# How do you find all zeros of the function f(x) = x^3 - 4x^2-9x+36?

1st Method

Check if the divisors of the constant term $36$ are roots of the equation.In our case $4 , - 3 , 3$ are roots of the equation hence

$f \left(x\right) = {x}^{3} - 4 {x}^{2} - 9 x + 36 = \left(x - 4\right) \left(x - 3\right) \left(x + 3\right)$

2nd Method

We can easily factorise this as follows

${x}^{3} - 4 {x}^{2} - 9 x + 36 = x \left({x}^{2} - 9\right) - 4 \left({x}^{2} - 9\right) = \left({x}^{2} - 9\right) \left(x - 4\right) = \left(x - 3\right) \left(x + 3\right) \left(x - 4\right)$