Jen23 said:
Yes it does become more complicated finding the roots! Thanks for showing how we got x=0.6448517208
Sometimes it is useful to know something about possible roots before you try to find them. Let us look at the two suggested versions of your problem:
(1) ##p_1(x) = 3 x^3 + 2x^2 + 75x - 50 = 0## (the original); and
(2) ##p_2(x) = 3 x^3 -2x^2 + 75 x - 50 = 0## (the suggested alternative).
Here are two helpful general results:
(I) Descarte's Rule of Signs; and (II) The Rational Root Theorem.
(I) The rule of signs states that for a polynomial ##p(x) = a_n p^n + a_{n-1} x^{n-1} + \cdots + a_0## the number of roots in the region ##\{ x \geq 0\}## is either ##k## or ##k-2## or ##k-4## ##\cdots##. Here, ##k## is the
number of sign changes in the coefficients ##a_n, a_{n-1}, \ldots, a_0##.
For the polynomial ##p_1## the coefficients are ##+3,+2,+75, -50## with one sign change, so there is exactly one root ##x > 0##. In fact, since ##p_1(0)=-50<0## and ##p_1(1) = 30>0## there must be a root between 0 and 1; and there are no other positive roots. For negative ##x## we can set ##x = -t## where ##t > 0##, and get ##p_1(x) = p_1(-t) = -3 t^3 + 2 t^2 - 75 t - 50##, with coefficients ##-3,+2,-75,-50## having two sign changes. Thus the number of roots ##t > 0## is either 2 or 0. And it turns out, is actually 0, because the roots are complex, not real.
For the polynomial ##p_2## the coefficients are ##+3,-2, +75, -50##, with 3 sign changes; thus, the number of roots ##x > 0## is either 3 or 1. Again, since ##p_2(0) = -50<0 ## and ##p_1(1) = 26 >0## there is a root between 0 and 1. For ##x = -t < 0## we have ##p_2(x) = p_2(-t) = -3 t^3 - 2 t^2 - 75 t - 50##, whose coefficients have no sign changes; thus, there are no negative real roots of ##p_2##.
So, Descarte's Rule of signs has given us some useful information to help us along the path to a solution.
(II) Next, consider the Rational Root Theorem. This says that if the coefficients ##a_n, a_{n-1}, \ldots, a_0## are all integers (+ or -), then any rational root of the form ##x = p/q ## (with integer ##p,q##) must have the property that ##p## is a divisor of the constant coefficient ##a_0## and ##q## is a divisor of the leading coefficient ##a_n##. For both ##p_1## and ##p_2## we know there is a root between 0 and 1. If ##p_1## has a rational root ##p/q##, ##p## must be a divisor of 50 and ##q## must be a divisor of ##3##. That is, either ##q=1## or ##q=3##. The value ##q=1## will not work because that would make ##x=p/q## an integer, and we already know that ##x=0## and ##x=1## are not roots. So, any possible integer root would be > 1, and we already know there are none. Therefore, if ##p_1## has a rational root we must have ##q = 3##, so ##x = p/3##, where ##p## is a divisor of 50. The only two divisors of 50 that give a result < 1 are 1 and 2, giving the possibilities ##1/3## and ##2/3##. If you plug these into ##p_1## you will see that they both
fail; thus, ##p_1## has
no rational roots at all in the region ##\{ x > 0 \}##. That means there is no hope for a "nice" factorization; you will need square roots or cube roots, or other unpleasant objects in your factorization.
The same considerations apply to ##p_2##: if there is a rational positive root, it must be either ##1/3## or ##2/3##. (The rational root theorem guarantees that there are no other possibilities, except for irrational roots.) By direct substitution we find that ##x = 2/3## is a root, so that means that one of the factors is ##3x-2##, as Sammy has shown. Note, however, that Sammy's suggested method of "grouping" does not always work, but the rational root theorem always works when there are, in fact, rational roots; and when the theorem fails, you know, 100%, that there are no rational roots.
For more information on these results, see, eg.,
http://www.purplemath.com/modules/drofsign.htm or
https://en.wikipedia.org/wiki/Descartes'_rule_of_signs
for the Rule of Signs, and
http://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml or
https://en.wikipedia.org/wiki/Rational_root_theorem
for the rational root theorem.
In each case the first cited links give gentle introductions, while the second links are more demanding, even including proofs of the results.
