Factorisation Of Algebraic Expressions Class 8th Mathematics (new) MHB Solution

Factorisation Of Algebraic Expressions

Class 8th Mathematics (new) MHB Solution

Class 8th Mathematics (new) MHB Solution
Practice Set 6.1
  1. Factorise.x2 + 9x + 18
  2. Factorise.x2 – 10x + 9
  3. Factorise.y2 + 24y + 144
  4. Factorise.5y2 + 5y – 10
  5. Factorise.p2 – 2p – 35
  6. Factorise.p2 – 7p – 44
  7. Factorise.m2 – 23m + 120
  8. Factorise.m2 – 25m + 100
  9. Factorise.3x2 + 14x + 15
  10. Factorise.2x2 + x – 45
  11. Factorise.20x2 – 26x + 8
  12. Factorise.44x2 – x – 3
Practice Set 6.2
  1. Factorise.x3 + 64y3
  2. Factorise.125p3 + q3
  3. Factorise.125k3 + 27m3
  4. Factorise.2l3 + 432m3
  5. Factorise.24a3 + 81b3
  6. Factorise. y^{3} + {1}/{ 8y^{3} }
  7. Factorise. a^{3} + {8}/{ a^{3} }
  8. Factorise. 1 + { q^{3} }/{125}
Practice Set 6.3
  1. Factorise :y3 – 27
  2. Factorise :x3 – 64y3
  3. Factorise :27m3 – 216n3
  4. Factorise :125y3 – 1
  5. Factorise :8p3 – 27/p3
  6. Factorise :343a3 – 512b3
  7. Factorise :64x2 – 729y2
  8. Factorise :16 a3 – 128/b3
  9. Simplify :(x + y)3 – (x – y)3
  10. Simplify :(3a + 5b)3 – (3a – 5b)3
  11. Simplify :(a + b)3 – a3 – b3
  12. Simplify :p3 – (p + 1)3
  13. Simplify :(3xy – 2ab)3 – (3xy + 2ab)3
Practice Set 6.4
  1. Simplify: { m^{2} - n^{2} }/{ (m+11) } x frac { m^{2} + mn+n^{2} }/{ m^{3} - n^{3}…
  2. Simplify: { a^{2} + 10a+21 }/{ a^{2} + 6a-7 } x frac { a^{2} - 1 }/{a+3}…
  3. Simplify: { 8x^{3} - 27y^{3} }/{ 4x^{2} - 9y^{2} }
  4. Simplify: { x^{2} - 5x-24 }/{ (x+3) (x+8) } x frac { x^{2} - 64 }/{ (x-8)^{2} }…
  5. Simplify: { 3x^{2} - x-2 }/{ x^{2} - 7x+12 } / frac { 3x^{2} - 7x-6 }/{ x^{2} -…
  6. Simplify: { 4x^{2} - 11x+6 }/{ 16x^{2} - 9 }
  7. Simplify: { a^{3} - 27 }/{ 5a^{2} - 16a+3 } / frac { a^{2} + 3a+9 }/{ 25a^{2} -…
  8. 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

a


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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