What are all the possible rational zeros for #f(x)=2x^4-7x^3-2x^2-7x-4# and how do you find all zeros?

1 Answer
Oct 1, 2016

Answer:

#f(x)# has zeros #-1/2#, #4#, #+-i#

Explanation:

#f(x) = 2x^4-7x^3-2x^2-7x-4#

By the rational roots 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 #-4# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2, +-1, +-2, +-4#

Before we embark on trying these, note that the coefficients #2, -2, -4# almost cancel out if we reverse the sign of the middle one and the coefficients #-7, -7# would cancel out if the sign of one were opposite to the other. the only way that could heppen would be if #x^2 = -1#. So let's try #x=i# first...

#f(i) = 2i^4-7i^3-2i^2-7x-4 = 2+7i+2-7i-4 = 0#

So #x=i# is a zero and #(x-i)# a factor.

Since the coefficients of #f(x)# are all Real, #x=-i# is also a zero and #(x+i)# a factor.

Hence #(x-i)(x+i) = x^2+1# is a factor of #f(x)#...

#2x^4-7x^3-2x^2-7x-4 = (x^2+1)(2x^2-7x-4)#

We can factor the remaining quadratic using an AC method:

Find a pair of factors of #AC = 2*4 = 8# which differ by #B=7#.

The pair #8, 1# works.

Use this pair to split the middle term and factor by grouping:

#2x^2-7x-4 = (2x^2-8x)+(x-4)#

#color(white)(2x^2-7x-4) = 2x(x-4)+1(x-4)#

#color(white)(2x^2-7x-4) = (2x+1)(x-4)#

Hence the other two zeros are:

#x=-1/2# and #x = 4#