AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.3 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.3

Question 1.

Carry out the following divisions

(i) 48a^{3} by 6a

(ii) 14x^{3 }by 42x^{3}

(iii) 72a^{3}b^{4}c^{5} by 8ab^{2}c^{3}

(iv) 11xy^{2}z^{3} by 55xyz

(v) -54l^{4}m^{3}n^{2} by 9l^{2}m^{2}n^{2}

Solution:

(i) 48a^{3} by 6a

48a^{3} ÷ 6a

= \(\frac{6 \times 8 \times a \times a^{2}}{6 \times a}\)

= 8a^{2}

(ii) 14x^{3 }by 42x^{3}

= 14x^{3} ÷ 42x^{3}

(iii) 72a^{3}b^{4}c^{5} by 8ab^{2}c^{3}

(iv) 11xy^{2}z^{3} by 55xyz

11xy^{2}z^{3} ÷ 55xyz

(v) -54l^{4}m^{3}n^{2} by 9l^{2}m^{2}n^{2}

-54l^{4}m^{3}n^{2} ÷ 9l^{2}m^{2}n^{2}

= -6l^{2}m

Question 2.

Divide the given polynomial by the given monomial

(i) (3x^{2} – 2x) ÷ x

(ii) (5a^{3}b – 7ab^{3}) ÷ ab

(iii) (25x^{5} – 15x^{4}) ÷ 5x^{3}

(iv) (4l^{5} – 6l^{4} + 8l^{3}) ÷ 2l^{2}

(v) 15 (a^{3}b^{2}c^{2} – a^{2}b^{3}c^{2} + a^{2}b^{2}c^{3} ) ÷ 3abc

(vi) 3p^{3}– 9p^{2}q – 6pq^{2}) ÷ (-3p)

(vii) (\(\frac{2}{3}\) a^{2} b^{2} c^{2}+ \(\frac{4}{3}\) a b^{2} c^{3}) ÷ \(\frac{1}{2}\)abc

Solution:

(i) (3x^{2} – 2x) ÷ x

(ii) (5a^{3}b – 7ab^{3}) ÷ ab

(iii) (25x^{5} – 15x^{4}) ÷ 5x^{3}

= 5x^{2} – 3x (or) x(5x – 3)

(iv) (4l^{5} – 6l^{4} + 8l^{3}) ÷ 2l^{2}

= 2l^{2} – 3l^{2} + 4l = l(2l^{2} – 3l + 4)

(v) 15 (a^{3} b^{2} c^{2} – a^{2} b^{3} c^{2} + a^{2} b^{2} c^{3} ) ÷ 3abc

= 5[a x abc – b x abc + c x abc ]

= 5abc [a – b + c]

(vi) 3p^{3}– 9p^{2}q – 6pq^{2}) ÷ (-3p)

= -[p^{2} – 3pq – 2q^{2}]

= 2^{2} + 3pq – p^{2}

(vii) (\(\frac{2}{3}\) a^{2}b^{2}c^{2}+ \(\frac{4}{3}\) ab^{2}c^{3}) ÷ \(\frac{1}{2}\)abc

Question 3.

Workout the following divisions:

(i) (49x -63) ÷ 7

(ii) 12x (8x – 20,) ÷ 4(2x – 5)

(iii) 11a^{3} b^{3} (7c – 35) ÷ 3a^{2} b^{2} (c – 5)

(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)

(v) 36(x + 4)(x^{2} + 7x + 10) ÷ 9(x + 4)

(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)

Solution:

(i) (49x -63) ÷ 7

(ii) 12x (8x – 20,) ÷ 4(2x – 5)

(iii) 11a^{3} b^{3} (7c – 35) ÷ 3a^{2} b^{2} (c – 5)

(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)

(v) 36(x + 4)(x^{2} + 7x + 10) ÷ 9(x + 4)

4 ( x^{2} + 7x + 10)

= 4 ( x^{2} + 5x + 2x + 10)

= 4 [x( x + 5) +2(x + 5)]

= 4( x + 5) (x + 2)

(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)

= ( a + 1)(a + 2)

Question 4.

Factorize the expressions and divide them as directed:

(i) (x^{2} + 7x + 12) ÷ (x + 3)

(ii) (x^{2} – 8x + 12) ÷ (x – 6)

(iii) (p^{2} + 5p + 4,) (p + l)

(iv) 15ab(a^{2} – 7a + 10) ÷ 3b(a – 2)

(v) 151m (2p^{2} – 2q^{2}) ÷ 3l(p + q)

(vi) 26z^{3}(32z^{2} – 18,) ÷ 13z^{2} (4z – 3)

Solution:

(i) (x^{2} + 7x + 12) ÷ (x + 3)

(x^{2} + 7x + 12) ÷ (x + 3)

x^{2} + 7x + 12 = x^{2} + 3x + 4x + 12

= x(x + 3) + 4(x + 3)

= (x + 3) (x + 4)

(ii) (x^{2} – 8x + 12) ÷ (x – 6)

(x^{2} – 8x + 12) ÷ (x – 6)

x^{2} – 8x + 12 = x^{2} – 6x – 2x + 12

= x(x – 6) – 2(x – 6)

= (x – 6) (x – 2)

∴ (x^{2} – 8x + 12) 4 (x – 6)

= \(\frac{(x-6)(x-2)}{(x-6)}\) = x – 2

(iii) (p^{2} + 5p + 4,) (p + 1)

p^{2} + 5p + 4 = p^{2} + p + 4p + 4

= p(p + 1) + 4(p + 1)

= (p + 1) (p + 4)

(p^{2} + 5p + 4) ÷ (p + 1)

= \(\frac{(p+1)(p+4)}{(p+1)}\) = p + 4

(iv) 15ab(a^{2} – 7a + 10) ÷ 3b(a – 2)

15ab (a^{2} – 7a + 10) ÷ 3b (a – 2)

15ab (a^{2} – 7a + 10) = 15ab (a^{2} – 5a – 2a + 10)

= 15ab [(a^{2} – 2a) – (5a -10)]

= 15ab [a(a – 2) – 5(a – 2)]

= 15ab(a – 2)(a – 5)

∴ 15ab (a^{2} – 7a + 10) ÷ 3b (a – 2)

(v) 151m (2p^{2} – 2q^{2}) ÷ 3l(p + q)

15lm (2p^{2} – 2q^{2}) ÷ 3l (p + q)

15lm (2p^{2} – 2q^{2}) = 15lm x 2(p^{2} – q^{2})

= 30lm (p + q) (p – q)

∴ 15lm(2p^{2} – 2q^{2}) ÷ 3l(p + q)

(vi) 26z^{3}(32z^{2} – 18,) ÷ 13z^{2} (4z – 3)

26z^{3}(32z^{2} – 18) ÷ 13z^{2} (4z – 3)

26z^{3}(32z^{2} – 18) = 26z^{3} (2 x 16z^{2} – 2 x 9)

= 26z^{3} x 2 [16z^{3} – 9]

= 52z^{3} [(4z)^{3} – (3)^{3}]

= 52z^{3} (4z + 3) (4z – 3)

∴ 26z^{3} (32z^{2} – 18) ÷ 13z^{2} (4z – 3)