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Find a cubic function with the given zeros. 7, -3, 2
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i keep getting x^3+6x^2-13x-42
but the answer choices r f(x) = x3 - 6x2 - 13x - 42 f(x) = x3 - 6x2 + 13x + 42 f(x) = x3 - 6x2 - 13x + 42 f(x) = x3 + 6x2 - 13 + 42
x^3+6x^2-13x-42 solves to -7, 3 and -2
Instead of 7, -3, 2
f(x) = (x-7)(x+3)(x-2) you have to turn the zeros into factors: x=7 => x-7 = 0 x-7 is a factor
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just have to multiply out to get the cubic you desire
ohh so i would really be (x-7) (x+3) (x-2)
im still not getting the answer
\((x-7)(x+3) =x^2-4x-21\) then multiply by (\(x-2)\)
so would is get f(x) = x3 - 6x2 + 13x + 42
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nope, check your multiplication
sorry i was wrong f(x) = x3 - 6x2 - 13x + 42
there you go
thank you so much!!!
you're welcome
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