Raymond B. answered 01/02/26
Math, microeconomics or criminal justice
actual real zeros = x intercepts = about .2 and 2.7
-2 is lower bound
2 is the upper bound if there were rational zeros
take integer factors of the constant term divided by integer factors of the coefficient of the highest degree term
by Descartes rule of signs there are 0, 2 or 4 positive real zeros and 0 negative real zeros. 0 would be the lower bound as there are no negative zeros
x f(x) = x^4 -2x^3 +x^2 -9x +2
0 2
1 1-2+1--+2 -9 +2 = -5
2 16 -16 +4 -18+2 = -12
3 81 -54 +9 -27 +2 = 11
at least one zero is between x=2 and 3, since f(x) changes sign from - to +
3 is the upper bound for real zeros