How do you use the rational roots theorem to find all possible zeros of $f (x) = 2 x^{3} + x^{2} - 13 x + 6$ $f(x)=2x^3+x^2-13x+6$ ?

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How do you use the rational roots theorem to find all possible zeros of $f (x) = 2 x^{3} + x^{2} - 13 x + 6$ $f(x)=2x^3+x^2-13x+6$ ?

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Answer 1 · 2016-03-23T19:12:14

$2 x^{3} + x^{2} - 13 x + 6 = (2 x - 1) (x + 3) (x - 2)$ $2x^3+x^2-13x+6 = (2x-1)(x+3)(x-2)$

Explanation:

$f (x) = 2 x^{3} + x^{2} - 13 x + 6$ $f(x) = 2x^3+x^2-13x+6$

By the rational roots theorem, any rational zeros of $f (x)$ $f(x)$ must be expressible in the form $\frac{p}{q}$ $p/q$ for integers $p$ $p$ and $q$ $q$ where $p$ $p$ is a divisor of the constant term $6$ $6$ and $q$ $q$ a divisor of the coefficient $2$ $2$ of the leading term.

That means that the only possible rational zeros are:

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

Let us try each in turn:

$f (\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} - \frac{13}{2} + 6 = 0$ $f(1/2) = 1/4+1/4-13/2+6 = 0$

So $x = \frac{1}{2}$ $x=1/2$ is a zero and $(2 x - 1)$ $(2x-1)$ a factor:

$2 x^{3} + x^{2} - 13 x + 6 = (2 x - 1) (x^{2} + x - 6)$ $2x^3+x^2-13x+6 = (2x-1)(x^2+x-6)$

We could continue simply trying the other possible zeros, but it is quicker to note that $3 \times 2 = 6$ $3xx2 = 6$ and $3 - 2 = 1$ $3-2 = 1$ , so:

$x^{2} + x - 6 = (x + 3) (x - 2)$ $x^2+x-6 = (x+3)(x-2)$

Putting it all together, we find:

$2 x^{3} + x^{2} - 13 x + 6 = (2 x - 1) (x + 3) (x - 2)$ $2x^3+x^2-13x+6 = (2x-1)(x+3)(x-2)$

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How do you use the rational roots theorem to find all possible zeros of $f (x) = 2 x^{3} + x^{2} - 13 x + 6$ $f(x)=2x^3+x^2-13x+6$ ?

1 Answer

Explanation:

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How do you use the rational roots theorem to find all possible zeros of f(x)=2x3+x2−13x+6f(x)=2x^3+x^2-13x+6?

1 Answer

Explanation:

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How do you use the rational roots theorem to find all possible zeros of $f (x) = 2 x^{3} + x^{2} - 13 x + 6$ $f(x)=2x^3+x^2-13x+6$ ?