**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> a ^{3}+8**

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

= (a+2) (a^{2}-2a+4)

**2> 8x ^{3}+343**

= (2x)^{3}+(7)^{3}

= (2x+7)(4x^{2}-14x+49)

**3> 8a ^{4}+27ab^{3}**

= 8a (8a^{3}+27b^{3})

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

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

**4> 8x ^{3}+1**

= (2x)^{3}+(1)^{3+}

= (2x+1)(4x^{2}-2x+1)

**5> 64a ^{3}+125b^{3}**

= (4a)^{3 }– (5b)^{3}

= (4a-5b) (16a^{2}+20ab+25b^{2})

**6> 729a ^{3}-64b^{3}c^{6}**

= (9a)^{3}-(4bc^{2})^{3}

= (9a-4bc^{2}) (81a^{2}+36abc^{2}+16b^{2}c^{4})

**7> 27a ^{3}b^{3 }+ 64b^{3}c^{3}**

= b^{3 }(27a^{3}+64c^{3})

= b^{3 }{(3a)^{3}+(4c)^{3}}

= b^{3}{(3a+4c) (9a^{2}-12ac+16c^{2})

**8> 56x ^{3 }– 189y^{3}**

= 7 (8x^{3}-27y^{3})

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

= 7 {(2x-3y) (4x^{2}-6xy+9y^{2})

**9> 3x-75x ^{3}**

= 3x (1-25x^{2})

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

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

**10> 4x ^{2}-y^{2}**

= (2x)^{2 }– y^{2}

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

**11> 3ay ^{2}+48a**

= a (3y-48)

= 3a (y^{2}-16)

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

**12> a ^{2}+2ab+b^{2}-p^{2}**

= (a-b)^{2}-p^{2}

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

**13> 16y ^{2}-a^{2}-6a-9**

= (4y)^{2}– (a+3)^{2}

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

**14> 8a+ap ^{3}**

= a(8+p^{3})

= a{(2)^{3}+p^{3}}

= a (2+p)(2^{2}-2p+p^{2})

= a (2+p)(4-2p+p^{2})

**15>2a ^{3}+16b^{3}**

= 2 (a^{3}+8b^{3})

= 2 [(a)^{3}+(2b)^{3}]

= 2 (a+2b)(a^{2}-2ab+4b^{2})

**16> x ^{2}+y^{2}-2xy-1 **

= x^{2}-2xy+y^{2}-1

= (x-y)^{2}-(1)^{2}

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

**18> x ^{4}-2x^{2}+1**

= x^{4}+1-2x^{2}

= (x^{2})^{2}+(1)^{2}-2x^{2}

= (x^{2}+1)(x-1)^{2}

**19> 36-12x+x ^{2}**

= x^{2}-12x+36

= x^{2}-6x-6x+36

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

= (x-6)^{2 }

**20> x ^{6}-y^{6}**

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

= (x^{3}+y^{3})(x^{3}-y^{3})

= (x+y) (x^{2}-xy+y^{2})(x-y)(x^{2}+xy+y^{2})

**21> (x-y) ^{3}+ z^{3}**

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

= (x-y+z)(x^{2}-2xy+y^{2}-xz+yz+z^{2})

**22> 64x ^{3}-8y^{3}**

= (4x)^{3}-(2y)^{3+}

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

= (4x-2y) (16x^{2}+8xy+4y^{2})

= 8 (2x-y) (4x^{2}+2xy+y^{2})

**23> x ^{2 }+14x+40**

= x^{2}+10x+4x+40

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

= (x+10)(x+4)

**24> x ^{2}+7x-120**

= x^{2}+15x-8x-120

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

= (x+15)(x-8)

**25> x ^{2}-51x+650**

= x^{2}-26x-25x+650

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

= ( x-26)(x-25)

**26> a ^{2}+7ab+12b^{2}**

= a^{2}+4ab+3ab+12b^{2}

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

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

**27> p ^{2}+2pq-80q^{2}**

= p^{2}+10pq-8pq-80q^{2}

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

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

**28> x ^{2}-3xy-40y^{2}**

= x^{2}-8xy+5xy-40y^{2}

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

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

**29> (x ^{2}-x)^{2}+3(x^{2}-x)-40**

Let, x^{2}-x= a

A^{2}-3a-40

= a^{2}-8a+5a-40

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

= (a-8)(a+5)

**30> (a ^{2}+b^{2})^{2}-18(a^{2}+b^{2})-88**

Let, a^{2}+b^{2}= x

Therefore, x^{2}-18x-88

= x^{2}-22x+4x-88

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

= (x-22)(x+4)

= (a^{2}+b^{2}-22)(a^{2}+b^{2}+4)

**31> (a ^{2}+7a)^{2}-8(a^{2}+7a)-180**

Let, a^{2}+7a=x

X^{2}-8x-180

= x^{2}-18x+10x-180

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

= (x-18)(x+10)

= (a^{2}+7a-18)(a^{2}+7a+10)

**33> 6x ^{2}-x-15**

= 6x^{2}-10x+9x-15

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

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

**35> 3x ^{2}+11x-4**

= 3x^{2}+12x-x-4

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

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

**36> 3x ^{2}-16x-12**

= 3x^{2}-18x+2x-12

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

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

**37> 2x ^{2}-9x-35 **

= 2x^{2}-14x+5x-35

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

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

**38> 2x ^{2}-5xy+2y^{2}**

= 2x^{2}-4xy-xy+2y^{2}

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

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

**40> 10p ^{2}+11pq-6q^{2}**

= 10p^{2}+15pq-4pq-6q^{2}

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

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

**41> 2(x+y) ^{2}-3(x+y)-2**

Let, x+y= a

2a^{2}-3a-2

= 2a^{2}-4a+a-2

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

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

**43> 15x ^{2}-11xy-12y^{2}**

= 15x^{2}-20xy+9xy-12y^{2}

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

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

**44> a ^{3}+3a^{2}b+3ab^{2}-2b^{3}**

= (a^{3}-3a^{2}b+3ab^{2}-b^{3}-b^{3})

= (a-b)^{3 }– b^{3}

=(a-b-b) {(a-b)^{2}+(a-b).b+b^{2})

= (a-2b) (a^{2}-2ab+b^{2}+ab-b^{2}+b^{2})

= (a-2b)(a^{2}-ab+b^{2})

Very good 39 downi

It helps me very much

Where is the 17th numder ?

32 number missing

Where is the 17th no. math?

nice but some maths are missing.

Thanks. But, some maths are missing.

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