How do you write a polynomial function in standard form with zeros at -6, 2, and 5?

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Answer 1 · 2018-08-02T08:45:16

$x^{3} - x^{2} - 32 x + 60$ $x^3-x^2-32x+60$

If $α, β$ $alpha,beta$ and $γ$ $gamma$ are the zeros,

the polynomial function is $(x - α) (x - β) (x - γ)$ $(x-alpha)(x-beta)(x-gamma)$

Hence, for zeros $- 6, 2$ $-6,2$ and $5$ $5$ polynomial function is

$(x - (- 6)) (x - 2) (x - 5)$ $(x-(-6))(x-2)(x-5)$

= $(x + 6) (x^{2} - 5 x - 2 x + 10)$ $(x+6)(x^2-5x-2x+10)$

= $(x + 6) (x^{2} - 7 x + 10)$ $(x+6)(x^2-7x+10)$

= $x^{3} - 7 x^{2} + 10 x + 6 x^{2} - 42 x + 60$ $x^3-7x^2+10x+6x^2-42x+60$

= $x^{3} - x^{2} - 32 x + 60$ $x^3-x^2-32x+60$