How do you find all the zeros of $x^{3} - 4 x^{2} - 44 x + 96$ $x^3-4x^2-44x+96$ ?

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How do you find all the zeros of $x^{3} - 4 x^{2} - 44 x + 96$ $x^3-4x^2-44x+96$ ?

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Answer 1 · 2016-08-14T13:32:48

This cubic has zeros: $2$ $2$ , $8$ $8$ and $- 6$ $-6$

Explanation:

$f (x) = x^{3} - 4 x^{2} - 44 x + 96$ $f(x) = x^3-4x^2-44x+96$

By the rational roots theorem, any rational zeros of $f (x)$ $f(x)$ are expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $96$ $96$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 32, \pm 48, \pm 96$ $+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-16, +-24, +-32, +-48, +-96$

Trying each in turn, we find:

$f (2) = 8 - 16 - 88 + 96 = 0$ $f(2) = 8-16-88+96 = 0$

So $x = 2$ $x=2$ is a zero and $(x - 2)$ $(x-2)$ a factor:

$x^{3} - 4 x^{2} - 44 x + 96 = (x - 2) (x^{2} - 2 x - 48)$ $x^3-4x^2-44x+96 = (x-2)(x^2-2x-48)$

To factor the remaining quadratic find a pair of factors of $48$ $48$ which differ by $2$ $2$ . The pair $8, 6$ $8, 6$ works. Hence:

$x^{2} - 2 x - 48 = (x - 8) (x + 6)$ $x^2-2x-48 = (x-8)(x+6)$

Hence zeros: $8$ $8$ and $- 6$ $-6$

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How do you find all the zeros of $x^{3} - 4 x^{2} - 44 x + 96$ $x^3-4x^2-44x+96$ ?

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How do you find all the zeros of x3−4x2−44x+96x^3-4x^2-44x+96?

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How do you find all the zeros of $x^{3} - 4 x^{2} - 44 x + 96$ $x^3-4x^2-44x+96$ ?