AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.2

Question 1.

Factorise the following expression

i) a^{2} + 10a +25

ii) l^{2} – 16l + 64

iii) 36x^{2} + 96xy + 64y^{2}

iv) 25x^{2} + 9y^{2} – 30xy

v) 25m^{2}– 40mn + 1 6n^{2}

vi) 81x^{2} – 198 xy + 12ly^{2}

vii) (x+y)^{2} – 4xy

(Hint : first expand ( x + y)^{2} )

viii) l^{4} + 4l^{2}m^{2} + 4m^{4}

Solution:

i) a^{2} + 10a +25

= (a)^{2} + 2 × a × 5 + (5)^{2}

It is in the form of a^{2} + 2ab + b^{2}

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

∴ a^{2} + 10a + 25 = (a + 5)^{2} = (a + 5) (a + 5)

ii) l^{2} – 16l + 64

l^{2} – 16l + 64

= (l)^{2} – 2 × l × 8 + (8)^{2}

It is in the form of a^{2} – 2ab + b^{2}

a^{2} – 2ab + b^{2} = (a – b)^{2}

∴ l^{2} – 16l + 64 = (l – 8)^{2} = (l – 8) (l – 8)

iii) 36x^{2} + 96xy + 64y^{2}

36x^{2} + 96xy + 64y^{2}

= (6x)^{2} + 2 × 6x × 8y + (8y)^{2}

It is in the form of a^{2} + 2ab + b^{2}

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

∴ 36x^{2} + 96xy + 64y^{2}

= (6x + 8y)^{2} = (6x + 8y) (6x + 8y)

iv) 25x^{2} + 9y^{2} – 30xy

25x^{2} + 9y^{2} – 30xy

= (5x)^{2} + (3y)^{2} – 2 × 5x × 3y

It is in the form of a^{2} + b^{2} – 2ab

a^{2} + b^{2} – 2ab = (a – b)^{2}

∴ 25x^{2} + 9y^{2} – 30xy

= (5x – 3y)^{2} = (5x – 3y) (5x – 3y)

v) 25m^{2}– 40mn + 1 6n^{2}

25m^{2} – 40mn + 16n^{2}

= (5m)^{2} – 2 × 5m × 4n + (4n)^{2}

It is in the form of a^{2} – 2ab + b^{2}

a^{2} – 2ab + b^{2} = (a – b)^{2}

∴ 25m^{2} – 40mn + 16n^{2}

= (5m – 4n)^{2}

= (5m – 4n) (5m – 4n)

vi) 81x^{2} – 198 xy + 12ly^{2}

81x^{2} – 198xy + 121y^{2}

= (9x)^{2} – 2 × 9x × 11y + (11y)^{2}

It is in the form of a^{2} – 2ab + b^{2}

a^{2} – 2ab + b^{2} = (a – b)^{2}

∴ 81x^{2} – 198xy + 121y^{2}

= (9x – 11y)^{2} – (9x – 11y) (9x – 11y)

vii) (x+y)^{2} – 4xy

(Hint : first expand ( x + y)^{2} )

= (x + y)^{2} – 4xy

= x^{2} + y^{2} + 2xy – 4xy

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

viii) l^{4} + 4l^{2}m^{2} + 4m^{4}

l^{4} + 4l^{2}m^{2} + 4m^{4}

= (l^{2})^{2} + 2 × l^{2} × 2m^{2} + (2m^{2})^{2}

It is in the form of a^{2} + 2ab + b^{2}

a^{2} + 2ab + b^{2} = (a – b)^{2}

∴ l^{4} + 4l^{2}m^{2} + 4m^{4}

= (l^{2} + 2m^{2})^{2} = (l^{2} + 2m^{2}) (l^{2} + 2m^{2})

Question 2.

Factorise the following

i) x^{2} – 36

ii) 49x^{2} – 25y^{2}

iii) m^{2} – 121

iv) 81 – 64x^{2}

v) x^{2}y^{2} – 64

vi) 6x^{2} – 54

vii) x^{2} – 81

viii) 2x -32 x^{5}

ix) 81x^{4} – 121x^{2}

x) (p^{2} – 2pq + q^{2})-r^{2}

xi) (x+y)^{2} – (x-y)^{2}

Solution:

i) x^{2} – 36

x^{2} – 36

⇒ (x)^{2} – (6)^{2} is in the form of a^{2} – b^{2}

a^{2} – b^{2} = (a + b) (a – b)

∴ x^{2} – 36 = (x + 6) (x – 6)

ii) 49x^{2} – 25y^{2}

= (7x)^{2} – (5y)^{2}

= (7x + 5y) (7x – 5y)

iii) m^{2} – 121

m^{2} -121

= (m)^{2} – (11)^{2}

= (m + 11) (m – 11)

iv) 81 – 64x^{2}

81 – 64x^{2}

= (9)^{2} – (8x)^{2}

= (9 + 8x) (9 – 8x)

v) x^{2}y^{2} – 64

= (xy)^{2} – (8)^{2}

= (xy + 8)(xy – 8)

vi) 6x^{2} – 54

6x^{2} – 54

= 6x^{2} – 6 x 9 ‘

= 6(x^{2} – 9)

= 6[(x)^{2} – (3)^{2}]

= 6(x + 3) (x – 3)

vii) x^{2} – 81

x^{2} – 81

= x^{2} – 9^{2}

= (x + 9 )(x – 9)

viii) 2x – 32 x^{5}

2x – 32 x^{5}

= 2x – 2x x 16x^{4}

= 2 x (1 – 16x^{4})

= 2x [1^{2}) – (4x^{2})^{2}]

= 2x (1 + 4x^{2}) (1 – 4x^{2})

= 2x (1 + 4x^{2}) [(1^{5} – (2x)^{2}]

= 2x (1 + 4x^{2}) (1 + 2x) (1 – 2x)

ix) 81x^{4} – 121x^{2}

81x^{4} – 121x^{2}

– x^{2} (81^{2} – 121)

= x^{2}[(9x)^{2} – (11)^{2}]

= x^{2} (9x + 11) (9x -11)

x) (p^{2} – 2pq + q^{2})-r^{2}

(p^{2} – 2pq + q^{2}) – r^{2}

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

= (p – q + r) (p – q – r)

xi) (x + y)^{2} – (x – y)^{2}

(x + y)^{2} – (x – y)^{2}

It is in the form of a^{2} – b^{2}

a = x + y, b = x- y

∴ a^{2} – b^{2} =(a + b)(a-b)

= (x + y + x – y) [x + y- (x – y)]

= 2x [x + y-x + y]

= 2x x 2y = 4xy

Question 3.

Factorise the expressions

(i) lx^{2} + mx

(ii) 7y^{2} + 35Z^{2}

(iii) 3x^{4} + 6x^{3}y + 9x^{2}Z

(iv) x^{2} – ax – bx + ab

(v) 3ax – 6ay – 8by + 4bx

(vi) mn + m + n + 1

(vii) 6ab – b^{2} + 12ac – 2bc

(viii) p^{2}q – pr^{2} – pq + r^{2}

(ix) x (y + z) -5 (y + z)

(i) lx^{2} + mx

lx^{2} + mx

= l × x × x + m × x = x(lx + m)

(ii) 7y^{2} + 35z^{2}

7y^{2}+ 35z^{2}

= 7 × y^{2} + 7 × 5 × z^{2}

= 7(y^{2} + 5z^{2})

(iii) 3x^{4} + 6x^{3}y + 9x^{2}Z

3x^{4} + 6x^{3}y + 9x^{2}Z

= 3 × x^{2} × x^{2} + 3 × 2 × x × x^{2} × y + 3 × 3 × x^{2} × z

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

(iv) x^{2} – ax – bx + ab

x^{2} – ax – bx + ab

= (x^{2} – ax) – (bx – ab)

= x(x – a) – b(x – a)

= (x – a) (x – b)

(v) 3ax – 6ay – 8by + 4bx

3ax – 6ay – 8by + 4bx

= (3ax – 6ay) + (4bx – 8by)

= 3a (x – 2y) + 4b (x – 2y)

= (x – 2y) (3a + 4b)

(vi) mn + m + n + 1

mn + m + n + 1

= (mn + m) + (n + 1)

= m (n + 1) + (n + 1)

= (n + 1) (m + 1)

(vii) 6ab – b^{2} + 12ac – 2bc

6ab – b^{2} + 12ac – 2bc

= (6ab – b^{2}) + (12ac – 2bc)

= (6 × a× b – b × b) + (6 × 2 × a × c – 2 × b × c)

= b [6a – b] + 2c [6a – b]

= (6a – b) (b + 2c)

(viii) p^{2}q – pr^{2} – pq + r^{2}

p^{2}q – pr^{2} – pq + r^{2}

= (p^{2}q – pr^{2}) – (pq – r^{2})

= (p × p × q – p × r × r) – (pq – r^{2})

= P(pq – r^{2}) – (pq – r^{2}) × 1

= (pq – r^{2})(p – 1)

(ix) x (y + z) -5 (y + z)

= x(y + z) – 5(y + z)

= (y + z) (x – 5)

Question 4.

Factorise the following

(i) x^{4} – y^{4}

(ii) a^{4} – (b + c)^{4}

(iii) l^{2} – (m – n)^{2}

(iv) 49x^{2} – \(\frac{16}{25}\)

(v) x^{4} – 2x^{2}y^{2} + y^{4}

(vi) 4 (a + b)^{2} – 9 (a – b)^{2}

Solution:

= (x^{2})^{2} – (y^{2})^{2} is in the form of a^{2} – b^{2}

a^{2} – b^{2} = (a + b) (a – b)

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

= (x^{2} + y^{2})(x + y)(x – y)

(ii) a^{4} – (b + c)^{4}

a^{4} – (b + c)^{4}

= (a^{2})^{2} – [(b + c)^{2}]^{2}

= [a^{2} + (b + c)^{2}] [a^{2} – (b + c)^{2}] ,

= [a^{2} + (b + c)^{2}] (a + b + c) [a – (b + c)]

= [a^{2} + (b + c)^{2}] (a + b + c) (a – b – c)

(iii) l^{2} – (m – n)^{2}

l^{2} – (m – n)^{2}

= (l)^{2} – (m – n)^{2}

= [l + m – n] [l – (m – n)]

= [l + m -n] [l – m + n]

(iv) 49x^{2} – \(\frac{16}{25}\)

= (7x)^{2} – (\(\frac{4}{5}\))^{2}

= (7x+ (\(\frac{4}{5}\)) (7x – (\(\frac{4}{5}\))

(v) x^{4} – 2x^{2} y^{2} + y^{4}

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

It is in the form of a^{2} – 2ab + b^{2}

a^{2} – 2ab + b^{2} = (a – b)^{2}

∴ x^{4} – 2x^{2} y^{2} + y^{4} = (x^{2} – y^{2} )^{2}

= [(x)^{2} – (y)^{2} ]^{2}

= [(x + y) (x – y)]^{2}

= (x + y)^{2} (x – y)^{2}

[∵ (ab)^{m} = a ^{m} . b^{n} ]

(vi) 4 (a + b)^{2} – 9 (a – b)^{2}

4 (a + b)^{2} – 9 (a – b)^{2}

= [2(a + b)]^{2} – [3(a – b)]^{2}

= [2(a + b) + 3(a- b)] [2(a + b)-3(a- b)]

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

= (5a – b) (5b – a)

Question 5.

Factorise the following expressions

(i) a^{2}+ 10a + 24

(ii) x^{2} +9x + 18

(iii) p^{2} – 10q + 21

(iv) x^{2} – 4x – 32

Solution:

(i) a^{2}+ 10a + 24

a^{2} + 10a + 24 .

= a^{2} + 6a + 4a + 24

= a x a + 6a + 4a + 6 × 4

= a(a + 6) + 4(a + 6)

= (a + 6) (a + 4) (or)

a^{2} + 10a + 24

∴ a^{2} + 10a + 24 = (a + 6) (a + 4)

(ii) x^{2} + 9x + 18

x^{2} + 9x + 18

= (x + 3) (x + 6)

∴ x^{2} + 9x + 18 = (x + 3) (x + 6)

(iii) p^{2} – 10q + 21

p^{2} – 10p + 21

= (P – 7) (p – 3)

∴ p^{2} – 10p + 21 = (p – 7)(p – 3)

(iv) x^{2} – 4x – 32

x^{2} – 4x – 32

= (x – 8) (x + 4)

∴ x^{2} – 4x – 32 = (x – 8) (x + 4)

Question 6.

The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.

Solution:

Perimeter of a triangle

= AB + BC + CA = 48

⇒ c + a + b = 48

The solutions of Harmeet, Rosy are wrong.

∴ Srikar had done it correctly.

⇒ 21 + a + b = 48

⇒ a + b = 48 – 21 = 27

∴ The lengths of a, b should be 10, 17

∴ a + b > c [the sum of any two sides of a triangle is greater than the 3rd side]

∴ 10 + 17 > 2

27 > 21 (T).

∴ The length of the shortest side is 10 cm.

Question 7.

Find the values of ‘m’ for which x^{2} + 3xy + x + my – in has two linear factors in x and y, with integer coefficients.

Solution:

Given equation is x^{2} + 3xy + x + my – m ……….(1)

Let the two linear equations in x and y be (x + 3y + a) and (x + 0y + b).

Then (x + 3y + a) (x + 0y + b)

= x^{2} + 0xy + bx + 3xy + 0y^{2} + 3by + ax + 0y + ab

= x^{2} + bx + ax + 3xy + 3by + ab ………….. (2)

Comparing equation (2) with (1),

x^{2} + 3xy + x + my – m

= x^{2} + (a + b)x + 3xy + 3by + ab

Equating the like terms on both sides,

ab = – m ………….. (3)

(a + b)x = x ⇒ a + b = 1 ……………. (4)

3by = my ⇒ 3b = m ⇒ b = \(\frac{\mathrm{m}}{3}\)

Substitute ‘b’ value in equation (4),

a = \(1-\frac{m}{3}=\frac{3-m}{3}\)

ab = -m

[ ∵ from (3)]

put a & b value then ,

\(\left(\frac{3-m}{3}\right)\left(\frac{m}{3}\right)\) = -m

\(\frac{3 \mathrm{~m}-\mathrm{m}^{2}}{9}\)= -m

⇒ 3m – m^{2} = – 9m

⇒ m^{2} – 12m = 0

⇒ m(m – 12) = 0

⇒ m = 0 (or) m = 12

lf m = 12

∴ b = \(\frac{12}{3}\) = 4&a = \(\frac{3-\mathrm{m}}{3}=\frac{3-12}{3}\)

= \(\frac{-9}{3}\) = -3

∴ Linear factors are (x + 3y – 3), (x + 4) If m = 0

b = \(\frac{0}{3}\) = 0 & a = \(\frac{3-0}{3}=\frac{3}{3}\) = 1

∴ Linear factors are (x + 3y + 1), x.