4p + 3p + 2p + p – 9p = 0
(vi) 3x + 2y = 6xy
(vii) (3x)2 + 4x +7 = 3x2 + 4x +7
(viii) (2x)2 + 5x = 4x + 5x = 9x
(ix) (2a + 3)2 = 2a2 + 6a +9
(x) Substitute x -3 in
(a) x2 + 7x + 12 (- 3)2 + 7(-3) + 12 = 9 + 4 + 12 = 25
(b) x2 – 5x + 6(-3)2 – 5(-3) + 69 – 15 + 6 = 0
(c) x2 +5x = (-3)2 + 5(3) + 6 = -9 – 15 = -24
(xi) (x – 4)2 = x2 – 16
(xii) (x + 7)2 = x2 +49
(xiii) (3a + 4b)(a – b)= 3a2 – 4a2
(xiv) (x + 4) (x + 2) = x2 + 8
(xv) (x – 4) (x – 2) = x2– 8
(xvi) 5x3 ÷ 5 x3 = 0
(xvii) 2x3 + 1 ÷ 2x3 = 1
(xviii) 3x + 2 ÷ 3x = \(\frac{2}{3 x}\)
(xix) 3x + 5 ÷ 3 = 5
(xx) \(\frac{4 x+3}{3}\) = x + 1
Solution:
(i) 3(x – 9) = 3x – 9
3(x – 9) = 3x – 9
⇒ 3x – 3 x 9 = 3x – 9
⇒ 3x – 27 = 3x – 9
⇒ – 27 ≠ – 9
∴ The given sentence is wrong. Correct sentence is 3(x – 9) = 3x – 27.

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4

(ii) x(3x+2) = 3x2 + 2
x(3x + 2) = 3x2 + 2
⇒ x × 3x + x × 2 = 3x2 + 2
⇒ 3x2 + 2x ≠ 3x2 + 2
∴ The given sentence is wrong.
Correct sentence is x(3x + 2) = 3x2 + 2x.

(iii) 2x+3x = 5x2
2x + 3x = 5x2
⇒ 5x = 5x2
⇒ x ≠ x2
∴ The given sentence is wrong. Correct sentence is 2x + 3x = 5x.

(iv) 2x + x + 3x = 5x
2x + x + 3x = 5x
⇒ 6x = 5x
⇒ 6 ≠ 5
∴ The given sentence is wrong. Correct sentence is 2x + 3x = 5x.

(v) 4p + 3p + 2p + p – 9p = 0
4p + 3p + 2p + p – 9p = 0
⇒ 10p – 9p = 0
⇒ p = 0
It is not possible
∴ The given sentence is wrong. Correct sentence is
4p + 3p + 2p + p – 9p – p = 0

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4

(vi) 3x + 2y = 6xy
3x + 2y = 6xy
a + b ≠ ab
∴ The given sentence is wrong.
Correct sentence is 3x x 2y = 6xy.

(vii) (3x)2 + 4x +7 = 3x2 + 4x +7
(3x)2 + 4x +7 = 3x2 + 4x +7
⇒ (3x)2 = 3x2
⇒ 9x2 = 3x2
⇒ 9 = 3
It is not possible
∴ The given sentence is wrong. Correct sentence is
(3x)2+ 4x + 7 = 9x2 + 4x + 7.

(viii) (2x)2 + 5x = 4x + 5x = 9x
(2x)2 + 5x = 4x + 5x = 9x
⇒ 4x2 + 5x = 4x + 5x
⇒ 4x2 = 4x
⇒ x2 = x
⇒ x ≠ √x
∴ The given sentence is wrong. Correct sentence is (2x)2 + 5x = 4x2 + 5x.

(ix) (2a + 3)2 = 2a2 + 6a +9
(2a + 3)2 = 2a2 + 6a +9
⇒ (2a)2 + 2 × 2a × 3 + 32 = 2a2 + 6a + 9
⇒ 4a2 + 12a + 9 = 2a2+ 6a + 9
⇒ 4a2 – 2a2 = 6a – 12a
⇒ 2a2 = – 6a
⇒ 2a ≠ 6
∴ The given sentence is wrong.
Correct sentence is
(2a + 3)2 = 4a2 + 12a + 9.

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4

(x) Substitute x -3 in
(a) x2 + 7x + 12 (- 3)2 + 7(-3) + 12 = 9 + 4 + 12 = 25
x2 + 7x + 12 = (- 3)2 + 7 (- 3) + 12
= 9 – 21 + 12
= 21 – 21
= 0 25 (False)

(b) x2 – 5x + 6(-3)2 – 5(-3) + 69 – 15 + 6 = 0
x2 – 5x + 6 = (-3)2 – 5 (- 3) + 6
= 9 + 15 + 6
= 30 ≠ 0 (False)

(c) x2 +5x = (-3)2 + 5(3) + 6 = -9 – 15 = -24
x2 + 5x = (- 3)2 + 5 (- 3)
= 9 – 15 = – 6 ≠ 24 (False)

(xi) (x – 4)2 = x2 – 16
(x – 4)2 = x2 – 16 = (x)2 – (4)2
(a – b)2 ≠ a2 – b2
∴ (x-4)2 ≠ (x)2 – (4)2
∴ The given sentence is wrong.
Correct sentence is (x – 4)2 = x2 – 8x + 16.

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4

(xii) (x + 7)2 = x2 +49
(x + 7)2 = x2 + 49 = (x)2 + (7)2
(a + b)2 ≠ a2 + b2
∴ (x+7)2 ≠ (x)2 – (7)2
∴ The given sentence is wrong.
Correct sentence is (x + 7)2 = x2 + 14x + 49.

(xiii) (3a + 4b)(a – b)= 3a2 – 4a2
3a(a – b) + 4b(a – b) = 3a2 – 42
3a2 – 3ab + 4ab – 4b2 = – a2
3a2 + ab – 4b2 ≠ a2
∴ The given sentence is wrong. Correct sentence is
(3a + 4b) (a – b) = 3a2 + ab – 4b2

(xiv) (x + 4) (x + 2) = x2 + 8
(x + 4) (x + 2) = x2 + 8
⇒ x2 + 6x + 8 = x2 + 8
⇒ 6x ≠ 0
Here ’6x’ term is missing in R.H.S.
∴ The given sentence is wrong. Correct sentence is
(x + 4)(x + 2) = x2 + 6x + 8.

(xv) (x – 4) (x – 2) = x2– 8
(x – 4) (x – 2) = x2 – 8
⇒ x2 – 6x + 8 ≠ x2 – 8
∴ The given sentence is wrong. Correct sentence is
(x – 4) (x – 2) = x2 – 6x + 8

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4

(xvi) 5x3 ÷ 5 x3 = 0
5x3 ÷ 5 x3 = 0
⇒ x3-3 = 0
⇒ x0 = 0
∴ 1 ≠ 0 (∵ but x° = 1)
∴ The given sentence is wrong. Correct sentence is 5x3 ÷ 5x3 = 1.
In the denominator the term T is missing. .•. The given sentence is wrong. Correct sentence is

(xvii) 2x3 + 1 ÷ 2x3 = 1
2x3 + 1 ÷ 2x3 = 1
⇒ \(\frac{2 x^{3}+1}{2 x^{3}}\) = 1
In the denominator the term T is missing.
∴ The given sentence is wrong. Correct sentence is
2x3 + 1 ÷ 2x3 = 1 + \(\frac{1}{2 \mathrm{x}^{3}}\)

(xviii) 3x + 2 ÷ 3x = \(\frac{2}{3 x}\)
3x + 2 ÷ 3x = \(\frac{2}{3 x}\)
⇒ \(\frac{3 x+2}{3 x}=\frac{2}{3 x}\)
⇒ 1 + \(\frac{2}{3 x}=\frac{2}{3 x}\) ⇒ 1 ≠ 0
∴ The given sentence is wrong. Correct sentence is 3x + 2 ÷ 3x = 1 + \(\frac{2}{3 x}\)

(xix) 3x + 5 ÷ 3 = 5
⇒ \(\frac{3 x+5}{3}\) = 5
⇒ \(\frac{3 x}{3}+\frac{5}{3}\) = 5 ⇒ x + \(\frac{5}{3}\) ≠ 5
∴ It is a wrong sentence.
Correct sentence is 3x + 5 ÷ 3 = x + \(\frac{5}{3}\)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4

(xx) \(\frac{4 x+3}{3}\) = x + 1
\(\frac{4 x+3}{3}\) = x + 1
⇒ \(\frac{4 \mathrm{x}}{3}+\frac{3}{3}\) = x + 1
⇒ \(\frac{4 \mathrm{x}}{3}\) + 1 ≠ x + 1
∴ It is a wrong sentence.
Correct sentence is \(\frac{4 x+3}{3}=\frac{4 x}{3}+1\)