Question

please send answer urgent please​

Answer 1

Topic: Polynomials(Vieta's formula)

Solution:-

Let the three known zero be alpha$(\alpha)$, beta$(\beta)$, gamma$(\gamma)$.

Let the remaining fourth zero be delta$(\delta)$.

Let the given polynomial be $f(x)$. Then it has α, β, γ, δ as zeroes.

By factor theorem,

$f(x)=(x-\alpha )(x-\beta )(x-\delta )(x-\gamma )$

Since the given zeroes are 0, 1, -1, the polynomial would be

$f(x)=x(x-1)(x+1)(x-\gamma )$

Let's partially simplify the term of degree 3.

$\implies -(0-1+1+\gamma)=2$

$\implies \gamma =-2$

By comparison of the coefficient we obtain $\gamma =-2$.

Hence the remaining zero is -2.

Answer:-

The required answer is 0, 1, -1, and -2.