Using the identities of algebraic expressions - find i 1002 - 1003ii 97 - 103- -2 MARKS-

Application: 1 Mark each (i) 1002 × 1003 = (1000 + 2)( 1000 + 3) ⇒1000^2 + ( 2 + 3) 1000 + (3)(2) = 1000000 + 5000 + 6 = 1005006 (ii) 97 × 103 = (100 3 ...

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Standard IX

Mathematics

Polynomial Expressions

Using the ide...

Question

Using the identities of algebraic expressions , find

$(i) 1002 \times 1003$

$(i i) 97 \times 103$ . [2 MARKS]

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Solution

Application: 1 Mark each

$(i) 1002 \times 1003$

$= (1000 + 2) (1000 + 3) \Rightarrow 1000^{2} + (2 + 3) 1000 + (3) (2)$

$= 1000000 + 5000 + 6$

$= 1005006$

$(i i) 97 \times 103$

$= (100 - 3) (100 + 3)$

$= [{(100)}^{2} - {(3)}^{2}]$

$= [10000 - 9]$

$= 9991$

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Using the identities of algebraic expressions , find (i) 1002×1003 (ii) 97×103. [2 MARKS]

Application: 1 Mark each (i)1002×1003 =(1000+2)(1000+3)⇒10002+(2+3)1000+(3)(2) =1000000+5000+6 =1005006 (ii)97×103 =(100−3)(100+3) =[(100)2−(3)2] =[10000−9] =9991

Using the identities of algebraic expressions , find

$(i) 1002 \times 1003$

$(i i) 97 \times 103$ . [2 MARKS]

Application: 1 Mark each

$(i) 1002 \times 1003$

$= (1000 + 2) (1000 + 3) \Rightarrow 1000^{2} + (2 + 3) 1000 + (3) (2)$

$= 1000000 + 5000 + 6$

$= 1005006$

$(i i) 97 \times 103$

$= (100 - 3) (100 + 3)$

$= [{(100)}^{2} - {(3)}^{2}]$

$= [10000 - 9]$

$= 9991$