WBBSE Class 7 Maths Solutions For Algebra Chapter 6 Factorisation

Algebra Chapter 6 Factorisation

Question 1. Choose the correct answers

1. The number of prime factors of 20 a²bc³

1. 7
2. 6
3. 9
4. 10

Solution:

Prime Factor 20a²bc³

20a²bc³

= 2 х 5 х 2 х а х a x b x c x c x c

= 9

∴ 20a²bc³ = 9

∴ 20a²bc³ = 9 option ‘3’ = 9 corret Answer.

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2. The sum of factors of (12a⁴ – 18a² + 30a) is

1. 2a⁴ – 2a² + 10
2. a⁴ – a² + 9
3. 2a³ – 2a – 10
4. 2a³ + 2a – 10

Solution:

Sum of factors (12a⁴ – 18a² + 30a)

12a⁴ – 18a² + 30a

6a(2a³ – 3a² + 5)

cubic polynomial ax³ + bx² + cx + d

sum of factors = –\(\frac{b}{a}\)

so for 2a³ – 3a² + 5 = -(\(\frac{-3}{2}\)) = \(\frac{3}{2}\)

Now let’s substitute the sum back to the original equation

6a x \(\frac{3}{2}\) = 2a³ – 2a + 10

∴ The sum of factors of 12a⁴ – 18a² + 30a is 2a³ – 2a + 10

∴ The option (1) 2a³ – 2a + 10 is correct answer

Class 7 Algebra Problems With Solutions

3. The Common Factor of 24a³b²c⁴, 36ab³c³ and 48a⁴bc² is

1. 6abc
2. 12abc²
3. 6a³bc²
4. 12a²b²c²

Solution:

24a³b²c⁴ = 2 x 12 x a x a x a x b x b x c x c x c x c

36ab³c³ = 3 x 12 x a x b x b x b x c x c x c

48a⁴bc² = 4 х 12 х а х а х ах а х b x c x с

Common Factor 12abc²

∴ Option (2) 12abc² is the correct answer.

4. The sum of factors of x(x²-q) is

1. x + x² – 29
2. x + 9
3. x – 9
4. 3х

Solution:

x(x²-9)

x(x²-(3)²)

x(x+3)(x-3)

Sum of factors x + x + 3 + x – 3 = 3x

∴ The option (4) 3x is the correct answer.

5. The difference of two factors of (4x² – 25) is

1. 4x
2. 4x – 10
3. 10
4. x – 5

Solution:

(4x² – 25)

{(2x)² – (5)²} [a²-b²=(a+b)(a-b)]

(2x+5)(2x-5)

The differences between the factors are

⇒ (2x + 5) – (2x – 5)

⇒ 2x + 5 – 2x + 5

⇒ 10

∴ Option (3) 10 is the correct answer.

Question 2. Write true or False

1. The number of prime factors of 6x is 3
Solution:

Prime factors of 6x  2 x 3 x x =3 → True

2. The sum of factors of (4-x²) is ‘2’
Solution:

⇒ ((2)² – x²)

⇒ (2 + x) (2 – x)

⇒2 + x + 2 – x

⇒ 4

∴  (4-x²) sum of factors is ‘2’

∴ False

Class 7 Maths Chapter 6 Solved Exercises

3. One of the factors of (a²-a-5a³) is (a²-1-5a)
Solution:

⇒ (a²-a-5a³)

⇒ a(a-1-5a²) is (a²-1-5a)

∴ False

4. The Prime Factors of 18ab²c⁵ are 2 x 3 x 3 x a, b, C
Solution:

18ab²c⁵

2 x 3 x 3 x a x b x b x c x c x c x c x c.

prime Factors of 18ab²c⁵ is 2 x 3 x 3 x a x b x b x c x c x c x c x c.

∴ False

5. The Common Factor of (2a² + 3a) and (4a – 7a²) is a
Solution:

(2a² + 3a), (4a – 7a²)

a(2a + 3) – a(-4-7a).

a + 2a + 3 – a – 4 + 7a

3a + 3

3(a+1)

and

(6a – 4)

2(30 + 2)

Here the common factor is ‘a’

∴  True

Question 3. Fill in the Blanks:

1. There is no common Prime factor of 5a²b and 9cd²
Solution: Prime

2. The sum of factors of (3x²-27) is
Solution:

(3x² – 27)

3(x² – 9)

3(x²-(3)²)

3(x+3)(x-3)

3 = x + 3 + x – 3

2x + 3

∴ The sum of factors of (3x² – 27) is (2x + 3)

3. The difference of factors of (9x² – 16)
Solution:

9x² – 16

⇒ (3x)² – (4)² [a² – b² = (a+b) (a-b)]

⇒ (3x + 4) (3x – 4)

Difference of factors we subtract the smaller

Factor from the larger one: (3x+4)- (3x-4)

Expanding this expression: 3x + 4 – 3x + 4 = 8

∴ The difference of factors of (9x² – 16) is 8

Algebra Formulas For Class 7 Wbbse

Question 4. If a² – 7a + 12 = (a – 4) (a + p). then find the value of ‘P’
Solution:

a² – 7a + 12 = (a – 4)(a + p)

⇒ a² – 7a +12 = a(a + p) – 4(a + p)

⇒ a² – 7a +12 = a² + ap – 4a – 4P

⇒ a² – 7a + 12 = a² + a(p-4) – 4p

∴ -7a = a(p-4)

⇒ \(\frac{7a}{a}\) = p – 4

-7 = p – 4

-7 + 4 = p

p = -3

and

12 = -4p

p = \(\frac{123}{-4}\)

p = -3

∴ a² – 7a + 12 = (a – 4)(a + p)

∴ p = -3

Question 5. The product of two expression is \(\left(\frac{81}{a^4}-\frac{b^4}{25}\right)\) if one expresion is \(\left(\frac{a}{a^2}-\frac{b^2}{5}\right)\) then Find the other expression.
Solution:

⇒ \(\left(\frac{81}{a^4}-\frac{b^4}{25}\right)\)

⇒ \(\left\{\left(\frac{a}{a^2}\right)^2-\left(\frac{b^2}{5}\right)^2\right\}\) (\(a^2-b^2=(a+b)(a-b)\))

⇒ \(\left(\frac{9}{a^2}+\frac{b^2}{5}\right),\left(\frac{a}{a^2}-\frac{b^2}{5}\right)\)

The other expression is \(\left(\frac{9}{a^2}+\frac{b^2}{5}\right)\)

Question 6. Find the value of ‘p’ in expression (x²-px-6) if one of the Factors is (x-3) Find also another factor.
Solution:

(x² – px – 6)

given that one factor is (x – 3) = 0

x = 3 Substitute in the above equation.

((3)² – p(3)-6)= 0

9 – 3p – 6 = 0

-3P = -9 + 6

-3p = -3

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

∴ p = 1

(x² – px – 6) = 0

(x² – (1)x – 6) = 0

x² – x – 6 = 0

x² + 2x – 3x – 6 = 0

x(x+2)-3(x+2)=0

∴ (x+2)=0, (x-3)=0

x = -2, x = 3

∴ The value of p is 1

The other factor is (x+2)=0

∴ x = -2, and x = 3

WBBSE Maths Study Material Class 7

Question 7. Express \(\left(\frac{a}{b}-\frac{b}{a}\right)\) as a product of two expressions such that the sum of the expression is \(\left(\frac{a}{b}+\frac{b}{a}\right)\)
Solution:

⇒ \(\left(\frac{a}{b}-\frac{b}{a}\right)\)

Some of the expression is \(\left(\frac{a}{b}+\frac{b}{a}\right)\)

This can be be written as 1 + \(\frac{b}{a}\) x (\(\frac{a}{b}\) + 1)

Now let’s verify if the given expression can be factored as the product of two expressions whose sum is \(\left(\frac{a}{b}+\frac{b}{a}\right)\):

⇒ \(\left(\frac{a}{b}-\frac{b}{a}\right)\)

= \(\frac{a^2-b^2}{a b}\)

= \(\frac{(a+b)(a-b)}{a b}\)

= \(\frac{a-b}{b} \times \frac{a+b}{a}\)

= \(\left(\frac{a}{b}-1\right)\left(1+\frac{b}{a}\right)\)

∴ So \(\left(\frac{a}{b}-\frac{b}{a}\right)\) can be expressed as the product of (\(\frac{a}{b}\)-1) and (1 + \(\frac{b}{a}\)), and their sum is \(\frac{a}{b}\) + \(\frac{b}{a}\)

Question 8. Find the sum of the Factors of the expression (3a² – b² – c² – 2ab – 2bc – 2ca)
Solution:

(3a² – b – c² – 2ab – 2bc – 2ca)

⇒ 3 + 3 + a + a – b – b – c – 2a + b − 2 + b + c – 2 + c + a

⇒ 6 + 4а – 6

⇒ 4a.

∴ The sum of factors of the expression (3a²-b²-c²-2ab-2bc-2ca) = 4a

Question 9. Resolve into factors.

1. \(\frac{a^4}{16}-\frac{81}{b^4}\)
Solution:

⇒ \(\frac{a^4}{16}-\frac{81}{b^4}\)

⇒ \(\left(\frac{a^2}{4}\right)^2-\left(\frac{9}{b^2}\right)^2\) (\(a^2-b^2=(a+b)(a-b)\))

⇒ \(\left(\frac{a^2}{4}+\frac{9}{b^2}\right)\left(\frac{a^2}{4}-\frac{9}{b^2}\right)\)

⇒ \(\left(\frac{a^2}{4}+\frac{9}{b^2}\right)\left(\left(\frac{a}{2}\right)^2-\left(\frac{3}{b}\right)^2\right)\)

⇒ \(\left(\frac{a^2}{4}+\frac{9}{b^2}\right)\left(\frac{a}{2}+\frac{3}{b}\right)\left(\frac{a}{2}-\frac{3}{b}\right)\)

∴ \(\frac{a^4}{16}-\frac{81}{b^4}=\left(\frac{a^2}{4}+\frac{9}{b^2}\right)\left(\frac{a}{2}+\frac{3}{b}\right)\left(\frac{a}{2}-\frac{3}{b}\right)\)

2. \(a^4-5 a^2+6 a\)
Solution:

⇒ \(a^4-5 a^2+6 a\)

⇒ \(a\left(a^3-5 a+6\right)\)

∴ \(a^4-5 a^2+6 a=a\left(a^3-5 a+6\right)\)

3. \(a^2-b^2+18 b-81\)
Solution:

⇒ \(a^2-b^2+18 b-81\)

⇒ \(a^2-\left(b^2-18 b+81\right)\)

⇒ \(a^2-(b-9)^2\)

(a+b-a)(a-b+9)

∴ \(a^2-b^2+18 b-81=(a+b-9)(a-b+9)\)

WBBSE Maths Study Material Class 7

4. 3x² – xy – 4y²
Solution:

3x² -xy – 4y²

⇒ 3x² + 3xy – 4xy – 4y²

⇒ 3x(x+y) -4(x+y)

⇒ (x+y) (3x-4)

∴ 3x² – xy – 4y² = (x+4) (3x-4)

5. 18ax² – 128a(x-2y)²
Solution:

18ax² – 128a(x-2y)²

a[18x² – 128(x-2y)²]

6. 18ax² – 128a(x – 2y)²
Solution:

⇒ \(2 a \cdot 9 x^2-2 a \cdot 64(x-2 y)^2\)

⇒ \(2 a \cdot 9 x^2-2 a \cdot 64\left(x^2-4 x y+4 y^2\right)\)

⇒ \(2 a\left(9 x^2-64 x^2+256 x y-256 y^2\right)\)

⇒ \(2 a\left(9 x^2-64 x^2\right)+2 a\left(256 x y-256 y^2\right)\)

⇒ \(2 a\left(-55 x^2+256 x y-256 y^2\right)\)

⇒ \(2 a\left(-11 x^2+16 x y-16 y^2\right)\)

Now Let’s re-arrange the terms:

⇒ \(2 a\left(16 x y-11 x^2-16 y^2\right)\)

⇒ \(2 a\left(16 y(11 x-16 y)-11 x^2\right)\)

⇒ \(2 a(11 x-16 y)(16 y-5 x)\)

So, \(18 a x^2-128 a(x-2 y)^2\) can be simplified to \(2 a(11 x-6 y)(16 y-5 x)\).

7. \(81 a^4+643^4\)
Solution:

⇒ \(81 a^4+64 b^4\)

⇒ (3a+2b)(3a-2b)(3a+2b)(3a-2b)

⇒ (3a+2b)\(^2(3 a-2 b)^2\)

⇒ \(\left(9 a^2+12 a b+4 b^2\right)\left(9 a^2-12 a b+4 b^2\right)\)

And we can simplify the expression by distributing the terms.

⇒ \(\left(9 a^2+12 a b+8 b^2\right)\left(9 a^2-12 a b+8 b^2\right)\)

So, \(81 a^4+64 b^4 \mathrm{can}\) indeed be expressed as

∴ \(\left(9 a^2+12 a b+8 b^2\right)\left(9 a^2-12 a b+8 b^2\right)\)

8. 4(a+3b-c)² – (4a-3b+2c)²
Solution:

Expand 4(a+3b-c)²

4(a+3b-c)(a+3b-c)

4(a² + 3ab – ac + 3ab + 9b² – 3bc – ac – 3bc + c²)

4(a² + 6ab – 2ac + 9b² – 6bc + c²)

Expand (4a – 3b + 2c)²

(4a-3b+2c) (4a-3b+2c)

⇒ (4a)² -2(4a) (3b) + 2(4a)(2c) + (-3b)2 – 2(-3b)(2c)+ (2c)²

⇒ 16a² – 24ab + 16ac + 9b² + 12bc + 4c²

Now let’s substitute these expansions back into the original expression:

4(a+3b-c)² – (4a-3b+2c)²

= 4(a² + 6ab- 2ac + 9b² – 6 bc + c²) – (16a² – 24ab + 16ac + 9b² + 12bc + 4c²)

= 4a² + 24ab – 8ac + 36b² – 24bc + 4c² – 16ac + 24ab – 16ac – 9b² – 12bc – 4c²

9. 4a² – 16a² + 24ab + 24ab – 8ac – 16ac + 36b² – 9b² – 24bc – 12bc + 4c² – 4c²
Solution:

= -12a² + 48ab – 24a + 27b² – 36bc

Now we can factor out common terms from this expression

= -12(a² – 4ab + 2ac – \(\frac{9}{4}\)b² + 3bc)

Now observe that a² – 4ab + 2ac – \(\frac{9}{4}\)b² + \(\frac{3}{2}\)bc

Can be factored as (a – 2b + \(\frac{3}{2}\)c) (a – 2b + \(\frac{3}{4}\)c)

Thus the expression becomes:

= -12(a – 2b + \(\frac{3}{2}\)c) (a-2b+\(\frac{3}{4}\)c)

Now let’s factor this expression further.

= −12(2a – 4b + 3c)(a − 2b + \(\frac{3}{4}\)c)

= 3(-4a + 8b – 6c)(2a – 4b + \(\frac{3}{2}\)c)

= 3(2a – 4b + \(\frac{3}{2}\)c)(-4a + 8b – 6c)

= 3(2a – 4b + \(\frac{3}{2}\)c)(9b – 2a – 4c)

So, 4(a + 3b – c)² – (4a – 3b + 2c)² can be simplified to 3(2a + b) (9b – 20 – 4c).

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

⇒ a² – b – c² + ď² + 2(ad +bc)

⇒ a² – b² – c² + d + 2ad + 2bc

⇒ a²+ 2ad + d² – (b² + c² – 2bc)

⇒ (a+d)² – (b-c)²

⇒ {(a+d) + (b-c)} {(a+d) – (b-c)}

⇒ (a+b-c+d) (a-b+c+d)

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

Class 7 Maths Algebra Solutions WBBSE

11. 7a(7a-2b) + (b+c)(b-c)
Solution:

7a(7a-2b) + (b+c) (b-c)

49a² – 14ab + b² – c²

((7a)² -2(7a)(b) + b²) – c²

(7a-b)² – c²

(7a-b+c)(7a-b-c)

∴ 7a(7a-2b)+(b+C)(b-c) = (7a-b+c) (7a-b-c)

12. (a+b)(x+cy) – (b+c)(x+ay)
Solution:

(a+b)(x+cy) – (b+c)(x+ay)

{a(x+cy) + b(x+cy)} – {b(x+ay) + c(x+ ay)}

{ax + ayc + bx + byc} – {bx + aby + cx + acy}

ax + ayc + bx + byc – bx – aby – cx – acy

ax + byc -aby -cx

(ax-cx)- by(a-c)

x(a-c)- by(a-c)

(a-c)(x-by)

∴ (a+b)(x+cy) – (b+c)(x+ay) = (a-c)(x-by)

Question 10. Factorize the following

1. a⁴ + a²b² + b⁴
Solution:

⇒ (a²)²+ a²b² + (b²)²

⇒ (a² + ab + b²) (a² – ab + b²)

∴ a⁴ + a²b² + b⁴ = (a² + ab + b²) (a²– ab + b²)

2. a⁴ – 2a²b² – 15b⁴
Solution:

⇒ a⁴ – 2a²b² – 15b⁴

⇒ a⁴ – 5a²b² + 3a²b² – 15b⁴

⇒ (a²-5b²) + 3b²(a²-5b²)

⇒ (a²-5b²) (a²+3b²)

∴ a⁴ – 2a²b² – 15b⁴ = (a²-5b²) (a²+3b²)

3. 16 (3x+2y)² – 9(x-2y)²
Solution:

⇒16(9x² + 4y² + 12xy) – 9(x² + 4xy² – 4xy)

⇒ (144x²+64y² + 192xy) (-9x²-36y²+36xy)

⇒ (144x -9x²)+(64y²-36y²) + (192xy+36xy)

⇒ 135x² + 28y² + 228xy

⇒ 135x² + 228xу + 28y²

⇒ 135x² + 210xy + 18xy + 28y²

⇒ 15x(9x+14y) + 2y(9x+14y)

⇒ (9x+14y) (15x+2y)

∴ 16(3x+2y)² – 9(x-2y)²= (9x+14y) (15x+2y)

Class 7 Maths Algebra Solutions WBBSE

4. x⁸-16y⁸
Solution:

(x⁴)² – {(4y)⁴}² (a²-b²=(a+b)(a-b))

⇒ {(x)⁴ + (4y)⁴} {(x)⁴-(4y)⁴}

⇒ {(x²)²+{(2y)²}²} {(x²)²-{(2y)²}²}

⇒ {(x² + 2y²)² – 2.x².2y²} {(x²+ 2y²)x(x²– 2y²)}

⇒ {(x²+2y²)² – 4x²y²} {(x²+2y²) × (x²-2y²)}

⇒ {(x²+2y²)²-(2xy)²} {(x²+2y²)x(x²-2y²)}

⇒ (x² + 2x² + 2xy)(x² + 2y² -2xy) (x²+2y²) x (x²-2y²)

∴ x⁸-16y⁸ = (x²+2y²+2xy) (x²+2y²-2xy) (x²+2y²)(x²-2y²)

5. (1+x)(y-z)+(1+z)(x−y)
Solution:

⇒ {1(y-z)+x(y-z)} + {1(x-y) + z(x-y)}

⇒ {y-2+xy-zx} + {x-y+2x-zy}

⇒ y-z+xy-zx+x-y+zx-zy

⇒ (xy-zy) + (x-z)

⇒ y((x-z)) + 1(x-z)

⇒ (x-z)(y+1)

∴ (1+x)(y-z) + (1+z)(x-y) = (x-z)(y+1)