How do you find the number of roots for f(x) = x^4 + 5x^3 +10x^2 + 20x + 24f(x)=x4+5x3+10x2+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: -22, -33, 2i2i, -2i2i.

Explanation:

f(x) = x^4+5x^3+10x^2+20x+24f(x)=x4+5x3+10x2+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.