Class 9 Maths


Polynomials

Exercise 2.5 Part 2

Question 2: Evaluate the following products without multiplying directly:

(i) `103 xx 107`

Answer: Given, `103 xx 107`
Given, expression can be written as:
`(100 + 3)(100 + 7)`
Let, x = 100, a = 3 and b = 7

Using identity, `(x + a)(x + b) = x^2 + (a + b) x + ab`, we get;
`(100 + 3)(100 + 7)`

`= 100^2+ (3 + 7)xx 100 + (3 xx 7)`

`= 10000 + (10 xx 100) + 21`

`= 10000 + 1000 + 21`

`= 11000 + 21= 11021` Answer

(ii) `95 xx 96`

Answer: Given, `95 xx 96`
Given, expression can be written as
`(90 + 5)(90 + 6)`
Let, `x = 90`, `a = 5` and `b = 6`

Using identity, `(x + a)(x + b) = x^2 + (a + b) x + ab`

`(90 + 5)(90 + 6)`

`= 90^2 + (5 + 6) xx 90 + (5 xx 6)`

`= 8100 + (11 xx 90) + 30`

`= 8100 + 990 + 30`

`= 9090 + 30= 9120` Answer

(iii) `104 xx 96`

Answer: Given, `104 xx 96`
Given expression can be written as
`(100 + 4)(100 – 4)`
Let, `x = 100` and `y = 4`

Therefore, using identity `(x + y)(x –y) = x^2 - y^2`

`(100 + 4)(100 – 4)`

`= 100^2 - 4^2`

`= 10000 – 16= 9984` Answer