Question

Verify that 1,-1 and +3 are the zeroes of the cubic polynomial x^3-3x^2-x+3check the relationship between zeroes and the coefficients.​

Answer 1

The polynomial p ( x ) is x³ - 3x² - x + 3.

• a = 1

• b = -3

• c = -1

• d = 3

Value of the polynomial x³ - 3x² - x + 3 When x = 1

p ( 1 ) = ( 1 )³ - 3 ( 1 )² - 1 + 3

= 1 - 3 - 1 + 3

= 0

So,1 is a zero of p ( x ).

On putting x = -1 in the cubic polynomial x³ - 3x² - x + 3.

p ( - 1 ) = ( -1 )³ - 3 × ( -1 )² - ( -1 ) + 3

= -1 - 3 × 1 + 1 + 3

= -1 - 3 + 1 + 3

= 0

So, -1 is a zero of p ( x ).

On putting x = 3 in the cubic polynomial x³ - 3x² - x + 3.

p ( 3 ) = ( 3 )³ - 3 × ( 3 )² - 3 + 3

= 27 - 3 × 9 - 3 + 3

= 27 - 27 - 3 + 3

= 0

So, 3 is a zero of p ( x ).

Hence, 1 , - 1 and 3 are the zeroes of the given polynomial.

Now,

★Sum of zeroes of p ( x ) :

α + β + γ = -b/a

⇒ 1 + ( - 1 ) + 3 = - ( -3 ) / 1

⇒ 3 = 3

★ Sum of zeroes in product of two by two of p ( x )

αβ + βγ + γα = c/a

⇒ 1 × ( - 1 ) + ( - 1 ) × 3 + 3 × 1 = -1 / 1

⇒ -1 - 3 + 3 = -1

⇒ -1 = -1

★Product of zeroes of p ( x ).

αβγ = -d/a

⇒ 1 × ( - 1 ) × 3 = -3 / 1

⇒ - 3 = - 3

Verified.