How do I find all the rational zeros of $p (x) = x^{3} - 12 x - 16$ $p(x)=x^3-12x-16$ ?

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Answer 1 · 2018-06-01T00:05:21

Check if any of the divisors of $- 16$ $-16$ is a root of $p (x)$ $p(x)$ in our case

$x = 4$ $x=4$ is a root hence $p (4) = 0$ $p(4)=0$

The polynomial can be written as

$p (x) = (x - 4) \cdot (x^{2} + b x + c)$ $p(x)=(x-4)*(x^2+bx+c)$

It is easy to proove that $b = 4$ $b=4$ and $c = 4$ $c=4$

Hence

$p (x) = (x - 4) {(x + 2)}^{2}$ $p(x)=(x-4)(x+2)^2$

So the zeros are $x = 4$ $x=4$ , $x = - 2$ $x=-2$ (double root)

How do I find all the rational zeros of p(x)=x^3-12x-16p(x)=x3−12x−16p(x)=x^3-12x-16?