What are the real zeroes of x^3+6x^2-9x-54

The answer is: 3,-3,-6

We need to find the values where the function tends to 0, so we need to make the equation, equal to 0:

Then, we have three factors multiplying each other, so we have to five three numbers which make at least one of those factors 0

So,

Trying with 3:

Trying with -3:

Trying with -6:

Have a nice day!

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

the correct answer is d

hope this