# 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. b
a
b
c
a
b
b
c
d
c
we didn't start the fire, it was always burning since the worlds been turning!

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2. Crazy and haha are correct, it is 100% accurate. Remember, it's not cheating if you're not caught!

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3. crazy’s right 100%

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4. To ease people who dont feel comfortable with just letters:
1) B. 1,3,9
2) A. 1,2,3,6,⅓,⅔
3) B. -i, 7-8i
4) C. 2i, _10^½
5) A. No positive real roots; Two or no
negative real roots
6) B. Two or no positive real roots; One
negative real root
7) B. Two or no positive real roots; One
negative real root
8) C. 5/2, -⅓
9) D. -⅔±(11/3i)^½
10) C. Complex solutions must appear
in conjugate pairs; Having and odd
number of them is impossible

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5. September 2021: Crazy La Paint is still right, just for anybody who’s still here, you’ll still get 100%

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6. as of september 2021 crazy la paint is still 100% right :)

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7. what's funny to me is that the questions and answers haven't changed in the 3 years since this was posted

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8. Thank you Crazy!

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9. Thanks

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10. 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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11. ^^^^All correct!!! 100%

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12. Thanks crazy la paint it was a 100%

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13. I istg I just saw mr D'arby on an English post...

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14. If you've got 11 questions the answer for the last one is B. Sometimes true.

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15. aiko is correct for all counts. thanks

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16. aiko is right

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17. 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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