How do you use the rational root theorem to list all possible roots for #2x^3+5x^2+4x+1=0#?

1 Answer
Jan 15, 2017

Answer:

The "possible" rational roots are: #+-1/2, +-1#

The actual roots are: #-1/2, -1, -1#

Explanation:

Given:

#f(x) = 2x^3+5x^2+4x+1#

Note that the rational root theorem will only tell us about rational zeros, not necessarily all zeros.

By the rational root 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 #1# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2#, #+-1#

In this particular example note that all of the coefficients are positive. Hence #f(x)# has no positive zeros, so let's check #f(-1)# first:

#f(-1) = -2+5-4+1 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

#2x^3+5x^2+4x+1 = (x+1)(2x^2+3x+1)#

Note that #-1# is also a zero of #2x^2+3x+1#, since #2-3+1 = 0#:

#2x^2+3x+1 = (x+1)(2x+1)#

So the third zero is #x=-1/2#

So in this particular example all of the roots were rational and therefore findable with the help of the rational roots theorem.