#question on how many solutions for a function
11 messages · Page 1 of 1 (latest)
hmm
lets imagine
we have the equation
(x-3)(x+2)(x-5)=0
this mean X can have 3 values where the equation equals zero
x=3
x=-2
x=5
and if you multiply all the parantheses you would get something like
ax^3 + bx^2 + cx + d = 0
so the number of solutions
normally is equal to the higher exponent