the function has a real zero in the given interval. approximate this solution correct to two decimal places. f(x)=x^4-x^3-7x^2+5x+10; (2,3)
I assume that (2,3) means that x=2 and x=3 would be solutions you found?
(but x=3 does not work, while x=2 does work)
by synthetic division by x-2 I got
x^3 + x^2 - 5x - 5
= x^2(x+1) - 5(x+1)
= (x+1)(x^2 - 5)
so the zeros of f(x) are
-1,2, ±√5