How do you find all the zeros of $P (x) = x^{3} + 5 x^{2} - 2 x - 24$ $P(x) = x^3 + 5x^2 - 2x - 24$ where one zero is 2?

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Answer 1 · 2016-08-07T10:50:41

$P (x)$ $P(x)$ has zeros $2$ $2$ , $- 3$ $-3$ and $- 4$ $-4$ .

$P (x) = x^{3} + 5 x^{2} - 2 x - 24$ $P(x) = x^3+5x^2-2x-24$

SInce we are told that $x = 2$ $x=2$ is one of the zeros, $(x - 2)$ $(x-2)$ must be a factor:

$x^{3} + 5 x^{2} - 2 x - 24$ $x^3+5x^2-2x-24$

$= (x - 2) (x^{2} + 7 x + 12)$ $=(x-2)(x^2+7x+12)$

The remaining quadratic can be factored by noticing that $3 + 4 = 7$ $3+4=7$ and $3 \times 4 = 12$ $3xx4=12$ . So:

$x^{2} + 7 x + 12 = (x + 3) (x + 4)$ $x^2+7x+12 = (x+3)(x+4)$

Hence the other zeros are $x = - 3$ $x=-3$ and $x = - 4$ $x=-4$

How do you find all the zeros of P(x)=x3+5x2−2x−24P(x) = x^3 + 5x^2 - 2x - 24 where one zero is 2?