f (x) =2X^3 + 2x^2 - 3x + 2
try values of x = ±1, ±2
f(1) = 2 + 2 - 3 + 2 ?0
f(-1) = -2 + 2 + 3 + 2 ? 0
f(2) = 16 + 8 - 6 + 2 ? 0
f(-2) = -16 + 8 + 6 + 2 = 0 , yeahhh
x+2 is a factor
Using synthetic division I got
(x+2)(2x^2 - 2x + 1) = 0
x = -2 or x = (2 ± ?-4)/2
x = -2 or x = 1 ± i
since you only wanted the reals,
x = -2
check:
http://www.wolframalpha.com/input/?i=plot+y+%3D+2x%5E3+%2B+2x%5E2+-+3x+%2B+2
notice the one real x-intercept at x = -2
f (x) =2X^3 + 2x^2 - 3x + 2
please help me set up to find all the real number zeros of the function
1 answer
1 answer