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

Question 1.

Find the errors and correct the following mathematical sentences

(i) 3(x – 9) = 3x – 9

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

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

(iv) 2x + x + 3x = sx

(v) 4p + 3p + 2p + p – 9p = 0

(vi) 3x + 2y = 6xy

(vii) (3x)^{2} + 4x +7 = 3x^{2} + 4x +7

(viii) (2x)^{2} + 5x = 4x + 5x = 9x

(ix) (2a + 3)^{2} = 2a^{2} + 6a +9

(x) Substitute x -3 in

(a) x^{2} + 7x + 12 (- 3)^{2} + 7(-3) + 12 = 9 + 4 + 12 = 25

(b) x^{2} – 5x + 6(-3)^{2} – 5(-3) + 69 – 15 + 6 = 0

(c) x^{2} +5x = (-3)^{2} + 5(3) + 6 = -9 – 15 = -24

(xi) (x – 4)^{2} = x^{2} – 16

(xii) (x + 7)^{2} = x^{2} +49

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

(xiv) (x + 4) (x + 2) = x^{2} + 8

(xv) (x – 4) (x – 2) = x^{2}– 8

(xvi) 5x^{3} ÷ 5 x^{3} = 0

(xvii) 2x^{3} + 1 ÷ 2x^{3} = 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.

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

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

⇒ x × 3x + x × 2 = 3x^{2} + 2

⇒ 3x^{2} + 2x ≠ 3x^{2} + 2

∴ The given sentence is wrong.

Correct sentence is x(3x + 2) = 3x^{2} + 2x.

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

2x + 3x = 5x^{2}

⇒ 5x = 5x^{2}

⇒ x ≠ x^{2}

∴ 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

(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 = 3x^{2} + 4x +7

(3x)^{2} + 4x +7 = 3x^{2} + 4x +7

⇒ (3x)^{2} = 3x^{2}

⇒ 9x^{2} = 3x^{2}

⇒ 9 = 3

It is not possible

∴ The given sentence is wrong. Correct sentence is

(3x)^{2}+ 4x + 7 = 9x^{2} + 4x + 7.

(viii) (2x)^{2} + 5x = 4x + 5x = 9x

(2x)^{2} + 5x = 4x + 5x = 9x

⇒ 4x^{2} + 5x = 4x + 5x

⇒ 4x^{2} = 4x

⇒ x^{2} = x

⇒ x ≠ √x

∴ The given sentence is wrong. Correct sentence is (2x)^{2} + 5x = 4x^{2} + 5x.

(ix) (2a + 3)^{2} = 2a^{2} + 6a +9

(2a + 3)^{2} = 2a^{2} + 6a +9

⇒ (2a)^{2} + 2 × 2a × 3 + 3^{2} = 2a^{2} + 6a + 9

⇒ 4a^{2} + 12a + 9 = 2a^{2}+ 6a + 9

⇒ 4a^{2} – 2a^{2} = 6a – 12a

⇒ 2a^{2} = – 6a

⇒ 2a ≠ 6

∴ The given sentence is wrong.

Correct sentence is

(2a + 3)^{2} = 4a^{2} + 12a + 9.

(x) Substitute x -3 in

(a) x^{2} + 7x + 12 (- 3)^{2} + 7(-3) + 12 = 9 + 4 + 12 = 25

x^{2} + 7x + 12 = (- 3)^{2} + 7 (- 3) + 12

= 9 – 21 + 12

= 21 – 21

= 0 25 (False)

(b) x^{2} – 5x + 6(-3)^{2} – 5(-3) + 69 – 15 + 6 = 0

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

= 9 + 15 + 6

= 30 ≠ 0 (False)

(c) x^{2} +5x = (-3)^{2} + 5(3) + 6 = -9 – 15 = -24

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

= 9 – 15 = – 6 ≠ 24 (False)

(xi) (x – 4)^{2} = x^{2} – 16

(x – 4)^{2} = x^{2} – 16 = (x)^{2} – (4)^{2}

(a – b)^{2} ≠ a^{2} – b^{2}

∴ (x-4)^{2} ≠ (x)^{2} – (4)^{2}

∴ The given sentence is wrong.

Correct sentence is (x – 4)^{2} = x^{2} – 8x + 16.

(xii) (x + 7)^{2} = x^{2} +49

(x + 7)^{2} = x^{2} + 49 = (x)^{2} + (7)^{2}

(a + b)^{2} ≠ a^{2} + b^{2}

∴ (x+7)^{2} ≠ (x)^{2} – (7)^{2}

∴ The given sentence is wrong.

Correct sentence is (x + 7)^{2} = x^{2} + 14x + 49.

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

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

3a^{2} – 3ab + 4ab – 4b^{2} = – a^{2}

3a^{2} + ab – 4b^{2} ≠ a^{2}

∴ The given sentence is wrong. Correct sentence is

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

(xiv) (x + 4) (x + 2) = x^{2} + 8

(x + 4) (x + 2) = x^{2} + 8

⇒ x^{2} + 6x + 8 = x^{2} + 8

⇒ 6x ≠ 0

Here ’6x’ term is missing in R.H.S.

∴ The given sentence is wrong. Correct sentence is

(x + 4)(x + 2) = x^{2} + 6x + 8.

(xv) (x – 4) (x – 2) = x^{2}– 8

(x – 4) (x – 2) = x^{2} – 8

⇒ x^{2} – 6x + 8 ≠ x^{2} – 8

∴ The given sentence is wrong. Correct sentence is

(x – 4) (x – 2) = x^{2} – 6x + 8

(xvi) 5x^{3} ÷ 5 x^{3} = 0

5x^{3} ÷ 5 x^{3} = 0

⇒ x^{3-3} = 0

⇒ x^{0} = 0

∴ 1 ≠ 0 (∵ but x° = 1)

∴ The given sentence is wrong. Correct sentence is 5x^{3} ÷ 5x^{3} = 1.

In the denominator the term T is missing. .•. The given sentence is wrong. Correct sentence is

(xvii) 2x^{3} + 1 ÷ 2x^{3} = 1

2x^{3} + 1 ÷ 2x^{3} = 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

2x^{3} + 1 ÷ 2x^{3} = 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 + \(\frac{5}{3}\)

(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 \(\frac{4 x+3}{3}=\frac{4 x}{3}+1\)