How do you use the rational roots theorem to find all possible zeros of #x^4-9x^2+20#?

1 Answer
Jul 28, 2016

Answer:

Zeros: #x = +-2# and #x = +-sqrt(5)#

Explanation:

Given #f(x) = x^4-9x^2+20#, the rational roots theorem tells you that any rational zeros must be expressible in the form #p/q# for integers #p#, #q# with #p# a divisor of the constant term #20# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-4, +-5, +-10, +-20#

Evaluating #f(x)# for each in turn, this allows you to find the two rational zeros:

#f(2) = f(-2) = 16-36+20 = 0#

You can then divide #f(x)# by #(x-2)(x+2) = x^2-4# to get #x^2-5#, which has zeros #x = +-sqrt(5)#.

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Alternatively:

Note that #x^4-9x^2+20# is a quadratic in #x^2#. That is:

#x^4-9x^2+20 = (x^2)^2-9(x^2)+20#

Note also that #4+5 = 9# and #4xx5 = 20#

So we find:

#x^4-9x^2+20#

#=(x^2-4)(x^2-5)#

#=(x^2-2^2)(x^2-(sqrt(5))^2)#

#=(x-2)(x+2)(x-sqrt(5))(x+sqrt(5))#

Hence zeros:

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