1or else use contradiction,
let the integer root be a
=> f(x) = (x-a)g(x)
=> f(2).f(7) = (2-a)(7-a)(blah-blah) = -15 ...{given}
but (2-a)(7-a) is always even
hence it is a contradiction and our assumption is wrong.
Suppose that f (x) is a polynomial with integer coefficients such that f (2) = 3 and f (7) = −5.
Show that f (x) has no integer roots.
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4 Answers
21again use the fact that
a-b| f(a) - f(b)
7-2| f(7) - f(2)
means 5| -8
which is not possible
so how can it be a polynomial with integral coefficients??
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