How do you solve $x^3 -3x^2 +16x -48 = 0$ ?

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Answer 1 · 2017-02-16T11:55:45

$3 and +-i4$ . See the Socratic graph of the cubic, making x-intercept 3..

From sign changes in the coefficients, the equation has utmost 3

positive roots,There are no changes in sign,

when x is changed to $-x$ . And so, there are no negative roots.

The cubic is 0 at x = 0. So, it becomes

$(x-3)(x^2+16)$

The other solutions are from by $x^2+16=0$ , giving $x =+-i4$

graph{(x-3)(x^2+16) [2, 4, -500, 500]} 0, -500, 500]}

Not to scale. It is large y vs small x, for approximating the solution.

How do you solve x^3 -3x^2 +16x -48 = 0x^3 -3x^2 +16x -48 = 0?