Using a suitable algebraic identity- find 41- 43- nbsp-1763

The correct option is A 1763We know that, nbsp; 41 × 43 can be written as (40 + 1) × (40 +3) So, using the algebraic identity, nbsp;(x + a) × (x +b)= x^2 + ...

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Standard VIII

Mathematics

Factorization of ax^2+bx+c

Using a suita...

Question

Using a suitable algebraic identity, find $41 \times 43$ .
1763

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Solution

The correct option is A 1763
We know that,
$41 \times 43$ can be written as $(40 + 1) \times (40 + 3)$
So, using the algebraic identity, $(x + a) \times (x + b) = x^{2} + (a + b) x + a b$

Here, $x = 40, a = 1, b = 3$
$(40 + 1) \times (40 + 3) = 40^{2} + (1 + 3) \times 40 + 1 \times 3$
$= 1600 + 160 + 3$
$= 1763$

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Using a suitable algebraic identity, find 41×43. 1763

The correct option is A 1763We know that, 41×43 can be written as (40+1)×(40+3) So, using the algebraic identity, (x+a)×(x+b)=x2+(a+b)x+ab Here, x=40,a=1,b=3 (40+1)×(40+3)=402+(1+3)×40+1×3 =1600+160+3 =1763

Using a suitable algebraic identity, find $41 \times 43$ .
1763