How do you use the rational roots theorem to find all possible zeros of #P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8 #?

1 Answer
Jun 7, 2016

Answer:

Use the rational roots theorem to find possible candidate rational zeros, any thereby find that it has none.

Explanation:

#P(x) = 12x^4+x^3+4x^2+7x+8#

By the rational roots theorem, any rational zeros of #P(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #8# and #q# a divisor of the coefficient #12# of the leading term.

So in our example, it means that the only possible rational zeros are:

#+-1/12#, #+-1/6#, #+-1/4#, #+-1/3#, #+-1/2#, #+-2/3#, #+-1#, #+-4/3#, #+-2#, #+-8/3#, #+-4#, #+-8#

That's rather a lot of possible zeros to try, but you can narrow it down a little by noting that the coefficients of #P(x)# are all positive. Hence it has no positive zeros, leaving the possible rational zeros:

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

Substituting each of these for #x# in #P(x)#, we find that none is a zero. So #P(x)# has no rational zeros.

That's all the rational roots theorem tells us.

In fact, this particular quartic only has non-Real Complex zeros.