Question

Use a suitable id
(i) (x + 3) (x + 3)
2.1) (1​

Answer 1

Answer:

(i) (x + 4) (x + 10)

: (x + a) (x + b) = x² + (a+b)x + ab

Here,

x = x , a = 4 , b = 10

=> (x)² + (4 + 10)x + 4*10

=> x² + 14x + 40

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(ii) (x + 8) (x - 10)

: (x + a) (x + b) = x² + (a+b)x + ab

Here,

x = x , a = 8 , b = (-10)

=> (x)² + { 8 + (-10) }x + 8*(-10)

=> x² + (8 - 10)x - 80

=> x² - 2x - 80

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(iii) (3x + 4) (3x - 5)

: (x + a) (x + b) = x² + (a+b)x + ab

Here,

x = 3x , a = 4 , b = (-5)

=> (3x)² + { 4 + (-5) }*3x + 4*(-5)

=> 9x² + (4 - 5)*3x - 20

=> 9x² + (-1)*3x - 20

=> 9x² - 3x - 20

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(iv) (y² + ) (y² - )

: (x + a) (x + b) = x² + (a+b)x + ab

Here,

x = y² , a =  , b = (-)

=> (y²)² + {  + (-  }*y² + *(- )

=>  + ( -  )*y² + (- )

=>  + 0*y² -  

=>  -  

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(v) (3 - 2x) (3 + 2x)

: (x + a) (x + b) = x² + (a+b)x + ab

Here,

x = 3 , a = (-2x) , b = 2x

=> (3)² + { (-2x) + 2x }*3 + (-2x)*2x

=> 9 + 0*3 + (-4x²)

=> 9 - 4x²

=> - 4x² + 9