Question

"Question 3 Simplify. (i) (x^2 − 5) (x + 5) + 25 (ii) (a^2 + 5) (b^3 + 3) + 5 (iii) (t + s^2) (t^2 − s) (iv) (a + b) (c − d) + (a − b) (c + d) + 2 (ac + bd) (v) (x + y) (2x + y) + (x + 2y) (x − y) (vi) (x + y) (x^2 − xy + y^2) (vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y (viii) (a + b + c) (a + b − c)

Class 8 Algebraic Expressions and Identities Page 148"

Answer 1

Algebraic expressions:

A combination of constants and variables connected by any or all of the four fundamental operations +, -,×,÷ is called an algebraic expression

Multiplication of algebraic expression:

The product of two factors with like signs is positive and the product of two factors with unlike signs is negative.

Multiplication of a binomial  and the binomial:

 

 Let  ( a+b)&  (c+d) are two Binomials. By using distributive law of multiplatinum over addition twice, we may find their product

( a+b)× (c+d)

= a(c+d) + b( c+d)
=(a ×  c+a ×  d)+(b ×  c+b × d)

= ac+ad+bc+bd

This method is known as the horizontal method.

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Solution:

i)  (x²-5)(x+5)+25

x²(x+5) -5(x+5)+25

x³+ 5x² - 5x - 25 + 25

= x³ + 5x² -5x

 

ii) (a²+5)(b³+3)+5

a²(b³+3) +5(b³+3)

a²b³ + 3a² + 5b³ + 15 + 5

= a²b³ + 5b³ + 3a² + 20

iii) (t+s²)(t²-s)

t(t²-s)+ s²(t²-s)

t³-ts+s²t²-s³

 

iv) (a+b)(c-d)+(a-b)(c+d)+2(ac+bd)

a(c-d)+b(c-d) + a(c+d)-b(c+d) +2(ac+bd)

 

(ac - ad + bc - bd) + (ac + ad - bc - bd) + (2ac + 2bd)

= ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd

= ac+ac+2ac -ad+ad+bc-bc-bd-bd+2bd

= 4ac

v) (x + y) (2x + y) + (x + 2y) (x − y)

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

 

=2x² + xy + 2xy + y² + x² - xy + 2xy - 2y²

= 2x²+x²+xy-xy+2xy+2xy+y²-2y²

= 3x² + 4xy - y²

 

vi) (x + y) (x² − xy + y²)  

x(x² − xy + y²)  +y(x² -xy+y²)

= x³- x²y + xy² + x²y – xy² +y³

= x³ -x²y+x²y+xy²-xy²+y³

= x³ + y³

 

vi) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y

= 1.5x(1.5x + 4y + 3) -4y(1.5x + 4y + 3) -4.5x+12y

 

=2.25x² + 6xy + 4.5x - 6xy - 16y² - 12y - 4.5x + 12y

= 2.25x² +6xy- 6xy+4.5x-4.5x-12y+12y-16y²

= 2.25x² - 16y²

 

vii) (a + b + c) (a + b − c)

 

= a(a + b − c)+b(a + b − c)+c(a + b − c)

= a² + ab - ac + ab + b² - bc + ac + bc - c²

= a² +ab +ab -ac+ac + b²- c² + bc-bc

= a²+2ab+b² - c²

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Hope this will help you....