ML Aggarwal Solution Class 9 Chapter 4 Factorisation Exercise 4.5

 Exercise 4.5

Question 1

(i) 8x3+y3

⇒(2x)3+y3

∴It is in the form of a3+b3=(a+b)(a2-ab+b2)

∴there a=2x ; b=7

∴⇒(2x)3+y3=(2x+y)[(2x)2-2x.y+y2]

=(2x+y)(4x2-2xy+y2)


(ii) 64x3-125y3

⇒(4a)3-(5y)3

⇒It is in the form of a3-b3=(a-b)(a2+ab+b2)

⇒here a=4x ; b=5y

∴⇒(4x-5y)[(4x)2+4x.5y+(5y)2]

⇒(4a-5y)(16x2+20xy+25y2)


Question 2

Sol :

(i) 64x3+1

⇒(4x)3+13

⇒It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=4x ; b=1

∴⇒(4x+1)[(4x)2-4x.1+12]

⇒(4x+1)(16x2-4x+1)


(ii) 7a3+56b3

⇒7(a3+8b3)

⇒7(a3+(2b)3)

It is in the form of a3+b3=(a+b)(a2-ab+b2)

∴Here a=a ; b=2b

∴⇒7(a+2b)(a2-a.2b+(2b)2)

⇒7(a+2b)(a2-2ab+4b2)


Question 3

(i) $\frac{x^6}{343}+\frac{343}{x^6}$

$\frac{x^6}{7^3}+\frac{7^3}{x^6}$

⇒It is in the form of a3+b3=(a+b)(a2-ab+b2)

∴Here $a=\frac{x^2}{7}$ ; $b=\frac{7}{x^2}$

⇒$\frac{(x^2)^3}{7^3}+\frac{7^3}{(x^2)^3}$

⇒$\left(\frac{x^2}{7}+\frac{7}{x^2}\right)\left[\left(\frac{x^2}{7}\right)^2-\frac{x^2}{7}\times \frac{7}{x^2}+\left(\frac{7}{x^2}\right)^2\right]$

⇒$\left(\frac{x^2}{7}+\frac{7}{x^2}\right)\left(\frac{x^4}{49}-1+\frac{49}{x^4}\right)$


(ii) $8x^3-\frac{1}{27y^3}$

⇒$(2x)^3-\frac{1}{(3y)}^3$

⇒It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=2x ; $b=\frac{1}{3y}$

⇒$\left(2x-\frac{1}{3y}\right)\left[(2x)^2+2x.\frac{1}{3y}+(3y)^2\right]$

⇒$\left(2x-\frac{1}{3y}\right)\left(4x^2+\frac{2x}{3y}+9y^2\right)$


Question 4

(i) x2+x5

⇒x2(1+x3)

⇒x3(13+x3)

It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=1 ; b=x

∴x2(1+x)(12-1.x+x2)

⇒x2(1+x)(1-x+x2)


(ii) 32x4-500

⇒4x(8x3-125)

⇒4x((2x)3-53)

It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=2x ; b=5

⇒4x(2x-5)((2x)2+2x.5+52)

⇒4x(2x-5)(4x2+10x+25)


Question 5

(i) 27x3y3-8

⇒(3xy)3-23

It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=3xy ; b=2

⇒(3xy-2)[(3xy)2+3xy.2+22]

⇒(3xy-2)[9x2y2+6xy+4]


(ii) 27(x+y)3+8(2x-y)3

⇒33(x+y)3+23(2x-y)3

⇒(3(x+y)3)+(2(x-y))3

It is in the form of a3+b3=(a+b)(a2-ab+b2)

⇒Here a=3(x+y) ; b=2(x-y)
⇒[3(x+y)+2(2x-y)][32(x+y)2-3(x+y)(2x-y).2+22(2x-y)2]
⇒(3x+3y+4x-2y)[9(x+y)2-6(x+y)(2x-y)+4(2x-y)2]
⇒(7x-y)[9(x2+y2+2xy)-6(2x2-xy+2xy-y2)+4(4x2+y2-4xy)]
⇒(7x-y)[9x2+9y2+18xy-12x2-6xy-6y2+16x2+4y2-16xy]
⇒(7x-y)[13x2-4xy+19y2]


Question 6

(i) a3+b3+a+b

⇒(a3+b3)+(a+b)

⇒(a+b)(a2-ab+b2)+(a+b)

⇒(a+b)(a2-ab+b2+1)


(ii) a3-b3-a+b

⇒(a3-b3)-(a-b)

⇒(a-b)(a2+ab+b2)-(a-b)

⇒(a-b)(a2+ab+b2-1)


Question 7

(i) x3+x+2

⇒x3+x+1+1

⇒(x3+1)+(x+1)

⇒(x+1)(x2-x+1)+(x+1)

⇒(x+1)(x2-x+1+1)

⇒(x+1)(x2-x+2)


(ii) a3-a-120

⇒a3-a-125+5

⇒a3-125-(a-5))

⇒(a3-53)-(a-5)

⇒(a-5)(a2+5a+52)-(a-5)

⇒(a-5)(a2+5a+25)-(a-5)

⇒(a-5)[a2+5a+25-1]

⇒(a-5)(a2+5a+24)


Question 8

(i) x3+6x2+12x+16

⇒x3+6x2+12x+8+18

⇒(x3+3.2.x2+3.22.x+23)+8

It is in the form of a3+b3=(a+b)(a2-ab+b2)

∴Here a=x ; b=2

∴(x+2)3+8

⇒(x+2)3+23

⇒(x+2+2)[(x+2)2-2.(x+2)+22]

⇒(x+4)(x2+4+4x-2x-4+4)

⇒(x+4)(x2+2x+4)


(ii) a3-3a2b+3ab2-2b3

⇒a3-3a2b+3ab2-b3-b3

⇒(a-b)3-b3

It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=a-b ; b=b

⇒(a-2b)((a-b)2+(a-b)b+b2)

⇒(a-2b)(a+b-2ab+ab-b2+b2)

⇒(a-2b)(a2+b2-ab)


Question 9

(i) 2a3+16b3-5a-10b

⇒a3+a3+8b3-5a-10b+8b3

⇒2(a3+(2b)3)-5(a+2b)

It is in the form of a3+b3=(a+b)(a2-ab+b2)

⇒2[(a+2b)(a2-2ab+(2b)a2)-5(a+2b)]

⇒2[(a+2b)(a2-2ab+4b2)]-5(a+2b)

⇒(a+2b)[2(a2-2ab+4b2)-5]

⇒(a+2b)[2a2-4ab+8b2-5]


(ii) $a^3-\frac{1}{a^3}-2a+\frac{2}{a}$

⇒a3-2a+a-a+$\frac{2}{a}+\frac{1}{a}-\frac{1}{a}-\frac{1}{a^3}$

⇒a3-3a+a+$\frac{3}{a}-\frac{1}{a}-\frac{1}{a^3}$

⇒a3-3a2.$\frac{1}{a}+3.a.\frac{1}{a^2}-\frac{1}{a^3}+\left(a-\frac{1}{a}\right)$

It is in the form of a3+b3=(a+b)(a2-ab+b2)

∴Here a=a ; $b=\frac{1}{a}$

∴$\left(a-\frac{1}{a}\right)^3+\left(a-\frac{1}{a}\right)$

⇒$\left(a-\frac{1}{a}\right)\left[\left(a-\frac{1}{a}\right)^2+1\right]$

⇒$\left(a-\frac{1}{a}\right)\left[a^2+\frac{1}{a^2}-2.a.\frac{1}{a}+1\right]$

⇒$\left(a-\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-2+1\right)$

⇒$\left(a-\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-1\right)$


Question 10

(i) a6-b6

⇒(a2)3-(b2)3

It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=a2 ; b=b2

∴(a2-b2)[(a2)2+a2-b2+(b2)2]

⇒(a2-b2)(a4+a2b2+b4)


(ii) x6-1

⇒(x2)3-13

⇒It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here x2=a ; b=1

∴(x2-1)[(x2)2+x2.1+12]

⇒(x2-1)(x4+x2+1)


Question 11

(i) 64x6-729y6

⇒(2x)6-(3y)6

⇒[(2x)2]3-[(3y)2]3

⇒It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=(2x)2 ; b=(3y)2

⇒[(2x)2-(3y)2][((2x)2)2+(2x)2.(3y)2+((3y)2)2]

⇒[4x2-9y2][16x4+4x2.9y2+((9y2)2]

⇒[4x2-9y2][16x4+36x2y2+81y4]

⇒[(2x)2-(3y)2][16x4+36x2y2+81y4]

⇒(2x+3y)(2x-3y)(16x4+36x2y2+81y4)


(ii) $x^2-\frac{8}{x}$

⇒$\left(x^2-\frac{8}{x}\right)\times \frac{x}{x}$

⇒$x\left(x^2-\frac{8}{x}\right)\frac{1}{x}$

⇒$\left(x^3-x.\frac{8}{x}\right)\frac{1}{x}$

⇒$(x^3-8)\frac{1}{x}$

⇒$\left(x^3-2^3\right).\frac{1}{x}$

It is in the form of a3+b3=(a+b)(a2-ab+b2)

a=x ; b=2

⇒$\frac{1}{x}.(x+2)(x^2+2x+4)$


Question 12

(i) 250(a-b)3+2

⇒$(250(a-b)^3+2)\frac{2}{2}$

⇒$2\frac{(250(a-b)^3+2)}{2}$

⇒2(125(a-b)3+1)

⇒2(53(a-b)3+1)

⇒2((5(a-b))3+13)

It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=5(a-b) ; b=1

⇒2[5(a-b)+1](5(a-b))2-5(a-b).1+12]

⇒2[(5a-5b+1)][25(a2+b2-2ab)-5a+5b+1]

⇒2(5a-5b+1)(25a2+25b2-50ab-5a+5b+1)


(ii) 32a2x3-8b2x3-4a2y3+b2y3

⇒x3(32a2-8b2)-y3(4a2-b2)

⇒8x3(4a2-b2)-y3(4a2-b2)

⇒(4a2-b2)(8x3-y3)

⇒((2a)2-(b)2)((2x)3-(y)3)

⇒(2a+b)(2a-b)[(2x-y)(2x)2+2xy+y]

⇒(2a+b)(2a-b)(2x-y)(4x2+2xy+y2)


Question 13

(i) x2+y2

⇒(x3)3+(y3)3

It is in the form of a3+b3=(a+b)(a2-ab+b2)

Here a=x3 ; b=y3

∴(x3+y3)[(x3)3-x3y3+(y3)2]

⇒(x3+y3)(x6-x3y3+(y3)2)

It is in the form of a3+b3=(a+b)(a2-ab+b2)

⇒Here a=x ; b=y

∴⇒(x+y)(x2-xy+y2)(x6-x3y3+y6)


(ii) x6-7x3-8

⇒(x2)3-7x3-x3-8

⇒(x2)3-8x3+x3-23

⇒[(x2)3-(2x)3]+(x3-22)

⇒(x2-2x)[(x2)2+x2.2x+(2x)2]+(x-2)(x2+2x+22)

⇒(x2-2x2)(x4+2x3+4x2)+(x-2)(x2+2x+4)

⇒x(x-2)x2(x2+2x+4)+(x-2)(x2+2x+4)

Taking common factor as (x-2) and x2+2x+4

∴(x-2)(x2+2x+4)(x+1)(x2-x+1)

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more