If c is a zero of f(x)=5x^4 - 4x^3 +3x^2 - 4x + 5, then show that (1/c) is also a zero?

Answer 1

f(c) = 0; f(c) = 5c^4 - 4c^3 +3c^2 - 4c + 5
so
5c^4 - 4c^3 +3c^2 - 4c + 5 = 0
If f(1/c) = 0 then
5(1/c)^4 - 4(1/c)^3 +3(1/c)^2 - 4(1/c) + 5 = 0
Then if we Multiply everything by c^4 and rearrange then we get:
5c^4 - 4c^3 +3c^2 - 4c + 5 = 0

Since they both are the same equation, then every zero will have it's inverse a zero too.

Answer 2

f(x) = 5x^4 - 4x^3 + 3x^2 - 4x + 5

f(c) = 5c^4 - 4c^3 + 3c^2 - 4c + 5

since c is zero of f(x)

5c^4 - 4c^3 + 3c^2 - 4c + 5 = 0

f(1/c) = 5/c^4 - 4/c^3 + 3/c^2 - 4/c + 5

multiply RHS with c^4/c^4

f(1/c) = (1/c^4)[5 - 4c + 3c^2 - 4c^3 + 5c^4]

but 5 - 4c + 3c^2 - 4c^3 + 5c^4 = 0 from eqn(1)

f(1/c) = (1/c^4)(0) = 0

so f(1/c) = 0