Solve this please... with explanation.
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hiii8
hiii
divide by 2 nd solve by hit nd trial method
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Hi ,
It is given that √3 and - √3 are
zeroes of p ( x ) = 2x³ + x² -6x- 3 .
( x - √3 ) and ( x + √3 ) are the factors
of p( x ).
Therefore ,
( x - √3 )( x + √3 ) = x² - 3 is a factor of
p( x )
x² - 3 ) 2x³ + x² - 6x - 3 ( 2x + 1
********2x³ + 0 - 6x
_________________
************x² -3
************x² - 3
_____________
************( 0 )
p( x ) = ( x² - 3 ) ( 2x + 1 )
= ( x - √3 ) ( x + √3 ) ( 2x + 1 )
Third zero of p ( x ) is
2x + 1 = 0
2x = -1
x = -1/2
Therefore ,
√3 , -√3 , ( -1/2 ) are three zeroes of
p( x ).
I hope this helps you.
:)
It is given that √3 and - √3 are
zeroes of p ( x ) = 2x³ + x² -6x- 3 .
( x - √3 ) and ( x + √3 ) are the factors
of p( x ).
Therefore ,
( x - √3 )( x + √3 ) = x² - 3 is a factor of
p( x )
x² - 3 ) 2x³ + x² - 6x - 3 ( 2x + 1
********2x³ + 0 - 6x
_________________
************x² -3
************x² - 3
_____________
************( 0 )
p( x ) = ( x² - 3 ) ( 2x + 1 )
= ( x - √3 ) ( x + √3 ) ( 2x + 1 )
Third zero of p ( x ) is
2x + 1 = 0
2x = -1
x = -1/2
Therefore ,
√3 , -√3 , ( -1/2 ) are three zeroes of
p( x ).
I hope this helps you.
:)
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