NCTB Class 7 Math Chapter Four Exercise 4.1 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 4.1 “Multiplication & Division of Algebraic Expressions” Exercise 4.1 Solution.

Board |
NCTB |

Class |
7 |

Subject |
Math |

Chapter |
4 |

Chapter Name |
“Multiplication & Division of Algebraic Expressions” |

Exercise |
4.1 Solution |

**Exercise:- 4.1**

**(1) Solution:-** (3abx 4^{3}a)

= 12a^{4} b

**(2) Solution:-** (5xyx 6az)

= 30axyz.

**(3) Solution:-** (5ax ^{2}X 3ax^{2} y)

= 15a^{3} x^{7} y

**(4) Solution:-** 8a^{2} b X (-2b^{2})

= -16a^{2} b^{3}

**(5) Solution:-** (-2abx^{2}) x (10bx^{3} y z)

= -20ab^{4} x^{3} y z

**(6) Solution:-** (-3p^{2} q^{3})x (-6p^{5} q^{4} )

= 18 p^{7}q^{7}

**(7) Solution:-** (-12m^{2} a^{2} x ^{2}) x (-2ma^{2} x^{2})

= 24 m^{3} a^{4} x^{5}

**(8) Solution:-** (7a^{3} bx^{5} y^{2}) x (-3x^{5} y^{5} a^{2} b^{2} )

= – 21a^{5} b^{3} x^{10}y^{5}

**(9) Solution:-** (2x+3y) (5xy)

= (10x^{2} y +15xy^{2})

**(10) Solution:-** (5x- 4xy) (9x y)

= 45x y – 36 x y

**(11) Solution:-** (2a^{2} – 3b^{2}+c^{2}) (a^{3} b^{2})

= 2a^{5} b^{2} – 3a^{3} b^{4}+ a^{3} b^{2} c^{2}

**(12) Solution:-** (x^{3} – y^{2} + 3xyz) (x^{4} y)

= x^{7} y- x^{4} y^{3} + 3x^{5} y z

**(13) Solution:-** (2a-3b) (3a+2b)

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

= 6a^{2} – 5ab- 6b^{2}

**(14) Solution:-** (a+ b) (a-b)

= a^{2} –b^{2}

**(15) solution:-** (x^{2}+1) (x^{2}-1)

= (x^{2})^{2} – (11)^{2}

= (x^{4}-1)

**(16) Solution:-** (a+ b) (a+ b)

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

**(17) Solution:-** (a^{2} –a b +b^{2}) (a +b)

= a^{3}+ a^{2} b – a^{2} b – a b^{2}+ a b^{2}+ b^{3}

= a^{3}+b^{3}

**(18) Solution:-** (x+2xy+y) 9x+y)

= x^{3}+yx^{2}+2x^{2} y+2x^{2} y+ xy^{2} +y^{3}

= x^{3}+ 3x^{2} y+ 3xy^{2}+ y^{3}

**(19) Solution:-** (x^{2}-2xy+y^{2}) (x-y)

= x^{2}-x^{2} y – 2x^{2} y+2x y^{2}+xy^{2}-y^{3}

= x^{3}– 3xy^{2}+ 3xy^{2}– y^{3}

**(20) Solution:-** (x+2x-3) (x+3)

= x^{3} + 3x^{2} + 2x^{2} +6x- 3x-9

= x^{3} + 5x^{2}+ 3x-9

**(21) Solution:- **(a^{2}+ab+ b^{2}) (b^{2}– ab+ a^{2})

= a^{2} b^{2} –a^{3}b+a^{4} +ab^{3}– a^{2} b^{2} +a^{3}b+ b^{4}-ab^{3}+a^{2} b^{2}

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

**(22) Solution:-** (a+ b +c) (a+ b +c)

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

= a^{2}+b^{2}+c^{2}+2ab+2bc+2ca

**(23) Solution:-** (x^{2}-x^{3}y +x^{2} y^{2}+ x^{3}y- x^{2} y^{2} + xy^{3}+ x^{2} y^{2} – xy^{3}+y^{4})

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

**(24) Solution:-** (y^{2}-y+1) (1+y+y^{2})

= (y^{2}+ y^{3}+ y^{4}-y- y^{2-} y^{3}+1+y+ y^{2})

= y^{4}+ y^{2}+1

**(25) Solution:-**

A= x^{2}+xy+y^{2}

B= x-y

Therefore, L.H.S, AB

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

= x^{3} – y3

= R.H.S

L.H.S=R.H.SS (proved).

**(26) Solution:-** AB.

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

= a^{3}+b^{3}

**(27) Solution:- L.H.S,**

(a^{2}+1) (a – 1) (a^{2}+1)

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

= (a^{2})^{2} – (1)^{2}

= a^{4} – 1

= R.H.S

**(28) Solution:-** **L.H.S,**

(x +y) (x-y) (x +y)

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

= (x^{2})^{2} – (y^{2})^{2}

= x^{4}-y^{4}

= R.H.S

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