SOLUTION: List the possible solutions: {{{P(x)=-3x^5-x^3-4x^4-5}}} I tried grouping but that was a dead end and I don't know what else to do with this because it has 4 terms. Could you ple

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: List the possible solutions: {{{P(x)=-3x^5-x^3-4x^4-5}}} I tried grouping but that was a dead end and I don't know what else to do with this because it has 4 terms. Could you ple      Log On


   

Question 1129875: List the possible solutions:
P%28x%29=-3x%5E5-x%5E3-4x%5E4-5
I tried grouping but that was a dead end and I don't know what else to do with this because it has 4 terms. Could you please help?
Found 2 solutions by rothauserc, MathLover1:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
Use The Rational Zeros Theorem since all the coefficients are integers
:
-3x^5 -x^3 -4x^4 -5 = 0
:
possible solutions or zeros are where f(x) crosses the x-axis
:
since the max degree of the given polynomial is 5, we expect to have 5 solutions
:
every rational zero of the given polynomial has the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient
:
constant term is -5 and the leading coefficient is -3
:
the factors of p are +or-1, +or-5
:
the factors of q are +or-1, +or-3
:
we have +or-1/1, +or-1/3, +or-5/1, +or-5/3
:

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=-3x%5E5-x%5E3-4x%5E4-5
Use the Rational Roots Test to Find All Possible Roots:

find every combination of ±p%2Fq where+p is a factor of the constant and q is a factor of the leading coefficient
The constant term of this polynomial is -5, with factors 1 and 5.
The leading coefficient is -3, with factors 1 and+3.


Tests yields the following possible solutions:
±1, ±, 1%2F3, ± 5, ± 5%2F3
test them:
f%281%29=-3%2A1%5E5-1%5E3-4%2A1%5E4-5=-3-1-4-5=-13-> does not work
f%28-1%29=-3%2A%28-1%29%5E5-%28-1%29%5E3-4%2A%28-1%29%5E4-5=3%2B1-4-5=-5-> does not work
f%281%2F3%29=-3%2A%281%2F3%29%5E5-%281%2F3%29%5E3-4%2A%281%2F3%29%5E4-5=-5.099-> does not work
f%28-1%2F3%29=-3%2A%28-1%2F3%29%5E5-%28-1%2F3%29%5E3-4%2A%28-1%2F3%29%5E4-5=-5-> does not work
f%285%29=-3%2A%285%29%5E5-%285%29%5E3-4%2A%285%29%5E4-5=-12005-> does not work
f%28-5%29=-3%2A%28-5%29%5E5-%28-5%29%5E3-4%2A%28-5%29%5E4-5=6995-> does not work
f%285%2F3%29=-3%2A%285%2F3%29%5E5-%285%2F3%29%5E3-4%2A%285%2F3%29%5E4-5=-2135%2F27-> does not work
f%28-5%2F3%29=-3%2A%28-5%2F3%29%5E5-%28-5%2F3%29%5E3-4%2A%28-5%2F3%29%5E4-5=595%2F81-> does not work

This polynomial has no rational roots that can be found using Rational Root Test.

try to find roots using Newton method:

If x%5Bn%5D+is an approximation a solution of f%28x%29=0 and if f′%28x%5Bn%5D%29%3C%3E0

the next approximation is given by, x%5Bn%2B1%5D=x%5Bn%5D-f%28x%5Bn%5D%29/f′%28x%5Bn%5D%29

f%28x%29=-3x%5E5-x%5E3-4x%5E4-5

f'%28x%29=+-x%5E2+%2815+x%5E2+%2B+16+x+%2B+3%29

%28-3x%5E5-x%5E3-4x%5E4-5%29%2F%28-x%5E2+%2815+x%5E2+%2B+16+x+%2B+3%29%29=0

x ≈ -1.46644


+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+-3x%5E5-x%5E3-4x%5E4-5%29+