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If the zeros of the quadratic equation x^2+25=0 are +-5 (plus-minus 5), what is the correct factored form?

(x+5)(x-5)=0
(x+5i)(x-5i)=0
(x+12.5i)(x-12.5i)=0
(x+12.5)(x-12.5)=0

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ANSWER

[tex](x + 5i)(x - 5i) = 0 [/tex]

EXPLANATION

The given function is

[tex] {x}^{2} + 25 = 0[/tex]

The zeros of this function are;

[tex]x = \pm5i[/tex]

Or

[tex]x = - 5i \: and \: x = 5i[/tex]

[tex]x + 5i = 0\: and \: x - 5i = 0[/tex]

Hence the factored form is:

[tex](x + 5i)(x - 5i) = 0 [/tex]

If the equation were:

[tex] {x}^{2} - 25 = 0[/tex]

Then the factored form is

[tex](x + 5)(x - 5) = 0 [/tex]

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