by Eigenvalue » Sun Oct 11, 2026 3:16 pm
f(x)=x³-5x²+4x-20
Possible real roots:±1,±2,±4,±5,±10,±20
Substitute the possible real roots; if the y-value evaluates to 0, then it is a root--roots are x-intercepts and have a y-coordinate of 0
f(1)=1³-5(1)²+4(1)-20=1-5+4-20=-20
f(2)=2³-5(2)²+4(2)-20=8-20+8-20=-24
f(5)=5³-5(5)²+4(5)-20=125-125+20-20=0 ✓
f(10)=10³-5(10)²+4(10)-20=1000-500+40-20=520
f(20)=20³-5(20)²+4(20)-20=8000-2000+80-20=6040
Divide x³-5x²+4x-20 by x-5
x³-5x²+4x-20=(x-5)(x²+4)
Factor x²+4
(x²+4)=(x+2i)(x-2i)
The complex roots are 5 and ±2i