The cubic polynomial p(x)=x³+x²–x–1 has two distinct zeros. State whether the given statement is true or false.
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Answered by
2
A polynomial of degree 1, 2 and 3 rae called linear, quadratic and cubic polynomial respectivey.
Now consider the polynomial given in the question p(x) = x³+x²–x–1.
x³+x²–x–1
or, x²(x+1)–1(x+1)
or, (x²-1)(x+1)
or, (x-1)(x+1)(x+1)
Therefore, it has two distinct zeroes: (-1. +1)
Thus the given statement is true.
Answered by
8
Hi ,
p( x ) = x³ + x² - x - 1
i ) If x = 1
p( 1 ) = 1 + 1 - 1 - 1
p( 1 ) = 0
Therefore ,
( x - 1 ) is a factor of p( x )
ii ) if x = -1 ,
p( - 1 ) = ( - 1 )³ + ( - 1 )² - ( - 1 ) - 1
p( - 1 ) = -1 + 1 + 1 - 1
p( - 1 ) = 0
Therefore ,
( x + 1 ) is a factor of p( x ).
( x - 1 ) ( x + 1 ) = x² - 1 is a factor of p( x )
x² - 1 ) x³ + x² - x - 1 ( x + 1
********x³ + 0 - x
______________
************x² - 1
************x² - 1
______________
****Remainder ( 0 )
p( x ) = ( x + 1 )( x - 1 )( x + 1 )
zeroes of p( x ) are -1 , 1 , -1
zeroes of p( x ) = { -1 , 1 }
It is true p( x ) has two distinct zeroes .
I hope this helps you.
: )
p( x ) = x³ + x² - x - 1
i ) If x = 1
p( 1 ) = 1 + 1 - 1 - 1
p( 1 ) = 0
Therefore ,
( x - 1 ) is a factor of p( x )
ii ) if x = -1 ,
p( - 1 ) = ( - 1 )³ + ( - 1 )² - ( - 1 ) - 1
p( - 1 ) = -1 + 1 + 1 - 1
p( - 1 ) = 0
Therefore ,
( x + 1 ) is a factor of p( x ).
( x - 1 ) ( x + 1 ) = x² - 1 is a factor of p( x )
x² - 1 ) x³ + x² - x - 1 ( x + 1
********x³ + 0 - x
______________
************x² - 1
************x² - 1
______________
****Remainder ( 0 )
p( x ) = ( x + 1 )( x - 1 )( x + 1 )
zeroes of p( x ) are -1 , 1 , -1
zeroes of p( x ) = { -1 , 1 }
It is true p( x ) has two distinct zeroes .
I hope this helps you.
: )
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