Question

If one zero of the cubic polynomial f(x)= 3x^3 - 5x^2- 11x -3 is -1/3 then find the other zeroes

Answer 1

Given that



f(x) = 3x³ - 5x² - 11x - 3

One of its zero = $\frac{-1}{3}$

To find

All the remaining zeroes = ?

Assumption

Let's assume that

The three zeroes of f(x) are :

$\alpha, \beta\:and\: \gamma$

As we know

There exists a relationship between the co-efficients and the zeroes of a polynomial.

As f(x) is a cubic polynomial (as the degree is 3) there will be maximum 3 zeroes and not more than that.

$\boxed{Note\::}$

In some rare cases, a cubic polynomial can also have only 1 zero or it can also have 2 zeroes.

The following are the relationship between the co-efficients and the zeroes of a cubic polynomial :

(i) $\alpha + \beta + \gamma = \frac{-b}{a}$

(ii) $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$

(iii) $\alpha \beta \gamma = \frac{-d} {a}$

Now, we'll find the zeroes of f(x).

If one zero is $\frac{-1}{3}$

$\therefore x = \frac{-1}{3}$

$(x + \frac{-1}{3})$

3x + 1

Divide f(x) by 3x + 1 to obtain other zeroes.

q(x) = x² + 2x - 3

r(x) = 0

Now, we've to factorise q(x) to obtain our zeroes.

x² + 2x - 3 = 0

The factors of - 3 which gives us the sum as 2 are

$\pm 1\:and\: \pm 3$

x² + 2x - 3 = 0

x² + 3x - x - 3 = 0

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

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

(x + 3) = 0

x + 3 = 0

$\boxed{\bold{x = - 3}}$

x - 1 = 0

$\boxed{\bold{x = 1}}$

Therefore, the zeroes of the f(x) are 1, - 3 and $\frac{-1}{3}$