Question 676320
{{{(x^4+4x^3-5x^2-12x+6)/(x^2-3)}}}
...
{{{x^2-3}}} goes into {{{x^4+4x^3}}} : x^2
to get
{{{x^4 - 3x^2}}}
subtracting
{{{x^4 - x^4 + 4x^3 - -3x^2}}}
yielding
{{{4x^3+3x^2}}}
...
{{{x^2-3}}} goes into {{{4x^3+3x^2-5x^2}}} which is
{{{4x^3-2x^2}}} : 4x
to get
{{{4x^3 - 12x}}}
subtracting
{{{4x^3 - 4x^3 - 2x^2 -  -12x}}}
yielding
{{{-2x^2+12x}}}
...
{{{x^2-3}}} goes into {{{-2x^2+12x-12x}}} which is 
{{{-2x^2}}} :-2
to get
{{{-2x^2 + 6}}}
subtracting
{{{-2x^2 -  -2x^2 - 6}}}
yielding
-6
...
{{{x^2-3}}} goes into -6-3, which is -9: 0
to get
0
subtracting
-9 - 0
yielding
-9 as the remainder.
...
{{{(x^4+4x^3-5x^2-12x+6)/(x^2-3)}}} = {{{(x^2+4x-2)((-9)/(x^2-3))}}}
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