How do you write a polynomial with roots 3, –3, and 3 ?

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Answer 1 · 2016-03-13T03:53:13

$x^{3} - 3 x^{2} - 9 x + 27 = 0$ $x^3-3x^2-9x+27=0$

From the given: The roots are

$x = 3$ $x=3$ and $x = - 3$ $x=-3$ and $x = 3$ $x=3$

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

Multiply them all and equate to zero

$(x - 3) (x + 3) (x - 3) = 0$ $(x-3)(x+3)(x-3)=0$

$(x - 3) (x^{2} - 9) = 0$ $(x-3)(x^2-9)=0$

$(x^{3} - 9 x - 3 x^{2} + 27) = 0$ $(x^3-9x-3x^2+27)=0$

rearanging

$x^{3} - 3 x^{2} - 9 x + 27 = 0$ $x^3-3x^2-9x+27=0$

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