WBBSE Solutions For Class 7 Maths Algebra Chapter 6 Factorisation Exercise 6 Solved Problems

Class 7 Math Solution WBBSE Algebra Chapter 6 Factorisation Exercise 6 Solved Problems

1. Factorisation of an algebraic expression is the process of finding two or more expressions whose product is the given expression.

⇒ These two or more expressions are called factors.
⇒ Factors of 6 are 1, 2, 3 and 6 [6 = 1 x 6 = 2 x 3]

⇒ The prime factors of 6 are 2, 3

2. Every number or expression is a factor of itself and 1 is a factor of any number or expression.

3. Generally 1 and the number itself may not be written to mention the factors of a number. The factors of 3ab is 3, a, and b where 3, a, and b are prime factors.

⇒ As 6x²y= 2 x 3 x x x x x y

Read and Learn More WBBSE Solutions for Class 7 Maths

Common factor: An expression that can divide all the terms of a polynomial is called the common factor of the polynomial.

Example: 4x³– 12x + 16x
= 4x x x² – 4x x 3 + 4x x 4
= 4x (x² – 3 + 4)

Here 4x is the common factor.

Method of factorization:

1. Dividing all the terms of the expression by the largest common factor which is written outside a bracket and putting the resulting quotient inside the bracket.

2.As we arrange the terms of the given algebraic expression in groups such that each group has a common factor. Then we factorise each group separately we take out factors which is common to each group.

3. If any expression is the difference of the square of two expressions, its factors will be the sum of those two expressions and their difference.
i.e., an expression of the form a²– b² can be expressed as the product of (a + b) and (a – b).

4. Some expressions which are not in the form a² – b², can be expressed as the difference of two squares by adding or subtracting some quantity to or from those expressions.

Wbbse Class 7 Maths Solutions

Question 1. Choose the correct answer 

1. The number of prime factors of 9x²y is

1. 2
2. 3
3. 4
4. 5

Solution: 9x²y= 3 x 3 x x x x x y
The number of prime factors of 9x²y is 5

⇒ So the correct answer is 4. 5

2. The common factor of 14ab² and 21ab is

1. 7ab
2. 7a
3. 7b
4. 7ab²

Solution: 14ab² = 2 x 7 x a x b x b
21 ab = 3 x 7 x a x b

The common factor of 14ab² and 21ab is (7 x a x b) or 7ab

⇒ So the correct answer is 1. 7ab

3. The sum of factors of (2x – 6) is

1. 2x-5
2. x – 1
3. x – 3
4. x-8

Solution: 2x – 6
= 2(x-3)

Sum of factors is 2 + x – 3 = x – 1

⇒ So the correct answer is 2. x – 1

4. If two factors of a expression are (x + 1) and (x 1), then the expression is

1. 2x
2. 2
3. x²-1
4. None of these

Solution: The expression is (x + 1)(x − 1) = x²– 1

⇒ So the correct answer is 3. x²-1

Wbbse Class 7 Maths Solutions

Question 2. Write ‘true’ or ‘false’

1. The common factor of 10x²y and 15xy² is 5xy

Solution: 10x²y=2 x 5 x x x x x y
15xy² =3 x 5 x x x y x y

The common factor is (5 x x x y) or 5xy

⇒ So the statement is true.

2. The sum of factors of (a³-a²+ a) is (a² + 1)

Solution: a³ – a² + a = a (a² – a + 1)

The sum of factors = a2 +1 = a2 + 1

⇒ So the statement is true.

3. Sum of factors of (3a² – 12a) is (4a + 4).

Solution: 3a² – 12a = 3a(a – 4)

Sum of factors= 3+a+a-4 = 2a – 1

⇒ So the statement is false.

Wbbse Class 7 Maths Solutions

Question 3. Fill in the blanks 

1. The common factor of 18x², 27x³ and 45x is _____

Solution:
18x² = 2 x 3 x 3 x x x x
27x³ = 3 x 3 x 3 x x x x x x
45x = 3 x 3 x 5 x x

⇒ The common factor is (3 x 3 x x) or 9x.

2. The sum of factors of (ax + bx- ay- by) is

Solution: ax + bx – ay – by
= x (a + b) -y (a + b)
=(a+b) (x-y)

⇒ Sum of factors = a+b+x-y.

3. The factors of (ab – 5b + a – 5) are ____

Solution: ab – 5b + a – 5
= b(a – 5) + 1(a-5)
= (a – 5)(b + 1)

⇒ The factors are (a – 5) and (b + 1).

Question 4. Find the number of prime factors of a(x² – y²).

Solution: a(x²– y²) = a(x + y)(x − y)

The prime factors are a, (x + y), and (x − y).
∴ The number of prime factor is 3.

Question 5. Find the sum of factors of (a³ – a).

Solution: a³ – a
= a(a² – 1)
= a(a + 1)(a− 1)

⇒ Sum of factors = a + a + 1+ a -1 = 3a

Wbbse Class 7 Maths Solutions

Question 6. Resolve into factors

1. a² – 1 + 2b – b²

2. x²– 2x – y²+ 2y

3. 81x⁴+4y⁴

4. 3x⁴ + 2x²y²– y⁴

5. a⁴ + a²b² + b⁴

6. \(a^2+\frac{1}{a^2}+1\)

7. xy(1 + z²) +z (x² +y²)

8. a² – 5a+6

9. 2a²b²+2b²c² + 2c²a²-a⁴-b⁴-c⁴

10. x² + 2x – 899

Solution:

1. a² – 1 + 2b – b²

2. x²– 2x – y²+ 2y

3. 81x⁴+4y⁴

4. 3x⁴ + 2x²y²– y⁴

5. a⁴ + a²b² + b⁴

6. \(a^2+\frac{1}{a^2}+1\)

7. xy(1 + z²) +z (x² +y²)

8. a² – 5a+6

9. 2a²b²+2b²c² + 2c²a²-a⁴-b⁴-c⁴

10. x² + 2x – 899

Wbbse Class 7 Maths Solutions

Question 7. Factorise the following

1. (a + b)⁴+ 4

2. a²(b-c)²-b²(c – a)²

3. 64ap²– 49a (p- 2q)²

4. a² – b² -6ap + 2bp + 8p²

5. 4a⁴ + 11a² + 25

Solution:

1. (a + b)⁴+ 4

2. a²(b-c)²-b²(c – a)²

3. 64ap²– 49a (p- 2q)²

4. a² – b² -6ap + 2bp + 8p²

5. 4a⁴ + 11a² + 25

 

Factorisation

Factorisation Exercise 6.1

Let us factorize the following :

1. 25xy = 5 × 5 × x × y

2. 18 × y² = 2 × 3 × 3 × x × y × y

3. 15q²r²= 3 × 5 × q × q× r × r

4. 10 × yz = 2 × 5 × x × y × z

5. 12 × yz = 2 × 2 × 3× x × y × z

Factorisation Exercise 6.2

Let us factorize the following :

1. 12x²y (x + 2)= 2 × 2 × 3 × x × x × y × (x + 2)

2. 1 8yz² (2y + 3z) = 2 × 3 × 3 × y × z × z × (2y + 3z)

3. 1 6 × yz (x + y)= 2 × 2 × 2×2 × x × y × z × (× + y)

4. 15pq² (p + 3q) = 3 × 5 × p × q × q × (p + 3q)

5. 14mn² (2m – n) = 2 × 7 × m × n × n × (2m-n)

Question 1. Let us factorize:

1. 2+14x
Solution:

2+14x  = 2(1 +7x)

2. 5x – 20y
Solution:

5x – 20y = 5 ( x- 4y)

3. 6x – 3y
Solution:

6x – 3y = 3 (2x – y)

4. 3a² – 12a
Solution:

3a² – 12a = 3a (a -4)

Question 2. Let us find the common factors of the following algebraic expressions.

1. 6a, 2a²
Solution:

6a, 2a²= 2, a, 2a:

2. 5x, 6xy
Solution:

5x, 6xy = x

3. 4xyz, 12yz
Solution:

4xyz, 12yz = 2, 4, yz, yz, 2y, 2z, 4y, 4z, 4yz.

4. 7a²b, 14abc
Solution:

7a²b, 14abc =  7, a, b, 7a, 7b, ab, 7ab

Class 7 Math Solution WBBSE

Factorization Exercise 6.3

Question 1. Let us Factorise the following 

1. xy + y + 3x + 3
Solution:

Given

xy + y + 3x + 3

xy + y + 3x + 3= y (x +1 ) + 3 (× + 1 )

= (x + 1 )(y + 3)

xy + y + 3x + 3 = (x + 1 )(y + 3)

2. pq – q + 2q – 2
Solution:

Given

pq – q + 2q – 2

pq – q + 2q – 2 = q (P – 1 ) + 2 (p – 1 )

= (p – 1 ) (q + 2)

pq – q + 2q – 2 = (p – 1 ) (q + 2)

3. 6xy + 3y + 4× + 2
Solution:

Given

6xy + 3y + 4x + 2

6xy + 3y + 4x + 2 = 3y (2x + 1) + 2 (2x + 1)

= (2x +1) (3y + 2)

6xy + 3y + 4x + 2 = (2x +1) (3y + 2)

4. 10xy + 2y + 5x + 1
Solution:

Given

10xy + 2y + 5x + 1

10xy + 2y + 5x + 1 = 2y (5x + 1 ) + 1 (5x + 1 )

= (5x + 1 ) (2y + 1 )

10xy + 2y + 5x + 1 = (5x + 1 ) (2y + 1 )

Question 2. Let us find the factors of the following algebraic expressions :

1. 7xy 
Solution:

7xy = 7 × x × y

2. 9 x²y
Solution:

9 x²y= 3 × 3 × x × x ×y

3. 16ab²c 
Solution:

16ab²c = 2 × 2 × 2 × 2 × a × b × b × c

4. -25lmn = 5 × 5 × | × m × n
Solution:

-25lmn = 5 × 5 × | × m × n

5. 12x (2 + x) 
Solution:

12x (2 + x) = 2 × 2 × 3 × a × (2 + x )

6. – 5 pq (p² + 8)
Solution:

– 5 pq (p² + 8) = -5 × p × q × (p² +8)

7. 21 xy² (3x – 2)
Solution:

21 xy² (3x – 2) = 3 × 7 × x × y × y × (3x -2)

8. 121mn (m² – n)
Solution: 

121mn (m² – n) = 11 × 11 × m × n × (m² -n)

Question 3. Let us find the common factors of the following algebraic expressions :

1. 22xy, 33xz
Solution:

22xy, 33xz

Common factors = 11 , x, 11x

2. 14ab² , 21ab
Solution: 

14ab² , 21ab

Common factors = 7, a, b, 7a, 7b, 7ab

3. – 16mnl, – 39nl²
Solution:

– 16mnl, – 39nl²

Common factors = – 1, – n, – 1, nl, – nl, n, I

4. 12a²b, 18ab², 24abc
Solution:

12a²b, 18ab², 24abc

Common factors = 2, 3, 6, a, b, 2a, 2b, 3a, 3b, 6a, 6b, ab, 2ab, 3ab, 6ab

5. 2xy, 4yz, 6xz
Solution:

2xy, 4yz, 6xz

Common factors = 2.

6. 18x², 27x³, – 45x
Solution:

18x², 27x³, – 45x

Common factors = 3, 9, x, 3x, 9x

7. 5mn, 6n²l², 7l³m²
Solution:

5mn, 6n²l², 7l³m²

Common factors = 1.

Question 4. Let us write two such algebraic e×pressions which have the following as their Common Factors 

1. x²
Solution:

⇒ x² =   3x², 4x³3y

2. 2xy
Solution:

2xy = 4x²y²q, 6xyq

3. 4a²
Solution:

4a²= 12a³y, 16a²x

4. (mn + 2)
Solution:

(mn + 2) =  5(mn + 2), 7(mn + 2)²

5. x (y + 2)
Solution:

x (y + 2) = 7x²(y + 2), 9x(p + q)(y + 2)

Question 5. Let us factorize the following 

1. 5 + 10x
Solution:

5 + 10x = 5(1 +2x)

2. 2x – 6
Solution:

2x – 6 = 2(x – 3)

3. 7m – 14n 
Solution:

7m – 14n = 7(m – 2n)

4. 18xy + 21 xz 
Solution:

18xy + 21 xz = 3x(6y + 7z)

5. 4xy + 6yz 
Soution:

4xy + 6yz = 2y(2x + 3z)

6. 7xyz – 6xy
Solution:

7xyz – 6xy = xy(7z – 6)

7. 7a² + 14a
Solution:

7a² + 14a = 7a(a + 2)

8. – 15m + 20 
Solution:

– 15m + 20 = 5(- 3m + 4)

9. 6a²b + 8a²b
Solution:

6a²b + 8a²b = 2ab(3a + 4b)

10. 3a² – ab²
Solution:

3a² – ab² = a (3a – b²)

11. abc – bcd =
Solution:

abc – bcd = bc(a – d)

12. 60xy³ – 4xy – 8 
Solution: 

60xy³ – 4xy – 8 = 4(15xy³ – xy – 2)

13.  x²yz + xy²z + xyz²
Solution:

x²yz + xy²z + xyz² = xyz(x + y + z)

14.  a³ – a² + a 
Solution:

a³ – a² + a = a(a² – a + 1 )

15. x²y²z² + x²y² + x²y²q² 
Solution:

x²y²z² + x²y² + x²y²q² = x²y²(z² +1 + q²)

Question 6. Let’s Factorise the following algebraic e×pressions 

1. xy + 2x + y + 2
Solution :

Given

xy + 2x + y + 2

= x(y + 2) + 1(y + 2) = (y + 2)(× + 1)

xy + 2x + y + 2 = (y + 2)(× + 1)

2. ab – 5b + a – 5
Solution:

Given

ab – 5b + a – 5

= b(a – 5) + 1 (a – 5) = (a – 5)(b + 5)

ab – 5b + a – 5 = (a – 5)(b + 5)

3. 6xy – 9y + 4x – 6
Solution :

Given

6xy – 9y + 4× – 6

= 3y(2x – 3) + 2(2x – 3) = (2x – 3)(3y + 2)

6xy – 9y + 4× – 6 = (2x – 3)(3y + 2)

4. 15m + 9 – 35mn – 21 n
Solution :

Given

15m + 9 – 35mn – 21 n

= 3(5m + 3) – 7n(5m + 3) = (3 – 7n)(5m + 3)

15m + 9 – 35mn – 21 n = (3 – 7n) (5m + 3)

5. ax + bx – ay – by 
Solution :

Given

ax + bx – ay – by

= x(a + b) – y(a + b) = (a +b)(x – y)

ax + bx – ay – by = (a +b)(x – y)

6. c – 9 + 9ab – abc
Solution :

Given

c – 9 + 9ab – abc

= – 1 (9 – c) + ab(9 – c) = (9 – c)(ab – 1)

c – 9 + 9ab – ABC = (9 – c)(ab – 1)

Class VII Math Solution WBBSE Factorization Exercise 6.4

Question 1. Let us factorize the following 

1. x² + 14× + 49
Solution :

Given

ײ + 14x + 49

= (x)² + 2.x.7 + (7)2 = (x + 7)² = (x + 7)(x+ 7)

ײ + 14x + 49 = (x + 7)(x+ 7)

2. 4m² – 36m + 81
Solution :

Given

4m² – 36m + 81

= (2m)² – 2.2m.9 + (9)²

= (2m – 9)² = (2m – 9)(2m – 9)

4m² – 36m + 81=(2m – 9)(2m – 9)

3. 25 x² + 30x + 9
Solution :

Given

25x² + 30x + 9

= (5x)² + 2.5 x .3 + (9)² = (5x + 3)² = (5x + 3) (5x + 3)

25x² + 30x + 9 = (5x + 3) (5x + 3)

4. 121b²– 88bc + 16c²
Solution :

Given

121b² – 88bc + 16c²

= (11b)² – 2.11 b.4c + (4c)² = (11 b – 4c)²

. = (11 b – 4c)(11 b – 4c)

121b² – 88bc + 16c² = (11 b – 4c)(11 b – 4c)

WBBSE Class 7 Math Solution

5. (xy)2 – 4x²y²
Solution :

Given

(x²y)² – 4x²y²

= x²y² – 4x²y² = .x²y²(x² – 4) = x²y²{(x)² – (2)2}

= x²y²(x + 2)(x – 2) = x.x.y.y.(x + 2)(x – 2)

(x²y)² – 4x²y² = x.x.y.y.(x + 2)(x – 2)

6. a⁴ + 4a²b² + 4b⁴
Solution :

Given

a⁴ + 4a²b² + 4b⁴

= (a²)² + 2.a².2b² + (2b)²

= (a² + 2b²)

= (a² + 2b²)(a² + 2b²)

a⁴ + 4a²b² + 4b⁴ = (a² + 2b²)(a² + 2b²)

7. 4x² – 1 6
Solution:

Given

4x²-16

= 4(x²– 4) = 4{(x)² – (2)²} = 2 × 2 × (x + 2)(x – 2)

4x²-16 = 2 × 2 × (x + 2)(x – 2)

8. 121 – 36x²
Solution :

Given

121 – 36x²

= (11)² – (6x)² = (11 + 6x)(11 – 6x)

121 – 36x² = (11 + 6x)(11 – 6x)

9. x²y² – p²q²
Solution :

Given

x²y² – p²q²

= (xy)² – (pq)²= (xy + pq)(xy – pq)

x²y² – p²q² = (xy + pq)(xy – pq)

10. 80m² -125
Solution :

Given

80m² -125

= 5(16m² – 25) = 5{(4m)² – (5)2} = 5(4m + 5)(4m – 5)

80m² -125 = 5(4m + 5)(4m – 5)

11. ax² – y²
Solution:

Given

ax² – y²

= ax² – ay² = a{(x)² – (y)²} = a (x + y)(x – y)

ax² – y²  = a (x + y)(x – y)

12.  I – (m + n)²
Solution :

Given

I – (m + n)²

= (I)² – (m + n)² = {I + (m + n)} {I – (m + n)}

= (I + m + n)(l – m – n)

I – (m + n)² = (I + m + n)(l – m – n)

13.  (2a – b – c)² – (a – 2b – c)²
Solution :

Given

(2a – b – c)² – (a – 2b – c)²

= {{2a – b – c) + (a – 2b – c)} {(2a – b – c) – (a – 2b – c)}

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

= (3a – 3b – 2c)(a + b)

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

14.  x² – 2xy – 3y²
Solution :

Given

x² – 2xy – 3y²

= x² – (3 – 1 )xy – 3y² = x² – 3xy + xy – 3y²

= x(x – 3y) + y(x – 3y) = (x – 3y)(x + y)(xv)

x² – 2xy – 3y² = (x – 3y)(x + y)(xv)

15. x²+ 9y² + 6xy – z²
Solution :

Given

x²+ 9y² + 6xy – z²

= x + 6xy + 9y² – z²

= (x)² + 2.x.3y + (3y)² – (z)²

= (x + 3y)² – (z)²

= (x + 3y + z)(x + 3y – z)

x²+ 9y² + 6xy – z²  = (x + 3y + z)(x + 3y – z)

16. a² – b² + 2bc – c²
Solution :

Given

a² – b² + 2bc – c²

= a² – (b² – 2bc + c²) = (a)² – (b – c)²

= {a + (b – c)} {a – (b – c)} = (a + b – c)(a – b + c)

a² – b² + 2bc – c² = (a + b – c)(a – b + c)

WBBSE Class 7 Math Solution

17. a²(b-c)²-b²(c-a)²
Solution :

Given

a²(b-c)²-b²(c-a)²

= {a (b – c)}² – {b(c – a)}² = (ab – ac)² – (bc – ab)²

= {(ab – ac) + (bc – ab)} {(ab – ac) – (bc – ab)}

= (ab – ac + bc – ab) (ab – ac – be + ab)

= (bc – ac) (2ab – ac – bc)

= c(b – a) (2ab – bc – ca)

a²(b-c)²-b²(c-a)² = c(b – a) (2ab – bc – ca)

18. x² – y² – 6yz – 9z²
Solution :

Given

x² – y² – 6yz – 9z²

= x² – (y ²+ 6yz + 9z²)

= (x)² – {(y)² + 2.y.3z + (3z)²}

= (x)² – (y + 3z)²

= (x + y + 3z) (x – y – 3z)

x² – y² – 6yz – 9z²  = (x + y + 3z) (x – y – 3z)

19. x² – y² + 4x – 4y
Solution :

Given

x² – y² + 4x – 4y

= {(x)² -(y)} ² + 4(x-y)

= (x + y)(x – y) + 4(x – y) = (x – y)(x + y + 4)

x² – y² + 4x – 4y = (x – y)(x + y + 4)

20.  a² -b² + c² – d² – 2(ac – bd)
Solution :

Given

a² -b² + c² – d² – 2(ac – bd)

= a² – b²+ c² – d² – 2ac + 2bd

= a² – 2ac + c² – b² + 2bd – d²

= {(a)² – 2,a.c + (c)²} – (b² – 2bd + d²)

= (a – c)² – (b – d)²

= {(a – c) + (b – d)} {(a – c) – (b – d)}

= (a – c + b – d) (a – c – b + d)

= (a + b – c – d) (a – b – c + d)

a² -b² + c² – d² – 2(ac – bd) = (a + b – c – d) (a – b – c + d)

21.  2ab – a² – b² + c²
Solution :

Given

2ab – a² – b² + c²

= c² – a² + 2ab – b²

= (c)² – (a² – 2ab + b²) = (c)² – (a – b)²

= (c + a – b)(c – a + b) = (a – b + c)(b + c – a)

2ab – a² – b² + c² = (a – b + c)(b + c – a)

22. 36x² – 16a² – 24ab – 9b²
Solution :

Given

36x² – 16a² – 24ab – 9b²

= 36x² – (16a² + 24ab + 9b²)

= 36x² – {(4a)² + 2.4a.3b + (3b)²}

= (6x²) – (4a + 3b)²

= {6x + (4a + 3b)} {6x – (4a + 3b)}

= (6x + 4a + 3b) (6x – 4a – 3b)

36x² – 16a² – 24ab – 9b² = (6x + 4a + 3b) (6x – 4a – 3b)

23. a² – 1 + 2b – b²
Solution :

Given

a² – 1 + 2b – b²

= a² – (1 – 2b + b)²

= (a)² – (1 – b) = (a +1 – b) {a (1 b)}

= (a – b + 1 )(a + b – 1 )

a² – 1 + 2b – b² = (a – b + 1 )(a + b – 1 )

WBBSE Class 7 Math Solution

24. a² – 2a – b² + 2b
Solution :

Given

a² – 2a – b² + 2b

= a² – b² + 2b – 2a

= (a + b) (a – b) – 2(a – b)

= (a b)(a + b – 2)

a² – 2a – b² + 2b = (a b)(a + b – 2)

25. (a² – b²)(c² – d²– 4abcd)
Solution :

Given

(a² – b²)(c² – d²– 4abcd

(a² – b²)(c² – d²) – 4abcd

= a²c² – b²c²– a²d² + b²d² – 4abcd

= a²c² + b²d² -.2abcd – a²d²– 2abcd – b²c²

= {(ac)² – 2abcd + (bd)²} – {a²d² + 2abcd + b²c²}

= (ac – bd)² – (ad + bc)²

= {(ac – bd) + (ad + bc)} {(ac – bd) – (ad + bc)}

= (ac – bd + ad + bc) (ac – bd – ad – bc)

(a² – b²)(c² – d²– 4abcd = (ac – bd + ad + bc) (ac – bd – ad – bc)

26. a² – b² – 4ac + 4bc
Solution :

Given

a² – b² – 4ac + 4bc

= {(a)² – (b)²} – 4c(a – b)

= (a + b) (a – b) – 4c(a – b) = (a – b)(a + b – 4c)

a² – b² – 4ac + 4bc = (a – b)(a + b – 4c)

27. a² – b² – c²+ d²)² – 4(ad – bc)²
 Solution:

Given

a² – b² – c²+ d²)² – 4(ad – bc)²

= (a² – b² – c²+ d2)² – {2(ad – bc)}²

= (a² – b² – c² + d² + 2ad – 2bc)

= {a² + 2ad + d²– b² – 2bc – c2}

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

= {(a + d)² – (b + c)²}

= {(a + d) + (b + c)} {(a + d) – (b + c)}

= (a + b + c + d) (a + d – b – c)

= (a + b + c + d)(a – b – c + d)

a² – b² – c²+ d²)² – 4(ad – bc)² = (a + b + c + d)(a – b – c + d)

28. 3x² – y² + z² – 2xy – 4xz
Solution :

Given

3x² – y² + z² – 2xy – 4xz

= 4x² – x² – y² + z² – 2xy – 4xz

= 4x² – 4xz + z² – x² – 2xy – y²

= {(2x)² – 2.2x.z + (z)²} – (x² + 2xy + y²)

= (2x – z)² – (x + y)²

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

= ( 2x – z + x + y) (2x – z – x – y)

= (3x + y – z) (x – y – z)

3x² – y² + z² – 2xy – 4xz = (3x + y – z) (x – y – z)

WBBSE Class 7 Math Solution Question 2. Let us factorize the following

1. 81x⁴ + 4v⁴
Solution :

Given

81x⁴+ 4y⁴

(9x²)² + (2y²)² + 2.9x².2y² – 36x2y²

= (9x² + 2y²)2 – (6xy)²

. = (9x² + 2y² + 6xy) (9x² + 2y² – 6xy)

= (9x² + 6xy + 2y²) (9x² – 6xy + 2y²)

81x⁴+ 4y⁴ = (9x² + 6xy + 2y²) (9x² – 6xy + 2y²)

2. p⁴ – 13p²q² + 4q⁴
Solution :

Given

p⁴ – 13p²q² + 4q⁴

= (p²)² – 2.p².2q² + (2q²)² -9p²q²

= (p² – 2q²)² – (3pq)² = (p² – 2q² + 3pq) (p²– 2q² – 3pq)

= (p² + 3pq – 2q²) (q² – 3pq – 2q²)

p⁴ – 13p²q² + 4q⁴ = (p² + 3pq – 2q²) (q² – 3pq – 2q²)

3. x⁸ – 16y⁸
Solution :

Given

x⁸ – 16y⁸

= (x⁴)² – (4y⁴)²

= (x⁴ + 4y⁴) (x⁴ – 4y⁴)

= {(x²)² + 2.x².2y² + (2y²)² – 4x²y²} {(x²)² – (2y)²}

= {(x² + 2y²)² – (2xy)²} (x² + 2y²) (x² – 2y²)

= (x² + 2y² + 2xy) (x² + 2y² – 2xy) (x² + 2y²) (x² – 2y²)

= (x² + 2xy + 2y²) (x² 2xy + 2y²) (x² + 2y²) (x² – 2y²)

x⁸ – 16y⁸= (x² + 2xy + 2y²) (x² 2xy + 2y²) (x² + 2y²) (x² – 2y²)

4. x⁴ + x²y² + y⁴
Solution :

Given

x⁴ + x²y² + y⁴

= (x²)² + 2x²y² + (y²)² – x²y²

= (x²+ y²)² – (xy)²

= (x² + y² + xy) (x² + y² – xy)

= (x² + xy + y²) (x² – xy + y²)

x⁴ + x²y² + y⁴ = (x² + xy + y²) (x² – xy + y²)

5. 3x⁴ + 2x²y² – y⁴
Solution:

Given

3x⁴ + 2x²y²– y⁴

= 3x⁴ + 3x²y² – x²y² – y⁴

= 3x²(x² + y²) – y²(x² + y²)

= (x² + y²) (3x² – y²)

3x⁴ + 2x²y²– y⁴ = (x² + y²) (3x² – y²)

6. x⁴ + x² + 1
Solution :

Given

x⁴ + x² + 1

= (x²)² + 2.x².1 +(1)²-x²

= (x² + 1 )² – (x)² = (x² + 1 +x) (x² + 1 – x)

= (x² + x + 1 ) (x² – x + 1 )

x⁴ + x² + 1 = (x² + x + 1 ) (x² – x + 1 )

7. x⁴ + 6x²y² + 8y⁴
Solution:

Given

x⁴ + 6x²y² + 8y⁴

= x⁴ + 4x²y² + 2x²y² + 8y²

= x²(x² + 4y²) + 2y²(x² + 4y²)

= (x² + 4y²)(x² + 2y²)

x⁴ + 6x²y² + 8y⁴ = (x² + 4y²)(x² + 2y²)

WBBSE Class 7 Math Solution 8. 3x² – y² + z² – 2xy – 4xz
Solution :

Given

3x² – y² + z² – 2xy – 4xz

= 4x² – x² – y² + z² – 2xy – 4xz

= 4x² – x² – y² + z² – 2xy – 4xz

= 4x² – 4xz + z² – x² – 2xy – y²

= (4x² – 4xz + z²) – (x² + 2xy + y²)

= {(2x)² – 2. 2x.z + z)²} – {(x)² + 2.x.y + (y)²}

= (2x – z)² – (x + y)²

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

= (2x – z + x + y) (2x – z – x – y)

= (3x + y – z) (x – y – z)

3x² – y² + z² – 2xy – 4xz = (3x + y – z) (x – y – z)

9. 3x⁴ – 4x²y² + y⁴
Solution :

Given

= 3x⁴ – 3x²y²– x²y² + y⁴

= 3x²(x² – y²) – y²(x² – y²)

= (x² – y²) (3x² – y²)

= {(x)² -(y)²} (3x² – y²)

= (x + y) (x – y) (3x² – y²)

3x⁴ – 3x²y²– x²y² + y⁴ = (x + y) (x – y) (3x² – y²)

10. p⁴ – 2p²q² – 15q²
Solution :

Given

p⁴ – 2p²q² – 15q²

= p⁴ – (5 – 3)p²q² – 15q⁵

= p⁴ – 5p² q² + 3p²q² – 15q²

= P²(P² 5q²) + 3q²(p² + 3q²)

= (p² – 5q²) (p² + 3q²)

p⁴ – 2p²q² – 15q² = (p² – 5q²) (p² + 3q²)

11. x⁸ + x⁴y⁴ + y⁸
Solution :

Given

x⁸ + x⁴y⁴ + y⁸

= x⁸ + 2x⁴y⁴ + y⁸ – x⁴y⁴

= {(x⁴)² + 2x⁴y⁴ + y⁴ + (y⁴)⁴} – (x⁴y⁴)⁴

= (x⁴ + y⁴)⁴ – (x⁴y⁴)⁴

= (x⁴ + y⁴ + x²y²) (x² + y⁴ – x²y²)

= {x⁴ + 2x²y² + y⁴ – x²y²} (x⁴ + y⁴ – x²y²)

= {(x²)² + 2x²y² + (y²)² – (xy)²} (x⁴ + y⁴ – x²y²)

= {(x² + y²) – (xy)²} (x⁴ + y⁴ – x²y²)

= (x² + y² + xy) (x² + y²– xy) (x⁴ + y⁴ – x²y²)

x⁸ + x⁴y⁴ + y⁸  = (x² + y² + xy) (x² + y²– xy) (x⁴ + y⁴ – x²y²)