# Alegbra 2

Use the rational root theorem to list all possible rational roots for the equation.
X^3+2x-9=0.

Use the rational root theorem to list all possible rational roots for the equation. 3X^3+9x-6=0.

A polynomial function P(x) with rational coefficients has the given roots. Find two additional roots of P(x)=o
-2i and the square root of 10

For the following determine what Descartes' rules of signs says about the number of positive and negative real roots

P(x)=x^2+5x+6
P(x)=9x^3-4x^2+10
P(x)=8x^3+2x^2-14x+5

Find all rational roots for P(x)=0
P(x)=6x^4-13x^3+13x^2-39x-15

Solve the equation.
3x^2+4x+5=0

A friend tells you that he has a cubic equation with exactly three complex roots. Determine which explanation best explains why this is impossible

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1. This looks like an assignment.
The usual procedure is that you do and show some of your work, then indicate where you are encountering difficulties.

I will do the first one for you.
X^3+2x-9=0
let f(x) = X^3+2x-9=0

since you rational factors only, consider only factors of -9
and try them
f(1) = 1 + 2 - 9 ≠ 0 , I really don't care what it is, all I care about is the zero
f(-1) ≠0
f(3) = 27 + 6 - 9 ≠ 0
f(-3) = -27 - 6 - 9 ≠ 0
f(±9) ≠ 0

there are no rational roots

btw, the 2nd one has no rational roots as well

some of these are nasty equations:
e.g. p(x) = 6x^4-13x^3+13x^2-39x-15
took a while to factor it, copped out and used Wolfram:
http://www.wolframalpha.com/input/?i=6x%5E4-13x%5E3%2B13x%5E2-39x-15%3D0

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2. A friend tells you that he has a cubic equation with exactly three complex roots. Determine which explanation best explains why this is impossible

Actually, this is quite possible.

However, if you have an equation with only real coefficients, it is impossible because the complex roots must occur in conjugate pairs. So, and odd number of complex roots cannot occur.

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