Which is the cubic polynomial in the standard form with roots 3, -6, and 0?

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Answer 1 · 2016-12-04T15:06:25

$x^3+3x^2-18x=0$

roots are:

$x=0;" "x=3;" "x=-6$

hence the corresponding linear factors are:

$x;" "(x-3);" "(x+6)$

the cubic is them formed by their products

$x(x-3)(x+6)=0$

multiply out.

$x(x^2+3x-18)=0$

$x^3+3x^2-18x=0$

Answer 2 · 2016-12-04T15:07:30

$x^3+3x^2-18x$

The simplest polynomial with zeros $3$ , $-6$ and $0$ is:

$f(x) = (x-3)(x+6)x$

$color(white)(f(x)) = x^3+3x^2-18x$

Any polynomial in $x$ with these zeros will be a multiple (scalar or polynomial) of this $f(x)$ .