Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Algebra 2 help!!!!!!!?
Use the rational theorem to list all the possible rational roots for each polynomial equation. then find the actual roots.
1.) x^3 - x^2 + 2x - 2 = 0
2.) x^3 + 2x^2 - 8x -16 = 0
find the roots of the polynomial equation
3.) 2x^4 - 5x^3 - 17x^2 + 41x - 21 = 0
10 points will be rewarded
thanks
1 Answer
- BieberLv 61 decade agoFavorite Answer
1) You don't need the theorem, just factor:
x^2(x - 1) + 2(x - 1) = 0
(x^2 + 2)(x - 1) = 0
x = 1, +/- i*sqrt(2)
2) Again, factorable:
x^2(x + 2) - 8(x + 2) = 0
(x^2 - 8)(x + 2) = 0
x = -2, +/- 2sqrt(2)
3. Here, we'll have to use the theorem:
+/- (21/2, 21, 7/2, 7, 3/2, 3, 1)
Let's see which of those work.
After plugging them in, I found 1 to be a solution, so divide out the factor (x - 1):
(2x^3 - 3x^2 - 20x + 21)(x - 1) = 0
Same possible solutions as the one before, so let's plug in all of them.
Turns out 7/2 is a solution, so let's divide (x - 7/2) from it:
(x - 7/2)(2x^2 + 4x - 6)(x - 1) = 0
Now let's use the quadratic equation:
(-4 +/- sqrt(16 + 48))/4 = x
x = -3, 1
So our solutions are:
x = -3, 1, and 7/2 (1 is a double root)
Have a great day! :-)