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put this solution on YOUR website! Third degree polynomial that has zeros when x = 2,4
ii) f(1) = 2
iii) f(x) > 0 : (-infinity,2)
iv) f(x) < 0 : (2,4)u(4,infinity)"
(iii) & (iv) means f(x) > 0 on(-oo,2)
and f(x) < 0 on (2,+oo) -{4}
This means f(x) and (x-2) have the different sign and
f(x) and (x-4)^2 also have the different sign.
Since f(x) > 0 or < 0 except 2 or 4, so
2 and 4 are the only two possible zeros(roots) of f.
Now f is a third degree polynomial with only two roots 2 & 4.
Case I) If f(x) = a(x-2)^2(x-4) for some real number a,
then f(1) = a (-1)^2 (-3) = -3a = 2 ,so a = -2/3.
but then f(3) = (-2/3)(1)^2 (-1) > 0, contradicts to (iv).
Case II) If f(x) = a(x-2)(x-4)^2 for some real number a,
By f(1) = a(-1)(-3)^2 = 2, so a = -2/9.
And , we obtain f(x) = -2/9(x-2)(x-4)^2
[You can check that both iii) & iv) are true for this f.]
Kenny
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