Question 1058365
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
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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.
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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.
{{{graph(300,300,-10,10,-100,100,x^3-5x^2+6x-30)}}}