What are the possible rational roots of #x^3-5x^2-2x+10 = 0# and how do you find all of the roots?

1 Answer
George C.
Dec 15, 2017

The rational roots theorem tells us that the "possible" rational roots are:

#+-1, +-2, +-5, +-10#

The actual roots are: #5#, #+-sqrt(2)#

Explanation:

Given:

#x^3-5x^2-2x+10 = 0#

By the rational roots theorem, any rational roots of the given cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #10# and #q# a divisor of of coefficient #1# of the leading term.

That means that the only possible rational roots are:

#+-1, +-2, +-5, +-10#

In fact, this cubic factors by grouping (as you can tell since the ratio of the first and second terms is the same as that between the third and fourth terms). So we find:

#0 = x^3-5x^2-2x+10#

#color(white)(0) = (x^3-5x^2)-(2x-10)#

#color(white)(0) = x^2(x-5)-2(x-5)#

#color(white)(0) = (x^2-2)(x-5)#

#color(white)(0) = (x^2-(sqrt(2))^2)(x-5)#

#color(white)(0) = (x-sqrt(2))(x+sqrt(2))(x-5)#

So roots:

#5# and #+-sqrt(2)#

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