How do you use the rational roots theorem to find all possible zeros of $P (x) = 12 x^{4} + x^{3} + 4 x^{2} + 7 x + 8$ $P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8$ ?

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How do you use the rational roots theorem to find all possible zeros of $P (x) = 12 x^{4} + x^{3} + 4 x^{2} + 7 x + 8$ $P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8$ ?

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Answer 1 · 2016-06-07T06:22:20

Use the rational roots theorem to find possible candidate rational zeros, any thereby find that it has none.

Explanation:

$P (x) = 12 x^{4} + x^{3} + 4 x^{2} + 7 x + 8$ $P(x) = 12x^4+x^3+4x^2+7x+8$

By the rational roots 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 $8$ $8$ and $q$ $q$ a divisor of the coefficient $12$ $12$ of the leading term.

So in our example, it means that the only possible rational zeros are:

$\pm \frac{1}{12}$ $+-1/12$ , $\pm \frac{1}{6}$ $+-1/6$ , $\pm \frac{1}{4}$ $+-1/4$ , $\pm \frac{1}{3}$ $+-1/3$ , $\pm \frac{1}{2}$ $+-1/2$ , $\pm \frac{2}{3}$ $+-2/3$ , $\pm 1$ $+-1$ , $\pm \frac{4}{3}$ $+-4/3$ , $\pm 2$ $+-2$ , $\pm \frac{8}{3}$ $+-8/3$ , $\pm 4$ $+-4$ , $\pm 8$ $+-8$

That's rather a lot of possible zeros to try, but you can narrow it down a little by noting that the coefficients of $P (x)$ $P(x)$ are all positive. Hence it has no positive zeros, leaving the possible rational zeros:

$- \frac{1}{12}$ $-1/12$ , $- \frac{1}{6}$ $-1/6$ , $- \frac{1}{4}$ $-1/4$ , $- \frac{1}{3}$ $-1/3$ , $- \frac{1}{2}$ $-1/2$ , $- \frac{2}{3}$ $-2/3$ , $- 1$ $-1$ , $- \frac{4}{3}$ $-4/3$ , $- 2$ $-2$ , $- \frac{8}{3}$ $-8/3$ , $- 4$ $-4$ , $- 8$ $-8$

Substituting each of these for $x$ $x$ in $P (x)$ $P(x)$ , we find that none is a zero. So $P (x)$ $P(x)$ has no rational zeros.

That's all the rational roots theorem tells us.

In fact, this particular quartic only has non-Real Complex zeros.

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How do you use the rational roots theorem to find all possible zeros of $P (x) = 12 x^{4} + x^{3} + 4 x^{2} + 7 x + 8$ $P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8$ ?

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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)=12x4+x3+4x2+7x+8P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8 ?

1 Answer

Explanation:

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How do you use the rational roots theorem to find all possible zeros of $P (x) = 12 x^{4} + x^{3} + 4 x^{2} + 7 x + 8$ $P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8$ ?