Question

The cubic polynomial p(x)=x³+x²–x–1 has two distinct zeros. State whether the given statement is true or false.

Answer 1

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.

Answer 2

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.

: )