How do you use the rational roots theorem to find all possible zeros of $P (x) = 2 x^{3} + 9 x^{2} + 11 x - 8$ $P(x)=2x^3+9x^2+11x-8$ ?

Question

How do you use the rational roots theorem to find all possible zeros of $P (x) = 2 x^{3} + 9 x^{2} + 11 x - 8$ $P(x)=2x^3+9x^2+11x-8$ ?

1 Answer

Impact of this question

1442 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-12T16:14:48

$P (x)$ $P(x)$ has zeros $\frac{1}{2}$ $1/2$ and $\frac{5}{2} \pm \frac{\sqrt{7}}{2} i$ $5/2+-sqrt(7)/2i$

Explanation:

$P (x) = 2 x^{3} + 9 x^{2} + 11 x - 8$ $P(x) = 2x^3+9x^2+11x-8$

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 $2$ $2$ of the leading term.

That means that the only possible rational zeros are:

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

We find:

$P (\frac{1}{2}) = 2 (\frac{1}{8}) + 9 (\frac{1}{4}) + 11 (\frac{1}{2}) - 8 = \frac{1 + 9 + 22 - 32}{4} = 0$ $P(1/2) = 2(1/8)+9(1/4)+11(1/2)-8 = (1+9+22-32)/4 = 0$

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

$2 x^{3} + 9 x^{2} + 11 x - 8$ $2x^3+9x^2+11x-8$

$= (2 x - 1) (x^{2} + 5 x + 8)$ $=(2x-1)(x^2+5x+8)$

$= (2 x - 1) ({(x - \frac{5}{2})}^{2} - \frac{25}{4} + 8)$ $=(2x-1)((x-5/2)^2-25/4+8)$

$= (2 x - 1) ({(x - \frac{5}{2})}^{2} + \frac{7}{4})$ $=(2x-1)((x-5/2)^2+7/4)$

$= (2 x - 1) ⎛ ⎝ {(x - \frac{5}{2})}^{2} - {(\frac{\sqrt{7}}{2} i)}^{2} ⎞ ⎠$ $=(2x-1)((x-5/2)^2-(sqrt(7)/2i)^2)$

$= (2 x - 1) (x - \frac{5}{2} - \frac{\sqrt{7}}{2} i) (x - \frac{5}{2} + \frac{\sqrt{7}}{2} i)$ $=(2x-1)(x-5/2-sqrt(7)/2i)(x-5/2+sqrt(7)/2i)$

Hence the other two zeros are:

$x = \frac{5}{2} \pm \frac{\sqrt{7}}{2} i$ $x = 5/2+-sqrt(7)/2i$

Science

Math

Humanities

... and beyond

How do you use the rational roots theorem to find all possible zeros of $P (x) = 2 x^{3} + 9 x^{2} + 11 x - 8$ $P(x)=2x^3+9x^2+11x-8$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you use the rational roots theorem to find all possible zeros of P(x)=2x3+9x2+11x−8P(x)=2x^3+9x^2+11x-8?

1 Answer

Explanation:

Related questions

Impact of this question

How do you use the rational roots theorem to find all possible zeros of $P (x) = 2 x^{3} + 9 x^{2} + 11 x - 8$ $P(x)=2x^3+9x^2+11x-8$ ?