Question

Find the complex zeros of the polynomial function
f(x)=x^(3)-3^(2)+7x-5

Answer 1

first, lets reduce this polynomial to a quadratic by finding one real root.
you can use the ratio test.

Answer 2

p : factors of 5 (the constant term)
q : factors of 1 (the coefficient of x^3)

Answer 3

or you can just recognize that 1+7 = 8, so a factor is (x-1)

Answer 4

so,
\[
p:\pm1\pm5\\
q:\pm1\\
\text{possible roots:}{p\over q}\quad \pm1,\pm5
\]

Answer 5

start checking the roots using the remainder theorem:
\[
x=-1\qquad f(-1)=(-1)^3-3(-1)^2+7(-1)-5=-6\ne0\implies \text{not a root}\\
f(1)=(1)^3-3(1)^2+7(1)-5=0\quad \text{YAY! root found}
\]

Answer 6

so,
\[x=1\implies x-1\]is a factor of the polynomial.
now, we use synthetic division to get the remaining quadratic factor.

are you familiar with synthetic division or long division?

Answer 7

I have been working on this problem for an hour, and I cant seem to get it right! I am somewhat familiar

Answer 8

did you follow this first part?

Answer 9

I follow you But I couldnt seem to get it myself.. Once I see it.. makes more sense!

Answer 10

it is all following the steps. these are just the rules to follow through and all pieces fall into their places

Answer 11

maybe you should define complexes first

Answer 12

now, divide the function by this factor
you can use the "draw button" below to show me your work so I can check it for ya

Answer 13

kapeesh?

Answer 14

Im trying to follow! ;/

Answer 15

if you followed the first part, we now have to divide. I will wait till you show the division