How do you find the number of roots for #f(x) = x^3 – 75x + 250# using the fundamental theorem of algebra?

1 Answer
George C.
Jun 18, 2016

The FTOA tells us that there are #3# zeros counting multiplicity.
Further investigation shows us that they are #x=5# with multiplcity #2# and #x=-10#.

Explanation:

The fundamental theorem of algebra tells you that any non-constant polynomial in one variable with Complex (possibly Real) coefficients has a zero which is a Complex (possibly Real) number.

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

So in our example:

#f(x) = x^3-75x+250#

is of degree #3#, so has exactly #3# zeros counting multiplicity.

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Bonus

What else can we find out about these zeros?

By the rational root theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #250# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are positive and negative factors of #250#:

#+-1#, #+-2#, #+-5#, #+-10#, #+-25#, #+-50#, #+-125#, #+-250#

We find:

#f(-10) = -1000+750+250 = 0#

So #x=-10# is a zero and #x+10# a factor:

#x^3-75x+250#

#=(x+10)(x^2-10x+25)#

#=(x+10)(x-5)(x-5)#

So the zeros are #x=-10# and #x=5# with multiplicity #2#.

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