How do you use the rational roots theorem to find all possible zeros of $x^{4} - 5 x^{3} - 5 x^{2} + 23 x + 10$ $x^4-5x^3-5x^2+23x+10$ ?

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Answer 1 · 2016-03-19T12:08:08

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Answer 2 · 2016-03-19T13:53:35

$- 2, 5, 1 \pm \sqrt{2}$ $-2,5, 1+-sqrt(2)$

By the rational root theorem the rational roots of the polynom can be:

$\pm 1, \pm 2 or \pm 5$ $+-1, +-2 or +-5$

If you try the roots, you will confirm that the real roots are -2 and 5:

Now you have to divide the polynom by :

$(x + 2) (x - 5) = x^{2} - 3 x - 10$ $(x+2)(x-5)=x^2-3x-10$

$\frac{x^{4} - 5 x^{3} - 5 x^{2} + 23 x + 10}{x^{2} - 3 x - 10} =$ $(x^4-5x^3-5x^2+23x+10)/(x^2-3x-10)=$

$x^{2} - 2 x - 1$ $x^2-2x-1$

Use the quadratic formula to solve this part:

$x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2 \sqrt{2}}{2}$ $x=(2+-sqrt(4+4))/2=(2+-sqrt(8))/2=(2+-2sqrt(2))/2$

$x = 1 \pm \sqrt{2}$ $x=1+-sqrt(2)$

How do you use the rational roots theorem to find all possible zeros of x4−5x3−5x2+23x+10x^4-5x^3-5x^2+23x+10?