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)