Factorisation Of Algebraic Expressions

Class 8th Mathematics (new) MHB Solution

Class 8^th Mathematics (new) MHB Solution

Practice Set 6.1

Factorise.x2 + 9x + 18
Factorise.x2 – 10x + 9
Factorise.y2 + 24y + 144
Factorise.5y2 + 5y – 10
Factorise.p2 – 2p – 35
Factorise.p2 – 7p – 44
Factorise.m2 – 23m + 120
Factorise.m2 – 25m + 100
Factorise.3x2 + 14x + 15
Factorise.2x2 + x – 45
Factorise.20x2 – 26x + 8
Factorise.44x2 – x – 3

Practice Set 6.2

Factorise.x3 + 64y3
Factorise.125p3 + q3
Factorise.125k3 + 27m3
Factorise.2l3 + 432m3
Factorise.24a3 + 81b3
Factorise. y^{3} + {1}/{ 8y^{3} }
Factorise. a^{3} + {8}/{ a^{3} }
Factorise. 1 + { q^{3} }/{125}

Practice Set 6.3

Factorise :y3 – 27
Factorise :x3 – 64y3
Factorise :27m3 – 216n3
Factorise :125y3 – 1
Factorise :8p3 – 27/p3
Factorise :343a3 – 512b3
Factorise :64x2 – 729y2
Factorise :16 a3 – 128/b3
Simplify :(x + y)3 – (x – y)3
Simplify :(3a + 5b)3 – (3a – 5b)3
Simplify :(a + b)3 – a3 – b3
Simplify :p3 – (p + 1)3
Simplify :(3xy – 2ab)3 – (3xy + 2ab)3

Practice Set 6.4

Simplify: { m^{2} - n^{2} }/{ (m+11) } x frac { m^{2} + mn+n^{2} }/{ m^{3} - n^{3}…
Simplify: { a^{2} + 10a+21 }/{ a^{2} + 6a-7 } x frac { a^{2} - 1 }/{a+3}…
Simplify: { 8x^{3} - 27y^{3} }/{ 4x^{2} - 9y^{2} }
Simplify: { x^{2} - 5x-24 }/{ (x+3) (x+8) } x frac { x^{2} - 64 }/{ (x-8)^{2} }…
Simplify: { 3x^{2} - x-2 }/{ x^{2} - 7x+12 } / frac { 3x^{2} - 7x-6 }/{ x^{2} -…
Simplify: { 4x^{2} - 11x+6 }/{ 16x^{2} - 9 }
Simplify: { a^{3} - 27 }/{ 5a^{2} - 16a+3 } / frac { a^{2} + 3a+9 }/{ 25a^{2} -…
Simplify: { 1-2x+x^{2} }/{ 1-x^{3} } x frac { 1+x+x^{2} }/{1+x}…

Practice Set 6.1

Question 1.

Factorise.

x² + 9x + 18

Answer:

On comparing with standard quadratic equation that is

we have,

a = 1, b = 9 and c = 18

Now here,

Product ac = 118 = 18

Factors of 18; 29 and 63

Sum should be b = + 9

From above factors (+ 6x + 3x)

will give + 9x sum

therefore + 9x is replaced by ( + 6x + 3x)

Now above eq. becomes

x² + 6x + 3x + 18

⇒ ; taking x common

⇒ (x + 3)(x + 6)

Question 2.

Factorise.

x² – 10x + 9

Answer:

On comparing with standard quadratic equation that is

we have,0

a = 1, b = – 10 and c = 9

Now here,

Product ac = 19 = 9

Factors of 9; 19 and 33

Sum should be b = – 10

From above factors ( – 1x – 9x)

will give – 10x sum

therefore – 10x is replaced by ( – 1x – 9x)

Now above eq. becomes

; taking x and – 9 common

(x – 1)(x – 9)

Question 3.

Factorise.

y² + 24y + 144

Answer:

On comparing with standard quadratic equation that is

we have,

a = 1, b = + 24 and c = + 144

Now here,

Product a × c = 1 × 144 = 144

Factors of 144; 12 × 12; 24 × 6;144 × 1;

48 × 3; 72 × 2

Sum should be b = 24

From above factors (12y + 12y)

will give + 24y sum

therefore + 24 is replaced by ( + 12y + 12y)

Now above eq. becomes

; taking y and + 12 common

(y + 1)(y + 12)

Note: Try to find all factors of “c”, then choose from it that combination whose sum or difference give “b”

Question 4.

Factorise.

5y² + 5y – 10

Answer:

On comparing with standard quadratic equation that is

we have,

a = 5, b = + 5 and c = – 10

Now here,

Product a × c = 5 × – 10 = – 50

Factors of 50; 5 × 10; 25 × 2;50 × 1

Sum should be b = + 5

From above factors ( – 5y + 10y)

will give + 5y sum

therefore + 5y is replaced by ( – 5y + 10y)

Now above eq. becomes

; taking 5y and + 10 common

(y – 1)(5y + 10)

5(y – 1)(y + 2); 5 common

Note: if given equation’s constant a,b,c have common multiple take it out and then factorize.

Question 5.

Factorise.

p² – 2p – 35

Answer:

On comparing with standard quadratic equation that is

we have,

a = 1, b = – 2 and c = – 35

Now here,

Product a × c = 1 × – 35 = – 35

Factors of 35; 1 × 35 and 7 × 5

Sum should be b = – 2

From above factors ( – 7p + 5p)

will give – 2p sum

therefore – 2p is replaced by ( – 7p + 5p)

Now above eq. becomes

; taking p and + 5 common

(p – 7)(p + 5)

Question 6.

Factorise.

p² – 7p – 44

Answer:

On comparing with standard quadratic equation that is

we have,

a = 1, b = – 7 and c = – 44

Now here,

Product a × c = 1 × – 44 = – 44

Factors of 44; 1 × 44; 2 × 22;4 × 11

Sum should be b = – 7

From above factors ( – 11p + 4p)

will give – 7p sum

therefore – 7p is replaced by ( – 11p + 4p)

Now above eq. becomes

; taking p and + 4 common

(p + 4)(p – 11)

Question 7.

Factorise.

m² – 23m + 120

Answer:

On comparing with standard quadratic equation that is

we have,

a = 1, b = – 23 and c = + 120

Now here,

Product a × c = 1 × + 120 = + 120

Factors of + 120; 1 × 120; 2 × 60; 4 × 30; 8 × 15; 24 × 5; 40 × 3

Sum should be b = – 23

From above factors ( – 15m – 8m)

will give – 23m sum

therefore – 23m is replaced by ( – 15m – 8m)

Now above eq. becomes

; taking m and – 8 common

(m – 15)(m – 8)

Question 8.

Factorise.

m² – 25m + 100

Answer:

On comparing with standard quadratic equation that is

we have,

a = 1, b = – 25 and c = 100

Now here,

Product a × c = 1 × 100 = 100

Factors of 100; 1 × 100; 2 × 50; 4 × 25;20 × 5

Sum should be b = – 25

From above factors ( – 20m – 5m)

will give – 25m sum

therefore – 25m is replaced by ( – 20m – 5m)

Now above eq. becomes

; taking m and – 5 common

(m – 5)(m – 20)

Question 9.

Factorise.

3x² + 14x + 15

Answer:

On comparing with standard quadratic equation that is

we have,

a = 3, b = + 14 and c = + 15

Now here,

Product a × c = 3 × 15 = + 45

Factors of 45; 1 × 45; 5 × 9;15 × 3

Sum should be b = + 14

From above factors ( + 9x + 5x)

will give + 14x sum

therefore + 14x is replaced by ( + 9x + 5x)

Now above eq. becomes

; taking x and + 5 common

(x + 9)(x + 3)

Question 10.

Factorise.

2x² + x – 45

Answer:

On comparing with standard quadratic equation that is

we have,

a = 2, b = 1 and c = – 45

Now here,

Product a × c = 2 × – 45 = 90

Factors of 90; 1 × 90; 2 × 45; 10 × 9; 30 × 3

Sum should be b = 1

From above factors ( + 10x – 9x)

will give + x sum

therefore + x is replaced by ( + 10x – 9x)

Now above eq. becomes

; taking 2x and – 9 common

(x + 5)(2x – 9)

Question 11.

Factorise.

20x² – 26x + 8

Answer:

On comparing with standard quadratic equation that is

we have,

a = 20, b = – 26 and c = 8

Now here,

Product a × c = 20 × 8 = 160

Factors of 160; 2 × 80; 4 × 40; 8 × 20; 16 × 10; 32 × 5

Sum should be b = – 26x

From above factors ( – 16x – 10x)

will give – 26x sum

therefore – 26x is replaced by ( – 16x – 10x)

Now above eq. becomes

; taking 4x and – 2 common

2(2x – 1)(5x – 4)

Question 12.

Factorise.

44x² – x – 3

Answer:

On comparing with standard quadratic equation that is

we have,

a = 44, b = – 1 and c = – 3

Now here,

Product a × c = – 132 = 44 × – 3

Factors of 132; 1 × 132; 2 × 66; 4 × 33;12 × 11

Sum should be b = – 1

From above factors ( – 12x – 11x)

will give – 1x sum

therefore – 1x is replaced by ( – 12x – 11x)

Now above eq. becomes

; taking x and – 9 common

(11x – 3)(4x – 1)

Practice Set 6.2

Question 1.

Factorise.

x³ + 64y³

Answer:

We know that

- - - - - (i)

Here a = 1x, b = 4y; putting values in eq.i

⇒

}

Note: Must memorize cubes upto 12

Question 2.

Factorise.

125p³ + q³

Answer:

We know that

- - - - - (i)

Here a = 5p, b = q; putting values in eq.i

⇒

}

Note: Must memorize cubes upto 12

Question 3.

Factorise.

125k³ + 27m³

Answer:

We know that

- - - - - (i)

Here a = 5k, b = 3m; putting values in eq.i

⇒

⇒ }

}

Note: Must memorize cubes upto 12

Question 4.

Factorise.

2l³ + 432m³

Answer:

We know that

- - - - - (i)

Taking 2 common, we get

Here a = l, b = 6m; putting values in eq.i

⇒

⇒ }]

}

Applying

}

Note: Must memorize cubes upto 12

Question 5.

Factorise.

24a³ + 81b³

Answer:

We know that

- - - - - (i)

Taking 3 as common, we get

3; solving only bracket term first,

Here a = 2a, b = 3b; putting values in eq.i

}

Applying

}

Ans: - 3}

Note: Must memorize cubes upto 12

Question 6.

Factorise.

Answer:

We know that

- - - - - (i)

Here a = y, b = ; putting values in eq.i

Applying

}

Note: Must memorize cubes upto 12

Question 7.

Factorise.

Answer:

We know that

- - - - - (i)

Here a = a, b = ; putting values in eq.i

Applying

}

Note: Must memorize cubes upto 12

Question 8.

Factorise.

Answer:

We know that

- - - - - (i)

Here a = 1, b = ; putting values in eq.i

Applying

}

Note: Must memorize cubes upto 12

Practice Set 6.3

Question 1.

Factorise :

y³ – 27

Answer:

We know that

on comparison with above, we get

a = y, b = 3

Note: Must memorize cubes upto 12

Question 2.

Factorise :

x³ – 64y³

Answer:

We know that

on comparison with above, we get

a = y, b = 3

Note: Must memorize cubes upto 12

Question 3.

Factorise :

27m³ – 216n³

Answer:

We know that

on comparison with above, we get

a = 3m, b = 6n

Note: Must memorize cubes upto 12

Question 4.

Factorise :

125y³ – 1

Answer:

We know that

on comparison with above, we get

a = 5y, b = 1

Note: Must memorize cubes upto 12

Question 5.

Factorise :

8p³ – 27/p³

Answer:

We know that

on comparison with above, we get

a = 2p, b = 3/p

Note: Must memorize cubes upto 12

Question 6.

Factorise :

343a³ – 512b³

Answer:

We know that

on comparison with above, we get

a = 7a, b = 8b

Note: Must memorize cubes upto 12

Question 7.

Factorise :

64x² – 729y²

Answer:

We know that

on comparison with above, we get

a = 4x, b = 9y

Note: Must memorize cubes upto 12

Question 8.

Factorise :

16 a³ – 128/b³

Answer:

We know that

taking 2 common from above given equation;

on comparison with above, we get

a = 2a, b = 4/b

Note: Must memorize cubes upto 12

Question 9.

Simplify :

(x + y)³ – (x – y)³

Answer:

We know that

On comparing with given equation we get,

a = (3a + 5b), b = (3a – 5b)

Applying and

Question 10.

Simplify :

(3a + 5b)³ – (3a – 5b)³

Answer:

We know that

On comparing with given equation we get,

a = (3a + 5b), b = (3a – 5b)

Applying and

Question 11.

Simplify :

(a + b)³ – a³ – b³

Answer:

We know that

On comparing with given equation we get

Question 12.

Simplify :

p³ – (p + 1)³

Answer:

We know that

On comparing with given equation we get

a = p, b = 1

Question 13.

Simplify :

(3xy – 2ab)³ – (3xy + 2ab)³

Answer:

We know that

On comparing with given equation we get,

a = (3xy – 2ab), b = (3xy + 2ab)

Applying and

Practice Set 6.4

Question 1.

Simplify:

Answer:

We know that

applying these equation in above expression, we get

= 1

Note: - Try to factorize that term which help in reducing expression.

Question 2.

Simplify:

Answer:

We know that

and factorization of numerator and denominator

= a + 1

Note: - Try to factorize that term which help in reducing expression.

Question 3.

Simplify:

Answer:

We know that

and

Note: - Try to factorize that term which help in reducing expression.

Question 4.

Simplify:

Answer:

Applying and factorization, we get

= 1

Note: - Try to factorize that term which help in reducing expression.

Question 5.

Simplify:

Answer:

Applying

and factorization, we get, also changing by reversing N and D

Note: - Try to factorize that term which help in reducing expression.

Question 6.

Simplify:

Answer:

Applying

and factorization, we get

= x – 2

Note: - Try to factorize that term which help in reducing expression.

Question 7.

Simplify:

Answer:

Applying

_, factorization and we get, also changing by reversing N and D

= 5a + 1

Note: - Try to factorize that term which help in reducing expression.

Question 8.

Simplify:

Answer:

Applying

and factorization, we get

Note: - Try to factorize that term which help in reducing expression.

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