How do you find the number of roots for #f(x) = x^4 + 5x^3 +10x^2 + 20x + 24# using the fundamental theorem of algebra?

1 Answer
May 27, 2016

The FTOA tells us that there are four zeros counting multiplicity.

Other methods help us find them: #-2#, #-3#, #2i#, #-2i#.

Explanation:

#f(x) = x^4+5x^3+10x^2+20x+24#

The fundamental theorem of algebra tells you that any non-constant polynomial in one variable with Real or Complex coefficients has a zero in the field of Complex numbers, #CC#.

A corollary of this, often stated as part of the FTOA, is that a polynomial with Complex coefficients of degree #n > 0# has exactly #n# Complex (possibly Real) zeros counting multiplicity.

In our example, #f(x)# is of degree #4#, so it has exactly four zeros counting multiplicity.

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What else can we find out about these zeros?

First note that all of the coefficients of #f(x)# are positive, so #f(x)# has no positive zeros.

If we reverse the signs on the terms of odd degree, then the pattern of signs of coefficients is: #+ - + - +#, which has #4# changes of sign. Since the coefficients of #f(x)# are Real, that means that it has #0#, #2# or #4# negative Real zeros.

Using the rational root theorem, the only possible rational zeros are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #24# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#-1, -2, -3, -4, -6, -8, -12, -24#

Trying each of these in turn, we find:

#f(-2) = 16-40+40-40+24 = 0#

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

#x^4+5x^3+10x^2+20x+24 = (x+2)(x^3+3x^2+4x+12)#

The remaining cubic factor factors by grouping as follows:

#x^3+3x^2+4x+12 = (x^3+3x^2)+(4x+12)#

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

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

#= (x-2i)(x+2i)(x+3)#

So the zeros are: #-2#, #-3#, #2i#, #-2i#.