All Rational zerai $$ 2 x^{5}+8 x^{3}+7 x^{2}-6 $$

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To find the rational zeros of the polynomial $$2x^5 + 8x^3 + 7x^2 - 6$$, we can use the Rational Root Theorem. This theorem states that any rational zero of the polynomial can be expressed in the form $$\frac{p}{q}$$, where $$p$$ is a factor of the constant term and $$q$$ is a factor of the leading coefficient.

In this case, the constant term is $$-6$$ and the leading coefficient is $$2$$. The factors of $$-6$$ are $$\pm1, \pm2, \pm3, \pm6$$, and the factors of $$2$$ are $$\pm1, \pm2$$. Therefore, the possible rational zeros are:

$$
\pm\frac{1}{1}, \pm\frac{1}{2}, \pm\frac{2}{1}, \pm\frac{2}{2}, \pm\frac{3}{1}, \pm\frac{3}{2}, \pm\frac{6}{1}, \pm\frac{6}{2}
$$

Simplifying these fractions, we get:

$$
\pm1, \pm\frac{1}{2}, \pm2, \pm3, \pm\frac{3}{2}, \pm6
$$

Now, we can test each of these possible zeros by substituting them into the polynomial and seeing if the result is zero. Let's start with the simplest ones:

For $$x = 1$$:

$$
2(1)^5 + 8(1)^3 + 7(1)^2 - 6 = 2 + 8 + 7 - 6 = 11 \neq 0
$$

For $$x = -1$$:

$$
2(-1)^5 + 8(-1)^3 + 7(-1)^2 - 6 = -2 - 8 + 7 - 6 = -9 \neq 0
$$

For $$x = 2$$:

$$
2(2)^5 + 8(2)^3 + 7(2)^2 - 6 = 2(32) + 8(8) + 7(4) - 6 = 64 + 64 + 28 - 6 = 150 \neq 0
$$

For $$x = -2$$:

$$
2(-2)^5 + 8(-2)^3 + 7(-2)^2 - 6 = 2(-32) + 8(-8) + 7(4) - 6 = -64 - 64 + 28 - 6 = -106 \neq 0
$$

For $$x = 3$$:

$$
2(3)^5 + 8(3)^3 + 7(3)^2 - 6 = 2(243) + 8(27) + 7(9) - 6 = 486 + 216 + 63 - 6 = 769 \neq 0
$$

For $$x = -3$$:

$$
2(-3)^5 + 8(-3)^3 + 7(-3)^2 - 6 = 2(-243) + 8(-27) + 7(9) - 6 = -486 - 216 + 63 - 6 = -645 \neq 0
$$

For $$x = \frac{1}{2}$$:

$$
2\left(\frac{1}{2}\right)^5 + 8\left(\frac{1}{2}\right)^3 + 7\left(\frac{1}{2}\right)^2 - 6 = 2\left(\frac{1}{32}\right) + 8\left(\frac{1}{8}\right) + 7\left(\frac{1}{4}\right) - 6 = \frac{1}{16} + 1 + \frac{7}{4} - 6 = \frac{1}{16} + \frac{4}{4} + \frac{7}{4} - \frac{96}{16} = \frac{12}{16} - \frac{96}{16} = -\frac{84}{16} = -\frac{21}{4} \neq 0
$$

For $$x = -\frac{1}{2}$$:

\[
2\left(-\frac{1}{2}\right)^5 + 8\left(-\frac{1}{2}\right)^3 + 7\left(-\frac{1}{2}\right)^2 - 6 = 2\left(-\frac{1}{32}\right) + 8\left(-\frac
The possible rational zeros are $$\pm1, \pm\frac{1}{2}, \pm2, \pm3, \pm\frac{3}{2}, \pm6$$.

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