The 1st equation, is the only one. PROOF:
f(x) = x⁴ - 2x² + 36x -80
1) f(2) = 2⁴ - 2(2)² + 36(2) - 80 = 0, hence 2 is a zero value (or a root)
2) f(-4) = (-4)⁴ - 2(-4)² + 36(-4) - 80 = 0, hence - 4 is another root.
Since x= 2 and x= -4, f(x) is then divisible by (x-2)(x+4) or if you proceed to this multiplication you will find:
(x-2)(x+4) = x² + 2x -8, that means f(x) is divisible by (x² + 2x -8)
IDivide (x⁴ - 2x² + 36x -80) / (x² + 2x -8) you will find:
(x⁴ - 2x² + 36x -80) / (x² + 2x -8) = x² -2x +10. Now let's find the zeros of
x² -2x +10; x' = [-b+√(b²-4ac)]/2a and x" = [-b-√(b²-4ac)]/2a, after plugging in the corresponding values you will find that:
x' = 1+3i and x" = 1 - 3i ==> x² -2x +10 = (x-1+3i)(x-1-3i)
and at last f(x) = x⁴ - 2x² + 36x -80 = (x-2)(x+4)(x-1+3i)(x-1-3i), with following zeros: -2 , -4 , 1-3i and 1+3i
