Question

Hiii there!!!

Plz solve the 48th question....

Plz answer sincerely!!

Answer 1

Hey Mate !

Here is your solution :

P( x ) = ax^4 + 2x^3 - 3x^2 + bx - 4

And,

g( x ) = x² - 4

= ( x² ) - ( 2 )²
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Using identity : |
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[ a² - b² = ( a + b ) ( a - b ) ] |
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= ( x + 2 ) ( x - 2 )

Hence, zeroes of g( x ) = 2 and -2.

Now,

If g( x ) be a factor of p( x ) , then remainder will be 0.

By Factor Theorem ,

When , x = 2

=> P( x ) = ax^4 + 2x^3 - 3x^2 + bx - 4

Substituting x = 2 ,

=> P( 2 ) = a( 2 )^4 + 2( 2 )^3 - 3( 2 )^2 + b( 2 ) - 4 = 0


=> a × 16 + 2 × 8 - 3 × 4 + 2b - 4 = 0

=> 16a + 16 - 12 + 2b - 4 = 0

=> 16a + 2b + 16 - 16 = 0

=> 16a + 2b + 0 = 0

Taking out 2 as common,

=> 2 ( 8a + b ) = 0

=> ( 8a + b ) = 0 ÷ 2

=> 8a + b = 0 ----------- ( 2 )

By Factor Theorem,

When , x = ( -2 ).

=> P( x ) = ax^4 + 2x^3 - 3x^2 + bx - 4

Substitute , x = -2

=> P( -2 ) = a( -2 )^4 + 2( -2 )^3 - 3( 2 )^2 + b( -2 ) - 4 = 0


=> a × 16 + 2 × ( -8 ) - 3 × 4 - 2b - 4 = 0

=> 16a - 16 - 12 - 2b - 4 = 0

=> 16a - 2b - 32 = 0

Taking out 2 as common ,

=> 2( 8a - b - 16 ) = 0

=> ( 8a - b - 16 ) = 0 ÷ 2

=> ( 8a - b - 16 ) = 0 --------- ( 2 )

Adding ( 1 ) and ( 2 ),

=> 8a + b + 8a - b - 16 = 0

=> 16a - 16 = 0

=> 16a = 16

=> a = 16 ÷ 16

•°• a = 1

Substituting the value of a in ( 1 ),

=> 8a + b = 0

=> 8( 1 ) + b = 0

=> 8 + b = 0

•°• b = -8


Hence, a = 1 and b = -8.

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Hope it helps !! ^_^