|
| quote: | Originally posted by ierxium
What's my prize? |
my respect
| quote: | x^2(x+1)-4(x+6)
So the zeros are 0, -1, and -6. please verify this TranceVanDyk.... |
the answers are x = -2(+/-)2i, 3
-----------------
find the zero's of x^3 + x^2 - 4x - 24
three roots. u find out through descartes rule that there is 1 positive zero, and 2 imaginary ones. so u know there are going to be imaginaries.
the hint was, (x-3) was a factor of the equation above. knowing that, u divide the equation by (x-3). the zero which is 3 b/c (3-3=0) is divided by the equation to get...
x^2 + 4x + 8
u cant factor it. so you use the quadratic equation to find the zeros. your answer should then be -2(+/-)2i.
answers are then x = 3, -2(+/-)2i
if u checked your answers 0, -1, and -6, they would not have equaled 0 in the original equation.
___________________
|
