How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of #f(x)=x^4+2x^3-12x^2-40x-32#?

1 Answer
Sep 23, 2017

Answer:

Possible rational zeros:

#+-1, +-2, +-4, +-8, +-16, +-32#

Descartes gives us that #f(x)# has one positive and #3# or #1# negative zeros.

Actual zeros: #-2, -2, -2, 4#

Explanation:

Given:

#f(x) = x^4+2x^3-12x^2-40x-32#

Rational roots theorem

By the rational roots theorem, 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 #-32# and #q# a divisor of the coefficient #1# of the leading term.

That means the the only possible rational zeros are:

#+-1, +-2, +-4, +-8, +-16, +-32#

Descartes' Rule of Signs

The pattern of signs of the coefficients of #f(x)# is #+ + - - -#. With one change of signs, Descartes' Rule of Signs tells us that #f(x)# has exactly one positive real zero.

The pattern of signs of coefficients of #f(-x)# is #+ - - + -#. With #3# changes of sign, Descartes' Rule of Signs allows us to deduce that #f(x)# has #3# or #1# negative real zero.

Bonus - Find the actual zeros

Note that the coefficient of #x^4# is odd, but all the other coefficients are even. Therefore any integer zero of #f(x)# must be even.

We find:

#f(-2) = (-2)^4+2(-2)^3-12(-2)^2-40(-2)-32 = 16-16-48+80-32 = 0#

So #-2# is a zero and #(x+2)# a factor:

#x^4+2x^3-12x^2-40x-32 = (x+2)(x^3-12x-16)#

We find that #-2# is also a zero of the remaining cubic expression:

#(-2)^3-12(-2)-16 = -8+24-16 = 0#

So #(x+2)# is a factor again:

#x^3-12x-16 = (x+2)(x^2-2x-8)#

Finally, to factor the remaining quadratic, note that #4*2 = 8# and #4-2=2#, so we find:

#x^2-2x-8 = (x-4)(x+2)#

So:

#f(x) = (x+2)^3(x-4)#

has zeros #-2# with multiplicity #3# and #4# with multiplicity #1#.