Math, asked by abi0411, 1 year ago

NEED HELP!!!!!!! (no spam answers)

Factorize -

1. 999^2-1

2. (10.2)^ 3

3. 1002 x 998



(using factor theorem)

Answers

Answered by gaurav2013c
10
Solution 1.)

999^2 - 1

= (999)^2 - (1)^2

= (999 + 1)(999-1)

= 1000 (998)

= 998000

Solution 2.)

(10.2)^3

= (10 + 0.2)^3

= (10)^3 + (0.2)^3 + 3(10)(0.2)(10.2)

= 1000 + 0.008 + 6 (10.2)

= 1000 + 0.008 + 61. 2

= 1061.208

Solution 3.)

1002 × 998

= (1000 + 2)(1000 - 2)

= (1000)^2 - (2)^2

= 1000000 - 4

= 999996
Answered by ɪᴛᴢᴛʀᴀɢɪᴄɢɪʀʟ
2

Solution 1.)

999^2 - 1

= (999)^2 - (1)^2

= (999 + 1)(999-1)

= 1000 (998)

= 998000

Solution 2.)

(10.2)^3

= (10 + 0.2)^3

= (10)^3 + (0.2)^3 + 3(10)(0.2)(10.2)

= 1000 + 0.008 + 6 (10.2)

= 1000 + 0.008 + 61. 2

= 1061.208

Solution 3.)

1002 × 998

= (1000 + 2)(1000 - 2)

= (1000)^2 - (2)^2

= 1000000 - 4

= 999996

Similar questions