Answer:
f(x)=(2x-7)^7(x+3)^4
When y=0, (2x-7)^7=0 and x=3.5 OR
(x+3)^4=0, and x=-3
Look at x=h or f(2)
That is 96-56-44+2 and it is not equal to 0. (x-2) is not a factor.
x^3-5x^2+6x-30
after trying 1,2, and 3, I tried 5
5/1===-5===6===-30
==1===0====6====30
(x^2-6) with no remainder
the factors are (x-5)(x^2+6)
the zeroes are 5, +/- i sqrt (6). Only one real 0.
hope this helps let me know in the comments if i'm wrong
