How do you use the rational roots theorem to find all possible zeros of $P (x) = x^{3} + 3 x^{2} - 4$ $P(x)= x^3 + 3x^2 - 4$ ?

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How do you use the rational roots theorem to find all possible zeros of $P (x) = x^{3} + 3 x^{2} - 4$ $P(x)= x^3 + 3x^2 - 4$ ?

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Answer 1 · 2016-08-09T18:53:31

$P (x)$ $P(x)$ has zeros $1, - 2, - 2$ $1, -2, -2$

Explanation:

$P (x) = x^{3} + 3 x^{2} - 4$ $P(x) = x^3+3x^2-4$

By the rational root theorem, any rational zeros of $P (x)$ $P(x)$ 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 $- 4$ $-4$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1, \pm 2, \pm 4$ $+-1, +-2, +-4$

We find:

$P (1) = 1 + 3 - 4 = 0$ $P(1) = 1+3-4 = 0$

So $x = 1$ $x=1$ is a zero and $(x - 1)$ $(x-1)$ a factor:

$x^{3} + 3 x^{2} - 4 = (x - 1) (x^{2} + 4 x + 4)$ $x^3+3x^2-4=(x-1)(x^2+4x+4)$

The remaining quadratic is a perfect square trinomial:

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

Hence the remaining zero is $- 2$ $-2$ with multiplicity $2$ $2$ .

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How do you use the rational roots theorem to find all possible zeros of $P (x) = x^{3} + 3 x^{2} - 4$ $P(x)= x^3 + 3x^2 - 4$ ?

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Explanation:

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How do you use the rational roots theorem to find all possible zeros of P(x)=x3+3x2−4P(x)= x^3 + 3x^2 - 4?

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How do you use the rational roots theorem to find all possible zeros of $P (x) = x^{3} + 3 x^{2} - 4$ $P(x)= x^3 + 3x^2 - 4$ ?