Question 681837
x = -5i
There will also be +5i as the conjugate pair
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{{{x^2 = (5i)^2}}}
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{{{x^2 = 25i^2}}}
where {{{i^2 = sqrt(-1)^2 = -1}}}
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{{{x^2 = 25(-1)}}}
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{{{x^2 = -25}}}
{{{x^2 + 25 = 0}}}
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{{{(x^3+5x^2+25x+125)/(x^2+25)}}}
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{{{x^2+25) goes into {{{x^3 +5x^2}}}: x
to get
{{{x^3 + 25x}}}
subtracting
{{{x^3 - x^3 + 5x^2 - 25x}}}
yielding
{{{5x^2 - 25x}}}
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{{{x^2+25}}} goes into {{{5x^2-25x+25x}}}: 5
to get
{{{5x^2+125}}}
subtracting
{{{5x^2 - 5x^2 - 125}}}
yielding
-125
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{{{x^2+25}}} goes into {{{-125+125}}}: 0
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{{{(x^3+5x^2+25x+125)/(x^2+25) = (x+5)}}}
Therefore, the factors are
{{{(x^2+25)(x+5)}}}=0
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