How do you write a polynomial in standard form given zeros 3, -4, 2?

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How do you write a polynomial in standard form given zeros 3, -4, 2?

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Answer 1 · 2016-03-25T08:21:20

$x^{3} - x^{2} - 14 x + 24 = 0$ $x^3-x^2-14x+24=0$

Explanation:

The method that I am used to is by equating each zero to the variable x

that is

$x = 3$ $x=3$ and $x = - 4$ $x=-4$ and $x = 2$ $x=2$

transposing the numbers to the left side, we have

$x - 3 = 0$ $x-3=0$ and $x + 4 = 0$ $x+4=0$ and $x - 2 = 0$ $x-2=0$

so that we have the factors

$(x - 3) \cdot (x + 4) \cdot (x - 2) = 0$ $(x-3)*(x+4)*(x-2)=0$

multiply them to obtain the standard form

$(x^{2} + x - 12) (x - 2) = 0$ $(x^2+x-12)(x-2)=0$
$(x^{3} + x^{2} - 12 x - 2 x^{2} - 2 x + 24) = 0$ $(x^3+x^2-12x-2x^2-2x+24)=0$
simplify to obtain the final answer

$x^{3} - x^{2} - 14 x + 24 = 0$ $x^3-x^2-14x+24=0$

God bless.... I hope the explanation is useful.

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How do you write a polynomial in standard form given zeros 3, -4, 2?

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