Question

If eqn $x^4-x^3+3x^2-7x+8=0$ has nots $\alpha, \beta, \gamma, \delta$ then find value of $(2-\alpha).(2-\beta).(2-\gamma).(2-\delta)$

Answer 1

14

Answer 2

Let the given polynomial be $P(x) = x^4-x^3+3x^2-7x+8$. Since $\alpha, \beta, \gamma, \delta$ are the roots of the equation $P(x)=0$, we can write the polynomial in factored form as: $P(x) = (x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$ The expression $(2-\alpha).(2-\beta).(2-\gamma).(2-\delta)$ is equivalent to $P(2)$. Substituting $x=2$ into $P(x)$: $P(2) = (2)^4 - (2)^3 + 3(2)^2 - 7(2) + 8$ $P(2) = 16 - 8 + 3(4) - 14 + 8$ $P(2) = 16 - 8 + 12 - 14 + 8$ $P(2) = 14$