let f(x) = x^3 + 2x^2 + 3x - 4
by trial: attempt x = ±1 , ±2, ±4
f(1) = 1+2+3-4 ≠ 0
f(-1) = -1+2-3-4 ≠ 0
f(2) = 8+8+6-4 ≠ 0
f(-2) = -8 + 8 - 6 - 4 ≠ 0
f(4) ≠ 0
f(-4) = -64 + 32 - 12 - 4 ≠ 0
There are no nice rational roots.
There is one real root at
appr x = .77605 , as well as two complex roots.
I suspect some type of typo.
The roots of the equation
x^3 + 2x^2 + 3x - 4 = 0
Are m, n , k
(a) write down the value of m + n + k
(b) write down the value of mn + nk + km
(b) find the value of m^2 + n^2 + k^2
please show detailed solutions
#thanks
1 answer
1 answer