How do you find all rational zeroes of the function using synthetic division #f(x)=2x^5+x^4-23x-16#?

1 Answer
Feb 2, 2018

Answer:

#f(x)# has three irrational real zeros and two non-real complex zeros.

Explanation:

Given:

#f(x) = 2x^5+x^4-23x-16#

By Descartes' Rule of Signs #f(x)# has exactly one positive real zero, since the signs of its coefficients #+ + - -# have only one change of signs.

#f(-x)# has signs in the pattern #- + + -#. With #2# changes of sign, we can deduce that #f(x)# has #0# or #2# negative real zeros.

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 #-16# 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, +-8, +-16#

We find:

#f(-2) = -64+16+46-16 = -18 < 0#

#f(-1) = -2+1+23-16 = 6 > 0#

#f(-1/2) = -1/16+1/16+23/2-16 = -9/2 < 0#

#f(1) = 2+1-23-16 = -36 < 0#

#f(2) = 64+16-46-16 = 18 > 0#

Hence #f(x)# has irrational zeros in #(-2, -1)#, #(-1, -1/2)# and #(1, 2)#

The remaining #2# roots are non-real complex.

We cannot immediately deduce it from what we have found, but actually none of the zeros are expressible in terms of #n#th roots (i.e. square, cube roots etc.).