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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Answer 1 · 2015-09-28T23:18:01

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

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

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

Doing same basic algebraic calculation we find that

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

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

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

The roots are $6, 2, - 3$ $6,2,-3$