i need help solving this

Question

The possible rational zeros of P(x) = 14x³ + 5x² − 18x − 34

 

Answer 1

14x^3 +5x^2 -18x -34

possible real zeros = +/-p/q where p =-34, q = 14

by the rational root theorem

that zeros in on possible real zeros +/- 1/1, 1/7, 1/14, 2/1, 2/7, 2/14, 17/1, 17/2, 17/7, 34/1, 34/2, 34/7, 34/14

that narrows things down a bit,

then do trial and error, trying out each of the 24 possibilities

check what you find with another method such as graphing & finding the x intercepts = actual real zeros if any

x = about 1.54749 an irrational zero

at least 1, 2 or 3 real zeros for any cubic

this cubic has only one real zero which is irrational. there are no rational zeroes

Answer 2

The possible rational zeros of this cubic polynomial are given by:

(±p)/(±q)

where p is a factor of the constant term (-34 in this case), and q is a factor of the leading coefficient (14 in this case).

± any nember of this set { 1, 1/2, 1/7, 1/14, 2, 2/7, 17, 17/2, 17/7, 17/14, 34, 34/7}

This is an application of the rational root theorem.

By the way, it turns out that there are no rational roots, one real irrational root, and one pair of complex conjugate roots.