NCTB Class 8 Math Chapter Four Exercise 4.3 Solution

NCTB Class 8 Math Chapter Four Exercise 4.3 Solutions by Math Expert. Bangladesh Board Class 8 Math Solution Chapter Four Algebraic Formulae and Applications Exercise 4.3 Solution.

Board

NCTB
Class

8

Subject

Mathematics
Chapter

4

Chapter Name

Algebraic Formulae and Applications
Exercise

4.3 Solution

Exercise – 4.3

Resolve into factors :

1> a3+8

= (a)3 + (2)3

= (a+2) (a2-2a+4)

2> 8x3+343

= (2x)3+(7)3

= (2x+7)(4x2-14x+49)

3> 8a4+27ab3

= 8a (8a3+27b3)

= a{(2a)3+(3b)3}

= a{(2a+3b)(4a2-6ab+9b2)}

4> 8x3+1

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

= (2x+1)(4x2-2x+1)

5> 64a3+125b3

= (4a)3 – (5b)3

= (4a-5b) (16a2+20ab+25b2)

6> 729a3-64b3c6

= (9a)3-(4bc2)3

= (9a-4bc2) (81a2+36abc2+16b2c4)

7> 27a3b3 + 64b3c3

= b3 (27a3+64c3)

= b3 {(3a)3+(4c)3}

= b3{(3a+4c) (9a2-12ac+16c2)

8> 56x3 – 189y3

= 7 (8x3-27y3)

= 7 {(2x)3-(3y)3}

= 7 {(2x-3y) (4x2-6xy+9y2)

9> 3x-75x3

= 3x (1-25x2)

= 3x {(1)2-(5x)2}

= 3x (1+5x)(1-5x)

10> 4x2-y2

= (2x)2 – y2

= (2x+y) (2x-y)

11>  3ay2+48a

= a (3y-48)

= 3a (y2-16)

= 3a (y+4)(y-4)

12> a2+2ab+b2-p2

= (a-b)2-p2

=(a-b+p)(a-b-p)

13> 16y2-a2-6a-9

= (4y)2– (a+3)2

= (4y+a+3)(4y-a-3)

14> 8a+ap3

= a(8+p3)

= a{(2)3+p3}

= a (2+p)(22-2p+p2)

= a (2+p)(4-2p+p2)

15>2a3+16b3

= 2 (a3+8b3)

= 2 [(a)3+(2b)3]

= 2 (a+2b)(a2-2ab+4b2)

16> x2+y2-2xy-1

= x2-2xy+y2-1

= (x-y)2-(1)2

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

18> x4-2x2+1

= x4+1-2x2

= (x2)2+(1)2-2x2

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

19> 36-12x+x2

= x2-12x+36

= x2-6x-6x+36

= x (x-6)-6(x-6)

= (x-6)2

20> x6-y6

=(x3)2-(y3)2

= (x3+y3)(x3-y3)

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

21> (x-y)3+ z3

= (x-y+z) {(x-y)2-(x-y)z+z2}

= (x-y+z)(x2-2xy+y2-xz+yz+z2)

22> 64x3-8y3

= (4x)3-(2y)3+

= (4x-2y) {(4x)2+4x.2y+(2y)2}

= (4x-2y) (16x2+8xy+4y2)

= 8 (2x-y) (4x2+2xy+y2)

23> x2 +14x+40

= x2+10x+4x+40

= x(x+10)+4(x+10)

= (x+10)(x+4)

24> x2+7x-120

= x2+15x-8x-120

= x(x+15)-8(x+15)

= (x+15)(x-8)

25> x2-51x+650

= x2-26x-25x+650

= x(x-26)-25(x-26)

= ( x-26)(x-25)

26> a2+7ab+12b2

= a2+4ab+3ab+12b2

=a(a+4b)+3b(a+4b)

= (a+3b)(a+4b)

27> p2+2pq-80q2

= p2+10pq-8pq-80q2

= p(p+10q)-8q(p+10q)

= (p+10q) (p-8q)

28> x2-3xy-40y2

= x2-8xy+5xy-40y2

= x(x-8y)+5y(x-8y)

= (x-8y)(x+5y)

29> (x2-x)2+3(x2-x)-40

Let, x2-x= a

A2-3a-40

= a2-8a+5a-40

= a (a-8)+5(a-8)

= (a-8)(a+5)

30> (a2+b2)2-18(a2+b2)-88

Let, a2+b2= x

Therefore, x2-18x-88

= x2-22x+4x-88

= x(x-22)+4(x-22)

= (x-22)(x+4)

= (a2+b2-22)(a2+b2+4)

31> (a2+7a)2-8(a2+7a)-180

Let, a2+7a=x

X2-8x-180

= x2-18x+10x-180

= x(x-18)+10(x-18)

= (x-18)(x+10)

= (a2+7a-18)(a2+7a+10)

33> 6x2-x-15

= 6x2-10x+9x-15

= 2x(3x-5)+3(3x-5)

= (3x-5)(2x+3)

35> 3x2+11x-4

= 3x2+12x-x-4

= 3x (x+4)-1(x+4)

= (3x-1)(x+4)

36> 3x2-16x-12

= 3x2-18x+2x-12

= 3x (x-6)+2(x-6)

= (x-6)(3x+2)

37> 2x2-9x-35

= 2x2-14x+5x-35

= 2x (x-7)+5(x-7)

=(2x+5)(x-7)

38> 2x2-5xy+2y2

= 2x2-4xy-xy+2y2

= 2x (x-2y)-y(x-2y)

= (x-2y)(2x-y)

40> 10p2+11pq-6q2

= 10p2+15pq-4pq-6q2

= 5p(2p+3q)-2q(2p+3q)

= (5p-2q)(2p+3q)

41> 2(x+y)2-3(x+y)-2

Let, x+y= a

2a2-3a-2

= 2a2-4a+a-2

= 2a (a-2)+1 (a-2)

= (a-2)(2a+1)

43> 15x2-11xy-12y2

= 15x2-20xy+9xy-12y2

= 5x (3x-4y)+3y (3x-4y)

= (5x+3y)(3x-4y)

44> a3+3a2b+3ab2-2b3

= (a3-3a2b+3ab2-b3-b3)

= (a-b)3 – b3

=(a-b-b) {(a-b)2+(a-b).b+b2)

= (a-2b) (a2-2ab+b2+ab-b2+b2)

= (a-2b)(a2-ab+b2)

Updated: March 24, 2021 — 1:23 pm

Leave a Reply

Your email address will not be published. Required fields are marked *