Find a polyomial of degree 3 with zeros x = – 1, 8, – 8, and f (1) = – 126.
f(x) = a(x+1)(x-8)(x+8), and f(1) =a(2)(-7)(9)= -126, or a(-126) = -126, so a=1, and therefore:
f(x) = (x+1)(x-8)(x+8)
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f(x) = a(x+1)(x-8)(x+8), and f(1) =a(2)(-7)(9)= -126, or a(-126) = -126, so a=1, and therefore:
f(x) = (x+1)(x-8)(x+8)