Question

find the zeros of the cubic polynomial 3 x cube minus 5 x square - 11 x minus 3​

Answer 1

GIVEN :

Find the zeroes of the cubic polynomial $3 x^3-5 x^2 - 11 x -3$

TO FIND :

The zeroes of the cubic polynomial $3 x^3-5 x^2 - 11 x -3$

SOLUTION :

Given that the cubic polynomial is $3 x^3-5 x^2 - 11 x -3$

Since the given polynomial is cubic hence it must have three zeroes.

By using the Synthetic Division method we can find the zeroes

-1 |  3     -5      -11      -3

     0      -3       8       3

  _________________

    3       -8       -3       0

⇒  x+1 is a factor of the given cubic polynomial $3 x^3-5 x^2 - 11 x -3$

x+1=0

∴ x=-1 is a zero of the cubic polynomial $3 x^3-5 x^2 - 11 x -3$

Now we have the quadratic equation

$3x^2-8x-3=0$

$3x^2-9x+x-3=0$

3x(x-3)+1(x-3)=0

(3x+1)(x-3)=0

3x+1=0 or x-3=0

$x=-\frac{1}{3}$ and x=3 are the other zeroes of the cubic polynomial $3 x^3-5 x^2 - 11 x -3$

∴ x=-1, $x=-\frac{1}{3}$ and x=3 are the three zeroes of the cubic polynomial $3 x^3-5 x^2 - 11 x -3$

Answer 2

Step-by-step explanation:

Given that:

find the zeros of the cubic polynomial 3 x cube minus 5 x square - 11 x minus 3

To find: Zeroes of cubic polynomial

Solution:

Given cubic polynomial is

$3 {x}^{3} - 5 {x}^{2} - 11x - 3$

Its one zero can be easily find using hit and trial method.

Put x= -1

$= 3( { - 1)}^{3} - 5( { - 1)}^{2} - 11( - 1) - 3 \\ \\ = - 3 - 5 + 11 - 3 \\ \\ = - 11 + 11 \\ \\ = 0$

On putting x=-1 cubic polynomial becomes zero,so

x= -1 is a zero of this polynomial.

Thus,

(x+1) is a factor.

We know that factor divides polynomial completely,So

$x + 1) \: 3 {x}^{3} - 5 {x}^{2} - 11x - 3 \: (3 {x}^{2} - 8x - 3 \\ \: \: \: \: \: \: \: \: \: \: \: \: 3 {x}^{3} + 3 {x}^{2} \\ \: \: \: \: \: \: \: \: ( - ) \: \: \: \: \: ( - ) \\ \: \: \: \: \: \: - - - - - - \\ \: \: \: \: \: \: \: \: \: \: - 8 {x}^{2} - 11x \\ \: \: \: \: \: \: \: \: \: \: - 8 {x}^{2} - 8x \\ \: \: \: \: \: \: \: \: \: \: ( + ) \: \: \: \: \: ( + ) \\ \: \: \: \: \: \: \: \: \: \: - - - - - - \\ \: \: \: \: \: \: \: \: \: \: - 3x - 3 \\ \: \: \: \: \: \: \: \: \: \: - 3x - 3 \\ \: \: \: \: \: \: \: \: \: \: ( + ) \: \: \: ( + ) \\ \: \: \: \: \: \: \: \: \: \: - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \\ \: \: \: \: \: \: \: \: \: \: - - - - - -$

Thus,factorize the quotient polynomial to find other two zeros of cubic polynomial

$\bold{3 {x}^{2} - 8x - 3 = 0} \\ \\ 3 {x}^{2} - 9x + x - 3 = 0 \\ \\ 3x(x - 3) + 1(x - 3) = 0 \\ \\ (3x + 1)(x - 3) = 0 \\ \\ 3x + 1 = 0 \\ \\ x = \frac{ - 1}{3} \\ \\ or \\ \\ (x - 3) = 0 \\ \\ x = 3$

Thus,

All the three Zeroes are

x= -1, 3, -1/3

Hope it helps you.