Question

B) Verify that 1.-1.-3 are the zeroes of the cubic polynomial x3+3x2

-x-3 and check

the relation between the zeroes and coefficients​

Answer 1

Solution :

It is given that :

• P(x) = x³ + 3x² - x - 3

P( 1 ) = ( 1 )³ + 3( 1 )² - ( 1 ) - 3

        = 1 + 3 - 1 - 3

        = 4 - 4

        = 0

Hence, 1 is the zero of the given polynomial.

Now,

P( - 1 ) = ( - 1 )³ + 3(- 1 )² - ( - 1 ) - 3

         = - 1 + 3 + 1 - 3

         = 2 - 2

 

        = 0

Hence, - 1 is the zero of the given polynomial.

Also,

P( - 3 ) = ( - 3 )³ + 3 (- 3 )² - ( - 3 ) - 3

          = - 27 + 3 × 9 + 3 - 3

          = - 27 + 27 + 0

         = 0

Hence,  - 3 is the zero of the given polynomial.

Now,

• α = 1
• β = -1
• λ = - 3

→ α + β +  λ = - coefficient of x² / coefficient of x³

Substitute the values.

We get,

1 + ( - 1 ) + ( - 3 ) =  - 3 / 1

⇒ - 3 = - 3

∴ L.H.S = R.H.S

→ αβ + βλ + λα = coefficient of x/ coefficient of x³

Substitute the given values.

We get,

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

⇒ - 1 + 3 - 3 = -1/1

⇒ - 1 + 0 = -1

⇒ - 1 = - 1

∴ L.H.S = R.H.S

→ αβλ =  Constant term/ coefficient of x³

Substitute the given values.

We get,

1 × ( - 1 ) × ( - 3 ) =  3/1

⇒ 3 = 3

∴ L.H.S = R.H.S

Hence proved!

$\rule{200}2$

Answer 2

It is given that :

P(x) = x³ + 3x² - x - 3

P( 1 ) = ( 1 )³ + 3( 1 )² - ( 1 ) - 3

= 1 + 3 - 1 - 3

= 4 - 4

= 0

Hence, 1 is the zero of the given polynomial.