AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

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) a² + 10a +25
ii) l² – 16l + 64
iii) 36x² + 96xy + 64y²
iv) 25x² + 9y² – 30xy
v) 25m²– 40mn + 1 6n²
vi) 81x² – 198 xy + 12ly²
vii) (x+y)² – 4xy
(Hint : first expand ( x + y)² )
viii) l⁴ + 4l²m² + 4m⁴
Solution:
i) a² + 10a +25
= (a)² + 2 × a × 5 + (5)²
It is in the form of a² + 2ab + b²
a² + 2ab + b²= (a + b)²
∴ a² + 10a + 25 = (a + 5)² = (a + 5) (a + 5)

ii) l² – 16l + 64
l² – 16l + 64
= (l)² – 2 × l × 8 + (8)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ l² – 16l + 64 = (l – 8)² = (l – 8) (l – 8)

iii) 36x² + 96xy + 64y²
36x² + 96xy + 64y²
= (6x)² + 2 × 6x × 8y + (8y)²
It is in the form of a² + 2ab + b²
a² + 2ab + b² = (a + b)²
∴ 36x² + 96xy + 64y²
= (6x + 8y)² = (6x + 8y) (6x + 8y)

iv) 25x² + 9y² – 30xy
25x² + 9y² – 30xy
= (5x)² + (3y)² – 2 × 5x × 3y
It is in the form of a² + b² – 2ab
a² + b² – 2ab = (a – b)²
∴ 25x² + 9y² – 30xy
= (5x – 3y)² = (5x – 3y) (5x – 3y)

v) 25m²– 40mn + 1 6n²
25m² – 40mn + 16n²
= (5m)² – 2 × 5m × 4n + (4n)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ 25m² – 40mn + 16n²
= (5m – 4n)²
= (5m – 4n) (5m – 4n)

vi) 81x² – 198 xy + 12ly²
81x² – 198xy + 121y²
= (9x)² – 2 × 9x × 11y + (11y)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ 81x² – 198xy + 121y²
= (9x – 11y)² – (9x – 11y) (9x – 11y)

vii) (x+y)² – 4xy
(Hint : first expand ( x + y)² )
= (x + y)² – 4xy
= x² + y² + 2xy – 4xy
= x² + y² – 2xy = (x – y)² = (x – y)(x – y)

viii) l⁴ + 4l²m² + 4m⁴
l⁴ + 4l²m² + 4m⁴
= (l²)² + 2 × l² × 2m² + (2m²)²
It is in the form of a² + 2ab + b²
a² + 2ab + b² = (a – b)²
∴ l⁴ + 4l²m² + 4m⁴
= (l² + 2m²)² = (l² + 2m²) (l² + 2m²)

Question 2.
Factorise the following
i) x² – 36
ii) 49x² – 25y²
iii) m² – 121
iv) 81 – 64x²
v) x²y² – 64
vi) 6x² – 54
vii) x² – 81
viii) 2x -32 x⁵
ix) 81x⁴ – 121x²
x) (p² – 2pq + q²)-r²
xi) (x+y)² – (x-y)²
Solution:
i) x² – 36
x² – 36
⇒ (x)² – (6)² is in the form of a² – b²
a² – b² = (a + b) (a – b)
∴ x² – 36 = (x + 6) (x – 6)

ii) 49x² – 25y²
= (7x)² – (5y)²
= (7x + 5y) (7x – 5y)

iii) m² – 121
m² -121
= (m)² – (11)²
= (m + 11) (m – 11)

iv) 81 – 64x²
81 – 64x²
= (9)² – (8x)²
= (9 + 8x) (9 – 8x)

v) x²y² – 64
= (xy)² – (8)²
= (xy + 8)(xy – 8)

vi) 6x² – 54
6x² – 54
= 6x² – 6 x 9 ‘
= 6(x² – 9)
= 6[(x)² – (3)²]
= 6(x + 3) (x – 3)

vii) x² – 81
x² – 81
= x² – 9²
= (x + 9 )(x – 9)

viii) 2x – 32 x⁵
2x – 32 x⁵
= 2x – 2x x 16x⁴
= 2 x (1 – 16x⁴)
= 2x [1²) – (4x²)²]
= 2x (1 + 4x²) (1 – 4x²)
= 2x (1 + 4x²) [(1⁵ – (2x)²]
= 2x (1 + 4x²) (1 + 2x) (1 – 2x)

ix) 81x⁴ – 121x²
81x⁴ – 121x²
– x² (81² – 121)
= x²[(9x)² – (11)²]
= x² (9x + 11) (9x -11)

x) (p² – 2pq + q²)-r²
(p² – 2pq + q²) – r²
= (p – q)² – (r)² [∵ p² – 2pq + q² = (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 a² – b²
a = x + y, b = x- y
∴ a² – b² =(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) lx² + mx
(ii) 7y² + 35Z²
(iii) 3x⁴ + 6x³y + 9x²Z
(iv) x² – ax – bx + ab
(v) 3ax – 6ay – 8by + 4bx
(vi) mn + m + n + 1
(vii) 6ab – b² + 12ac – 2bc
(viii) p²q – pr² – pq + r²
(ix) x (y + z) -5 (y + z)

(i) lx² + mx
lx² + mx
= l × x × x + m × x = x(lx + m)

(ii) 7y² + 35z²
7y²+ 35z²
= 7 × y² + 7 × 5 × z²
= 7(y² + 5z²)

(iii) 3x⁴ + 6x³y + 9x²Z
3x⁴ + 6x³y + 9x²Z
= 3 × x² × x² + 3 × 2 × x × x² × y + 3 × 3 × x² × z
= 3x² (x² + 2xy + 3z)

(iv) x² – ax – bx + ab
x² – ax – bx + ab
= (x² – 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 – b² + 12ac – 2bc
6ab – b² + 12ac – 2bc
= (6ab – b²) + (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) p²q – pr² – pq + r²
p²q – pr² – pq + r²
= (p²q – pr²) – (pq – r²)
= (p × p × q – p × r × r) – (pq – r²)
= P(pq – r²) – (pq – r²) × 1
= (pq – r²)(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) x⁴ – y⁴
(ii) a⁴ – (b + c)⁴
(iii) l² – (m – n)²
(iv) 49x² – \(\frac{16}{25}\)
(v) x⁴ – 2x²y² + y⁴
(vi) 4 (a + b)² – 9 (a – b)²
Solution:
= (x²)² – (y²)² is in the form of a² – b²
a² – b² = (a + b) (a – b)
x⁴ – y⁴ = (x² + y²)(x² – y²)
= (x² + y²)(x + y)(x – y)

(ii) a⁴ – (b + c)⁴
a⁴ – (b + c)⁴
= (a²)² – [(b + c)²]²
= [a² + (b + c)²] [a² – (b + c)²] ,
= [a² + (b + c)²] (a + b + c) [a – (b + c)]
= [a² + (b + c)²] (a + b + c) (a – b – c)

(iii) l² – (m – n)²
l² – (m – n)²
= (l)² – (m – n)²
= [l + m – n] [l – (m – n)]
= [l + m -n] [l – m + n]

(iv) 49x² – \(\frac{16}{25}\)
= (7x)² – (\(\frac{4}{5}\))²
= (7x+ (\(\frac{4}{5}\)) (7x – (\(\frac{4}{5}\))

(v) x⁴ – 2x² y² + y⁴
= (x² )² – 2x² y² + (y² )²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ x⁴ – 2x² y² + y⁴ = (x² – y² )²
= [(x)² – (y)² ]²
= [(x + y) (x – y)]²
= (x + y)² (x – y)²
[∵ (ab)^m = a ^m . bⁿ ]

(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) a²+ 10a + 24
(ii) x² +9x + 18
(iii) p² – 10q + 21
(iv) x² – 4x – 32
Solution:
(i) a²+ 10a + 24
a² + 10a + 24 .
= a² + 6a + 4a + 24
= a x a + 6a + 4a + 6 × 4
= a(a + 6) + 4(a + 6)
= (a + 6) (a + 4) (or)
a² + 10a + 24

∴ a² + 10a + 24 = (a + 6) (a + 4)

(ii) x² + 9x + 18
x² + 9x + 18
= (x + 3) (x + 6)

∴ x² + 9x + 18 = (x + 3) (x + 6)

(iii) p² – 10q + 21
p² – 10p + 21
= (P – 7) (p – 3)

∴ p² – 10p + 21 = (p – 7)(p – 3)

(iv) x² – 4x – 32
x² – 4x – 32
= (x – 8) (x + 4)

∴ x² – 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 x² + 3xy + x + my – in has two linear factors in x and y, with integer coefficients.
Solution:
Given equation is x² + 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)
= x² + 0xy + bx + 3xy + 0y² + 3by + ax + 0y + ab
= x² + bx + ax + 3xy + 3by + ab ………….. (2)
Comparing equation (2) with (1),
x² + 3xy + x + my – m
= x² + (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 – m² = – 9m
⇒ m² – 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.