Question

HELP ME PLEASE!!! Find the roots of the polynomial equation.

2x^3 + 2x^2 – 19x + 20 = 0 (1 point)

Answer 1

A. \[\frac{ 3+i }{ 2}, \frac{ 3-i }{ 2 }, -4\]

B. \[\frac{ -3+i }{ 2 }, \frac{ -3-i }{ 2}, 4\]

C. \[\frac{ -3+i }{ 2}, \frac{ -3-i }{ 2 },-4\]

D. \[\frac{ 3+2i }{ 2}, \frac{ 3-2i }{ 2}, 4\]

Answer 2

@amistre64 @agent0smith @Ashleyisakitty @grant330sims @OtonoGold @austinL anyone help ?

Answer 3

f(x) = 2x^3+2x^2-19x+20
Using rational root theorem to test root, you can find f(-4) = 0. So, one root is x = -4.
Now you can reduce the degree by synthetic division,
-4 | 2 2 -19 20
.......-8 24 -20
_________________
....2 -6..5 | 0
Solve 2x^2 - 6x + 5 = 0
x = (1/2)[3 +/- i] <==two imaginary roots

Answer 4

The answer is A. Im a 100 % sure

Answer 5

thank you :)

Answer 6

No problem :)