What are the possible rational roots of $x^{3} - 5 x^{2} - 4 x + 20 = 0$ $x^3-5x^2-4x+20=0$ and then determine the rational roots?

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What are the possible rational roots of $x^{3} - 5 x^{2} - 4 x + 20 = 0$ $x^3-5x^2-4x+20=0$ and then determine the rational roots?

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Answer 1 · 2016-12-12T04:29:38

The possible rational roots are $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$ $+-1,+-2,+-4,+-5,+-10,+-20$ . The rational roots are $x = - 2, x = 2, x = 5$ $x=-2, x=2, x=5$ .

Explanation:

$1 x^{3} - 5 x^{2} - 4 x + 20 = 0$ $color(blue)1x^3-5x^2-4x+color(red)(20)=0$

The possible rational roots are the factors of the constant $20$ $color(red)20$ divided by the factors of the leading coefficient $1$ $color(blue)1$ . The factors of the constant are called $p$ $p$ and the factors of the leading coefficient are called $q$ $q$ .

$\frac{p}{q} = \frac{\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20}{\pm 1} =$ $p/q=(+-1,+-2,+-4,+-5,+-10,+-20)/(+-1)=$

$\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$ $+-1,+-2,+-4,+-5,+-10,+-20$

The rational roots of this particular function can be found by factoring.

Factor by grouping.

First, group the first two terms and the second two terms.

$(x^{3} - 5 x^{2}) a + (- 4 x + 20) = 0$ $(x^3-5x^2)color(white)a+(-4x+20)=0$

Factor out a GCF from each group.

$x^{2} (x - 5) - 4 (x - 5) = 0$ $x^2(x-5)-4(x-5)=0$

Regroup.

$(x^{2} - 4) (x - 5) = 0$ $(x^2-4)(x-5)=0$

Factor $x^{2} - 4$ $x^2-4$ as the difference of squares.

$(x + 2) (x - 2) (x - 5) = 0$ $(x+2)(x-2)(x-5)=0$

Set each factor equal to zero and solve.

$x + 2 = 0 a a x - 2 = 0 a a x - 5 = 0$ $x+2=0color(white)(aa)x-2=0color(white)(aa)x-5=0$

$x = - 2 a a a x = 2 a a a x = 5$ $x=-2color(white)(aaa)x=2color(white)(aaa)x=5$

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What are the possible rational roots of $x^{3} - 5 x^{2} - 4 x + 20 = 0$ $x^3-5x^2-4x+20=0$ and then determine the rational roots?

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Explanation:

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What are the possible rational roots of $x^{3} - 5 x^{2} - 4 x + 20 = 0$ $x^3-5x^2-4x+20=0$ and then determine the rational roots?