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.