High School

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Determine the zeros of \( f(x) = x^3 - 3x^2 - 16x + 48 \).

Answer :

[tex] x^{2} - 3x^{2} - 16x + 48 = 0[/tex]

[tex] x^{2} (x - 3) - 16(x - 3) = 0[/tex]

[tex](x - 3) ( x^{2} - 16) = 0[/tex]

[tex](x - 3) (x - 4) (x + 4) = 0[/tex]

[tex]x = -4, 3, 4[/tex]

The zeros are: -4, 3, 4

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