Question

If sum and product of two zeroes of the polynomial x 3 + x 2 – 3x – 3 are 0 and 3 respectively, find all zeroes of the polynomial.

Answer 1

Answer:

The zeros of this polynomial are $\bold{1,-\sqrt{3},+\sqrt{3}}$

Step-by-step explanation:

$p(x)=x^{3}+x^{2}-3 x-3$

Given:

$\begin{array}{l}{\alpha+\beta=0, \Rightarrow \alpha=-\beta \ldots \ldots(i)} \\ {\alpha \beta=3}\end{array}$

Where,

$\alpha, \beta=r o o t s$

We know that,

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

Using (i)

$\begin{array}{l}{0+\gamma=-1} \\ {\gamma=-1, \therefore \text { first }_{-} \text {zero} \ldots(i i)}\end{array}$

Again,

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

From (i) & (ii)

$\begin{array}{l}{(-\beta)(\beta)(-1)=3} \\ {-\beta^{2} \cdot(-1)=3} \\ {\beta^{2}=3} \\ {\beta=\pm \sqrt{3}}\end{array}$

Case (i)

$\begin{aligned} \beta &=+\sqrt{3} \\ \alpha &=-\sqrt{3}(\text {from}(i i)) \end{aligned}$

Case (ii)

$\begin{array}{l}{\beta=-\sqrt{3}} \\ {\alpha=+\sqrt{3}} \\ {\therefore \text { zeros }=1,-\sqrt{3},+\sqrt{3}}\end{array}$

The zeros of a polynomial are the values of the variable in that variable for which the polynomial becomes equal to zero. It is that value of x that makes the polynomial equal to 0. In the above equation for x = $1,-\sqrt{3},+\sqrt{3}$  we will get p(x) = zero. Therefore, the zeros of this polynomial are $\bold{1,-\sqrt{3},+\sqrt{3}}$  . This is a classic way to calculate the zeros of a polynomial.

Answer 2

Answer:

Step-by-step explanation:

x^2 - 2x - x + 2

x(x - 2) - 1(x-2) = 0

(x-1)(x-2) = 0

x= 1 , 2

sum of zeroes = a +b = -b/a

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

product of zeroes = ab = c/a

= 1 x 2 = 2 = 2/1 = 2