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When the polynomial f(x)=px^3+qx+r, where p, q and r are constants, is divided by (x+3) and (x-2) the remainders are -12 in each case. If (x+1) is a factor of f(x), find f(x) and the zeros of f(x)

Respuesta :

Answer:

f(-3) = p(-3)³ - 3q + r = -12

f(2) = p(2³)+ 2q + r = -12

f(-1) = p(-1)³ - q + r = 0

-27p - 3q + r = -12

8p + 2q + r = -12

-p - q + r = 0

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p = 1, q = -7, r = -6

f(x) = x³ - 7x - 6 = (x + 2)(x + 1)(x - 3)

The zeroes of f(x) are -2, -1, and 3.

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