Maths question help

Show that the quadratic factor never equals 0 by considering it's discrimant. Compute the discriminant, what does the value of the discriminant tell you about the number of real and/orimaginary roots.

okay f(x)=x^3+4x-5
is f(x)=0
then x^3 + 4x - 5 = 0
which means
(x-1)(ur answer for ii)=0
if you do the discriminant of the quadratic (answer for ii) you should find that b^2-4ac < 0
and this means the quadratic (answer for ii) has no real roots/solutions
so the only root is x=1

I have completed question I) and Ii) but need help with Iiii)

To show that this has one real root b^2-4ac=0
However in this it is a number greater than zero

okay f(x)=x^3+4x-5
is f(x)=0
then x^3 + 4x - 5 = 0
which means
(x-1)(ur answer for ii)=0
if you do the discriminant of the quadratic (answer for ii) you should find that b^2-4ac < 0
and this means the quadratic (answer for ii) has no real roots/solutions
so the only root is x=1

Unfortunately not. f(x) already has one real root x - 1 = 0, you need to show it has no other real roots. To do this look at the discriminant of the quadratic factor. If it's coming out as greater than 0, either you're computing it wrong or your answer to (ii) is wrong.

Putting x^2-x+5 (answer to ii) intob^2-4ac gives -19<0 so no real roots therefore the only real root of fx is x-1

Yea I managed to work this out from my answer. Zero is greater than my answer which shows it has no real roots but why does the question asks to show that it has exactly one real root if it has none. Does this make sence?

The only real root is the solution to x-1=0, which is...

(A root to an equation is a number!)

Should be 1