# Given the function f(x)=x^3-5x^2-12x+36 and given that f(6)=0, what are the roots of the function?

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#### Explanation:

Hence $f \left(6\right) = 0$ that means that $x - 6$ is a factor of the polynomial

${x}^{3} - 5 {x}^{2} - 12 x + 36$ hence we can phrase it as

${x}^{3} - 5 {x}^{2} - 12 x + 36 = \left(x - 6\right) \cdot \left({x}^{2} + b x + c\right) + d$

Doing same basic algebraic calculation we find that

${x}^{3} - 5 {x}^{2} - 12 x + 36 = \left(x - 6\right) \cdot \left({x}^{2} + x - 6\right) + 0 = \left(x - 6\right) \left({x}^{2} + x - 6\right)$

but ${x}^{2} + x - 6 = \left(x - 2\right) \cdot \left(x + 3\right)$ hence we have that

${x}^{3} - 5 {x}^{2} - 12 x + 36 = \left(x - 6\right) \left(x - 2\right) \left(x + 3\right)$

The roots are $6 , 2 , - 3$