Mathematics
OpenStudy (anonymous):

solve each equation. state the number and type of roots: PROBLEM: x^(3)+6x+20=0 answer pwease and explanation... ^__^

OpenStudy (cwrw238):

using factor theorem let x = -2 them f(-2) = )-2(^3 - 12 + 20 = 0 so x = -2 is one root and x + 2 is a factor

OpenStudy (anonymous):

ohhhh okiii, thanks :D

OpenStudy (anonymous):

BUT,BUT why did u use 2?

OpenStudy (cwrw238):

just by observation - its not always possible of course

OpenStudy (cwrw238):

now divide x^(3)+6x+20 by x + 2 gives x^2 - 2x + 10 - solve this to get the other roots

OpenStudy (anonymous):

ahhh do i really need to solve them......tooo much work >.<"

OpenStudy (cwrw238):

x = -(-2) +- sqrt(4 - 4 *1 *10) / 2 - oh right - no. and type of roots there are 3 roots one real - thats the 2 and 2 complex roots as u can see from the above the sqrt is of a negative number

OpenStudy (anonymous):

ahhhh i get it now, thank you so much ^^

OpenStudy (cwrw238):

perhaps theres a simpler way of doing this which i dont know about you r welcome

OpenStudy (anonymous):

yea i think there is, cause my alg teacher taught us 2 ways, the long an short, in which i tottly forgot.... ^^" uhhhg meh, im so irresponsible, but thanks dude~