Q : 8           Using  (x+a)(x+b)=x^2+(a+b)x+ab , find 

                   (iii) 103 \times 98

Answers (1)

We know,

(x+a)(x+b)=x^2+(a+b)x+ab

Using this formula,

103 \times 98 = (100 + 3)(100 - 2) = (100 + 3){100 + (-2)}

Here x =100, a = 3, b = -2

\therefore 103 \times 98  = 100^2+(3 + (-2))100+(3\times(-2))

= 10000+100-6

= 10094

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