How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given #f(x)=x^3+7x^2+7x-15#?

1 Answer
Dec 13, 2016

Answer:

Use Descartes' Rule of Signs to find that #f(x)# has #1# positive Real zero and #0# or #2# negative zeros.

Further find all (rational) zeros: #x=1#, #x=-3# and #x=-5#

Explanation:

Given:

#f(x) = x^3+7x^2+7x-15#

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Descartes' Rule of Signs

The pattern of signs of the coefficients is #+ + + -#. With one change of signs, that means that this cubic has exactly #1# positive Real zero.

The pattern of signs of coefficients of #f(-x)# is #- + - -#. With two changes of sign, that menas that #f(x)# has #0# or #2# negative Real zeros.

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Rational Roots Theorem

Since #f(x)# is expressed in standard form and has integer coefficients we can apply the rational roots theorem to assert that any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-15# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-3, +-5, +-15#

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Sum of coefficients shortcut

The sum of the coefficients of #f(x)# is #0#. That is:

#1+7+7-15 = 0#

Hence #f(1) = 0# and #(x-1)# is a factor:

#x^3+7x^2+7x-15 = (x-1)(x^2+8x+15)#

Note that #3+5 = 8# and #3*5 = 15#, so we find:

#x^2+8x+15 = (x+3)(x+5)#

So the remaining two zeros are #x=-3# and #x=-5# as we might suspect.