Find a polynomial function f of degree 3 with leading coefficient 1 whose graph has x- intercepts at
(-1,0), (2,0), (4,0)
Two methods. I'll do both:
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Method 1:
Suppose the function is
then since it has x-intercept (-1,0) then when you substitute
-1 for x in that you get 0, so
which equals 0, so
or
and since it has x-intercept (2,0) then when you substitute
2 for x in that you get 0, so
which equals 0, so
or
and since it has x-intercept (4,0) then when you substitute
4 for x in that you get 0, so
which equals 0, so
or
So you have this system of 3 equations in 3 unknowns:
Solve that and get A=-5, B=2 and C=8
So substitute in
and get
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Method 2:
(-1,0), (2,0), (4,0)
If you had solved you would have
ended up with
x = -1, x = 2, x = 4
before which you would have had
x + 1 = 0, x - 2 = 0, x - 4 = 0
before which you would have used the zero-factor
principle on this:
which you would have had after factoring, so before
factoring you would have had that multiplied out,
so we multiply it out
and since that was the solution to , then
Same answer either way. Take your pick.
Edwin