If (y-1) is a factor of f(y), f(y)=0 when y=1. So if you find that f(1)=0, then (y-1) is a factor of f(y).
f(y)=y^3-9y^2+10y+5
f(1)=1-9+10+5=7
Since f(1)=7, (y-1) is not a factor.
13x + 13 + 2x = 2
15x + 13 = 2
15x = -11
When t = 0 the point on the graph is (0,36). Your equation is just a line and always has the same tangent line so your equation is the same s = 36 - t.
