Factorisation
Class 8th Mathematics Part I Karnataka Board Solution
- Resolve in to factors: (i) x^2 + xy (ii) 3x^2 - 6x (iii) (1.6)a^2 - (0.8)a (iv)…
- a^2 + ax + ab + bx Factorise:
- 3ac + 7bc - 3ad - 7bd Factorise:
- 3xy - 6zy - 3xt + 6zt Factorise:
- y^3 - 3y^2 + 2y - 6 - xy + 3x Factorise:
- 4a^2 - 25 Factorise:
- x^9 - 9/16 Factorise:
- x^4 - y^4 Factorise:
- (7 3/10)^2 - (2 1/10)^2 Factorise:
- (0.7)^2 - (0.3)^2 Factorise:
- (5a - 2b)^2 - (2a - b)^2 Factorise:
- pq = 18 and p + q = 11 In the following, you are given the product pq and the…
- pq = 32 and p + q = -12 In the following, you are given the product pq and the…
- pq = -24 and p + q = 2 In the following, you are given the product pq and the…
- pq = -12 and p + q = 11 In the following, you are given the product pq and the…
- pq = -6 and p + q = -5 In the following, you are given the product pq and the…
- pq = -44 and p + q = -7 In the following, you are given the product pq and the…
- x^2 + 6x + 8 Factorise:
- x^2 + 4x + 3 Factorise:
- a^2 + 5a + 6 Factorise:
- a^2 - 5a + 6 Factorise:
- a^2 - 3a - 40 Factorise:
- x^2 - x - 72 Factorise:
- x^2 + 14x + 49 Factorise:
- 4x^2 + 4x + 1 Factorise:
- a^2 - 10a + 25 Factorise:
- 2x^2 -24x + 72 Factorise:
- p^2 - 24p + 144 Factorise:
- x^3 - 12x^2 + 36x Factorise:
- 4a + 12b is equal toA. 4a B. 12b C. 4(a + 3b) D. 3a
- The product of two numbers is positive and their sum negative only whenA. both are…
- Factorising x^2 + 6x + 8, we getA. (x + 1)(x + 8) B. (x + 6)(x + 2) C. (x + 10)(x - 2)…
- The denominator of an algebraic fraction should not beA. 1 B. 0 C. 4 D. 7…
- If the sum of two integers is -2 and their product is -24, the numbers areA. 6 and 4…
- The difference (0.7)^2 - (0.3)^2 simplifies toA. 0.4 B. 0.04 C. 0.49 D. 0.56…
- (i) x^2 + 6x + 9 (ii) 1 - 8x + 16x^2 (iii) 4x^2 - 81y^2 (iv) 4a^2 + 4ab + b^2 (v) a^2…
- x^2 + 7x + 12 Foctorise the following:
- x^2 + x - 12 Foctorise the following:
- x^2 - 3x - 18 Foctorise the following:
- x^2 + 4x - 21 Foctorise the following:
- x^2 - 4x - 192 Foctorise the following:
- x^4 - 5x^2 + 4 Foctorise the following:
- x^4 - 13x^2 y^2 + 36y^4 . Foctorise the following:
- 2x^2 + 7x + 6 Foctorise the following:
- 3x^2 - 17x + 20 Foctorise the following:
- 6x^2 - 5x - 14 Foctorise the following:
- 4x^2 + 12xy + 5y^2 Foctorise the following:
- 4x^4 - 5x^2 + 1. Foctorise the following:
- x^8 - y^8 Factorise the following:
- ax^4 - ax^12 Factorise the following:
- x + x^2 + 1 Factorise the following:
- x^4 + 5x^2 + 9. Factorise the following:
- Factorise x^4 + 4y^4 . Use this to prove that 2011^4 + 64 is a composite number.…
Exercise 4.1
Question 1.Resolve in to factors:
(i) x2 + xy
(ii) 3x2 – 6x
(iii) (1.6)a2 – (0.8)a
(iv) 5 – 10m – 20n
Answer:
(i) x × x + x × y
taking x in common,
= x(x + y)
(ii) 3 × x × x–6 × x
taking 3 x in common,
= 3x(x–2)
(iii) 2 × 0.8 × a × a–0.8 × a
taking 0.8a in common,
= 0.8a(2a–1)
(iv) 5–5 × 2 × m–5 × 4 × n
taking 5 in common,
= 5(1–2m–4n)
Question 2.
Factorise:
a2 + ax + ab + bx
Answer:
= a(a + x) + b(a + x)
= (a + b)(a + x)
Question 3.
Factorise:
3ac + 7bc – 3ad – 7bd
Answer:
= c(3a + 7b)–d(3a + 7b)
= (c–d)(3a + 7b)
Question 4.
Factorise:
3xy – 6zy – 3xt + 6zt
Answer:
= 3y(x–2z)–3t(x–2z)
= (3y–3t)(x–2z)
= 3(y–t)(x–2z)
Question 5.
Factorise:
y3 – 3y2 + 2y – 6 – xy + 3x
Answer:
= y3 + 2y–xy–3y2 + 3x–6
= y(y2 + 2–x)–3(y2 + 2–x)
= (y–3)(y2 + 2–x)
Question 6.
Factorise:
4a2 – 25
Answer:
Using the identity,
x2–y2 = (x + y)(x–y)
so, (2a)2–(5)2
= (2a–5)(2a + 5)
Question 7.
Factorise:
Answer:
Using the identity,
x2–y2 = (x + y)(x–y)
so, x2–
= 
Question 8.
Factorise:
x4 – y4
Answer:
Using the identity,
x2–y2 = (x + y)(x–y)
so, (x2)2–(y2)2
= (x2–y2)(x2 + y2)
= (x–y)(x + y)(x2 + y2)
Question 9.
Factorise:
Answer:
: Using the identity,
x2–y2 = (x + y)(x–y)
so,
Question 10.
Factorise:
(0.7)2 – (0.3)2
Answer:
: Using the identity,
x2–y2 = (x + y)(x–y)
so, 0.72–0.32 = (0.7–0.3)(0.7 + 0.3)
= 0.4
1.0
= 0.4
Question 11.
Factorise:
(5a – 2b)2 – (2a – b)2
Answer:
: Using the identity,
x2–y2 = (x + y)(x–y)
so, (5a – 2b)2 – (2a – b)2 = (5a–2b–2a + b)(5a–2b + 2a–b)
= (3a–b)(7a–3b)
= 21a2–9ab–7ab + 3b2
= 21a2–16ab + 3b2
Exercise 4.2
Question 1.In the following, you are given the product pq and the sum p + q. Determine p and q:
pq = 18 and p + q = 11
Answer:
as, p + q = 11
⇒p = 11–q
putting the value of q in other equation,
⇒ pq = 18
⇒ (11–q)q = 18
⇒11q–q2 = 18
⇒q2–11q + 18 = 0
⇒q2–2q–9q + 18 = 0
⇒q(q–2)–9(q–2) = 0
⇒ (q–2)(q–9) = 0
So, q = 2 & q = 9
As, p = 11–q
Thus, p = 11–2 = 9 & p = 11–9 = 2
Question 2.
In the following, you are given the product pq and the sum p + q. Determine p and q:
pq = 32 and p + q = –12
Answer:
As, p + q = –12
⇒p = –12–q
putting the value of q in other equation,
⇒ pq = 32
⇒ (–12–q)q = 32
⇒–12q–q2 = 32
⇒q2 + 12q + 32 = 0
⇒q2 + 8q + 4q + 32 = 0
⇒q(q + 8) + 4(q + 4) = 0
⇒ (q + 4)(q + 8) = 0
so, q = –4 & q = –8
as, p = –12–q
thus, p = –12 + 4 & p = –12 + 8
p = –8 & p = –4
Question 3.
In the following, you are given the product pq and the sum p + q. Determine p and q:
pq = –24 and p + q = 2
Answer:
As, p + q = 2
⇒p = 2–q
putting the value of q in other equation,
⇒ pq = –24
⇒ (2–q)q = –24
⇒2q–q2 = –24
⇒q2–2q–24 = 0
⇒q2–6q + 4q–24 = 0
⇒q(q–6) + 4(q–6) = 0
⇒ (q + 4)(q–6) = 0
so, q = –4 & q = 6
as, p = 2–4
thus, p = 2 + 4 & p = 2–6
p = 6 & p = –4
Question 4.
In the following, you are given the product pq and the sum p + q. Determine p and q:
pq = –12 and p + q = 11
Answer:
As, p + q = 11
⇒p = 11–q
putting the value of q in other equation, ⇒
⇒ pq = –12
⇒ (11–q)q = –12
⇒11q–q2 = –12
⇒q2–11q–12 = 0
⇒q2 + q–12q–12 = 0
⇒q(q + 1)–12(q + 1) = 0
⇒ (q–12)(q + 1) = 0
So. q = 12 & q = –1
As, p = 11–q
thus, p = 11–12 & p = 11 + 1
p = –1 & p = 12
Question 5.
In the following, you are given the product pq and the sum p + q. Determine p and q:
pq = –6 and p + q = –5
Answer:
p + q = –5
⇒p = –5–q
putting the value of q in other equation,
⇒ pq = –6
⇒ (–5–q)q = –6
⇒–5q–q2 = –6
⇒q2 + 5q–6 = 0
⇒q2 + 6q–q–6 = 0
⇒q(q + 6)–1(q + 6) = 0
⇒ (q–1)(q + 6) = 0
so, q = 1 & q = –6
as p = –5–q
thus, p = –5–1 & p = –5 + 6
p = –6 & p = 1
Question 6.
In the following, you are given the product pq and the sum p + q. Determine p and q:
pq = –44 and p + q = –7
Answer:
p + q = –7
⇒p = –7–q
putting the value of q in other equation,
⇒ pq = –44
⇒ (–7–q)q = –44
⇒–7q–q2 = –44
⇒q2 + 7q–44 = 0
⇒q2 + 11q–4q–44 = 0
⇒q(q + 11)–4(q + 11) = 0
⇒ (q–4)(q + 11) = 0
so, q = 4 & q = –11
as, p = –7–q
thus, p = –7–4 & p = –7 + 11
p = –11 & p = 4
Question 7.
Factorise:
x2 + 6x + 8
Answer:
x2 + 4x + 2x + 8
= x(x + 4) + 2(x + 4)
= (x + 2)(x + 4)
Question 8.
Factorise:
x2 + 4x + 3
Answer:
x2 + x + 3x + 3
= x(x + 1) + 3(x + 1)
= (x + 1)(x + 3)
Question 9.
Factorise:
a2 + 5a + 6
Answer:
a2 + 2a + 3a + 6
= a(a + 2) + 3(a + 2)
= (a + 3)(a + 2)
Question 10.
Factorise:
a2 – 5a + 6
Answer:
a2–2a–3a + 6
= a(a–2)–3(a–2)
= (a–3)(a–2)
Question 11.
Factorise:
a2 – 3a – 40
Answer:
a2–8a + 5a–40
= a(a–4) + 5(a–4)
= (a + 5)(a–4)
Question 12.
Factorise:
x2 – x – 72
Answer:
x2–9x + 8x–72
= x(x–9) + 8(x–9)
= (x + 8)(x–9)
Question 13.
Factorise:
x2 + 14x + 49
Answer:
using (x + y)2 = x2 + 2xy + y2
= x2 + 2 × 7 × x + 7 × 7
= (x + 7)2
= (x + 7)(x + 7)
Question 14.
Factorise:
4x2 + 4x + 1
Answer:
using (x + y)2 = x2 + 2xy + y2
= (2x)2 + 2 × 2x × 1 + 12
= (2x + 1)2
= (2x + 1)(2x + 1)
Question 15.
Factorise:
a2 – 10a + 25
Answer:
using (x–y)2 = x2–2xy + y2
= a2–2 × a × 5 + 52
= (a–5)2
= (a–5)(a–5)
Question 16.
Factorise:
2x2–24x + 72
Answer:
using (x–y)2 = x2–2xy + y2
= 2[x2–12x + 36]
= 2[x2–2 × x × 6 + 62]
= 2(x–6)2
= 2(x–6)(x–6)
Question 17.
Factorise:
p2 – 24p + 144
Answer:
using (x–y)2 = x2–2xy + y2
= p2–2 × p × 12 + 122
= (p–12)2
= (p–12)(p–12)
Question 18.
Factorise:
x3 – 12x2 + 36x
Answer:
using (x–y)2 = x2–2xy + y2
= x[x2–12x + 36]
= x[x2–2 × x × 6 + 62]
= x(x–6)2
= x(x–6)(x–6)
Additional Problems 4
Question 1.4a + 12b is equal to
A. 4a
B. 12b
C. 4(a + 3b)
D. 3a
Answer:
⇒ 4a + 12b
⇒ 4(a + 3b)
Question 2.
The product of two numbers is positive and their sum negative only when
A. both are positive
B. both are negative
C. one positive the other negative
D. one of them equal to zero
Answer:
The product of two numbers is positive, when either both the numbers are positive or both the numbers are negative.
Sum of two positive numbers is positive and the sum of two negative numbers is negative.
∴ The product of two numbers is positive and their sum negative only when both are negative.
Question 3.
Factorising x2 + 6x + 8, we get
A. (x + 1)(x + 8)
B. (x + 6)(x + 2)
C. (x + 10)(x – 2)
D. (x + 4)(x + 2)
Answer:
⇒ x2 + 6x + 8
⇒ x2 + 4x + 2x + 8
⇒ x(x + 4) + 2(x + 4)
⇒ (x + 4)(x + 2)
Question 4.
The denominator of an algebraic fraction should not be
A. 1
B. 0
C. 4
D. 7
Answer:
The denominator of a algebraic function should not be 0.
∵ When denominator is 0, the fraction becomes undefined.
Question 5.
If the sum of two integers is –2 and their product is –24, the numbers are
A. 6 and 4
B. –6 and 4
C. –6 and –4
D. 6 and –4
Answer:
Let, the two integers = x and y
According to problem,
⇒ x + y = - 2 ……… (1)
and
⇒ xy = - 24 ……… (2)
∴ (x + y)2 = (-2)2
⇒ (x – y)2 + 4xy = 4
⇒ (x – y)2 + 4 × (-24) = 4
⇒ (x – y)2 = 4 + 96 = 100
⇒ x – y = ±10
Case 1.
x - y = 10 …… (a) and x + y = - 2 …… (b)
From (a) + (b) we get,
⇒ 2x = 8
⇒ x = 4
∴ 4 – y = 10 [from (a)]
⇒ y = 4 – 10
⇒ y = - 6
Case 2.
x – y = - 10 …… (c) and x + y = - 2 …… (d)
From (c) + (d) we get,
⇒ 2x = - 12
⇒ x = - 6
∴ -6 – y = - 10
⇒ y = 10 – 6
⇒ y = 4
∴ The numbers are 4 and – 6.
Question 6.
The difference (0.7)2 – (0.3)2 simplifies to
A. 0.4
B. 0.04
C. 0.49
D. 0.56
Answer:
⇒ (0.7)2 – (0.3)2
⇒ (0.7 + 0.3)(0.7 – 0.3)
⇒ 1 × 0.4
⇒ 0.4
Question 7.
Factorise the following:
(i) x2 + 6x + 9
(ii) 1 – 8x + 16x2
(iii) 4x2 – 81y2
(iv) 4a2 + 4ab + b2
(v) a2b2 + c2d2 – a2c2 – b2d2.
Answer:
(i) ⇒ x2 + 6x + 9
⇒ x2 + 2 × 3× x + 32
⇒ (x + 3)2
(ii) ⇒ 1 – 8x + 16x2
⇒ 12 – 2 × 1 × 4x + (4x)2
⇒ (1 – 4x)2
(iii) ⇒ 4x2 – 81y2
⇒ (2x)2 – (9y)2
⇒ (2x + 9y)(2x – 9y)
(iv) ⇒ 4a2 + 4ab + b2
⇒ (2a)2 + 2 × 2a × b + b2
⇒ (2a + b)2
(v) ⇒ a2b2 + c2d2 – a2c2 – b2d2
⇒ (a2b2 – a2c2) – (c2d2 – b2d2)
⇒ a2(b2 – c2) – d2(b2 – c2)
⇒ (b2 – c2) (a2 – d2)
Question 8.
Foctorise the following:
x2 + 7x + 12
Answer:
⇒ x2 + 7x + 12
⇒ x2 + 4x + 3x + 12
⇒ x(x + 4) + 3(x + 4)
⇒ (x + 4)(x + 3)
Question 9.
Foctorise the following:
x2 + x – 12
Answer:
⇒ x2 + x – 12
⇒ x2 + 4x - 3x - 12
⇒ x(x + 4) – 3(x + 4)
⇒ (x + 4)(x - 3)
Question 10.
Foctorise the following:
x2 – 3x – 18
Answer:
⇒ x2 - 3x - 18
⇒ x2 - 6x + 3x - 18
⇒ x(x - 6) + 3(x - 6)
⇒ (x - 6)(x + 3)
Question 11.
Foctorise the following:
x2 + 4x – 21
Answer:
⇒ x2 + 4x – 21
⇒ x2 + 7x - 3x – 21
⇒ x(x + 7) – 3(x + 7)
⇒ (x + 7)(x – 3)
Question 12.
Foctorise the following:
x2 – 4x – 192
Answer:
⇒ x2 - 4x – 192
⇒ x2 - 16x + 12x – 192
⇒ x(x – 16) + 12(x – 16)
⇒ (x – 16)(x + 12)
Question 13.
Foctorise the following:
x4 – 5x2 + 4
Answer:
⇒ x4 - 5x2 + 4
⇒ x4 - 4x2 – x2 + 4
⇒ x2(x2 - 4) - (x2 - 4)
⇒ (x2 - 4)(x2 - 1)
⇒ (x2 – 22)(x2 – 12)
⇒ (x – 2)(x + 2)(x – 1)(x + 1)
Question 14.
Foctorise the following:
x4 – 13x2y2 + 36y4.
Answer:
⇒ x4 - 13x2y2 + 36y4
⇒ x4 - 4x2 y2 - 9x2y2 + 36y4
⇒ x2(x2 – 4y2) - 9y2(x2 – 4y2)
⇒ (x2 – 4y2)(x2 – 9y2)
⇒ {x2 – (2y)2}{x2 – (3y)2}
⇒ (x – 2y)(x + 2y)(x – 3y)(x + 3y)
Since,
Question 15.
Foctorise the following:
2x2 + 7x + 6
Answer:
⇒ 2x2 + 7x + 6
⇒ 2x2 + 4x + 3x + 6
⇒ 2x(x + 2) + 3(x + 2)
⇒ (x + 2)(2x + 3)
Question 16.
Foctorise the following:
3x2 – 17x + 20
Answer:
⇒ 3x2 – 17x + 20
⇒ 3x2 - 12x - 5x + 20
⇒ 3x(x - 4) - 5(x - 4)
⇒ (x - 4)(3x - 5)
Question 17.
Foctorise the following:
6x2 – 5x – 14
Answer:
⇒ 6x2 - 5x – 14
⇒ 6x2 - 12x + 7x – 14
⇒ 6x(x – 2) + 7(x – 2)
⇒ (x - 2)(6x + 7)
Question 18.
Foctorise the following:
4x2 + 12xy + 5y2
Answer:
⇒ 4x2 + 12xy + 5y2
⇒ 4x2 + 10xy + 2xy + 5y2
⇒ 2x(2x + 5y) + y(2x + 5y)
⇒ (2x + 5y)(2x + y)
Question 19.
Foctorise the following:
4x4 – 5x2 + 1.
Answer:
⇒ 4x4 - 5x2 + 1
⇒ 4x4 - 4x2 – x2 + 1
⇒ 4x2(x2 - 1) - (x2 – 1)
⇒ (x2 - 1)(4x2 - 1)
⇒ (x – 1)(x + 1)(2x – 1)(2x + 1)
Question 20.
Factorise the following:
x8 – y8
Answer:
⇒ x8 – y8
⇒ (x4)2 – (y4)2
⇒ (x4 + y4)(x4 – y4)
⇒ (x4 + y4){(x2)2 –( y2)2}
⇒ (x4 + y4)(x2 + y2)(x2 – y2)
⇒ (x4 + y4)(x2 + y2)(x + y)(x – y)
Question 21.
Factorise the following:
ax4 – ax12
Answer:
⇒ ax4 – ax12
⇒ ax4(1 – x8)
⇒ ax4{1 – (x4)2}
⇒ ax4(1 + x4)(1 – x4)
⇒ ax4(1 + x4){1 – (x2)2}
⇒ ax4 (1 + x4)(1 + x2)(1 - x2)
⇒ ax4 (1 + x4)(1 + x2)(1 + x)(1 – x)
Question 22.
Factorise the following:
x + x2 + 1
Answer:
⇒ x2 + x4 + 1
⇒ x4 + 2x2 + 1 – x2
⇒ x4 + 2 × x2 × 1 + 12 – x2
⇒ (x2 + 1)2 – x2
⇒ (x2 + 1 + x)(x2 + 1 – x) 
Question 23.
Factorise the following:
x4 + 5x2 + 9.
Answer:
⇒ x4 + 5x2 + 9
⇒ x4 + 6x2 + 9 – x2
⇒ (x2)2 + 2 × 3 × x2 + 32 – x2
⇒ (x2 + 3)2 – x2
⇒ (x2 + 3 + x) (x2 + 3 – x)
Since,
Question 24.
Factorise x4 + 4y4. Use this to prove that 20114 + 64 is a composite number.
Answer:
⇒ x4 + 4y4
⇒ (x2)2 + (2y2)2
⇒ (x2 + 2y2)2 - 2× x2 × 2y2
⇒ (x2 + 2y2)2 - 4x2y2
⇒ (x2 + 2y2)2 – (2xy)2
⇒ (x2 + 2y2 + 2xy)(x2 + 2y2 – 2xy)
Now we get,
⇒ 20114 + 64
⇒ 20114 + 4 × 24
⇒ (20112)2 + (2×22)2
⇒ (20112 + 42)2 - 2× 20112 × 2 × 22
⇒ (20112 + 42)2 – 4 × 20112 × 22
⇒ (20112 + 42)2 – (2×2011×2)2
()
⇒ (20112 + 42 + 2×2011×2)(20112 + 42 – 2×2011×2)
⇒ (4044121 + 16 + 8044)(4044121 + 16 – 8044)
⇒ 4052181 × 4036093