How do you divide $(x^3 + 4x^2 - 3x -12)$ by $x^2 - 3$ ?

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Answer 1 · 2016-10-12T03:58:15

It would be helpful to notice that

$(x^3 + 4x^2 - 3x -12)/(x^2 - 3)$

$[x^3-3x+4x^2-12]/[x^2-3]$

$[x(x^2-3)+4(x^2-3)]/[x^2-3]$

$[(x+4)*(x^2-3)]/(x^2-3)$

$[(x+4)*cancel(x^2-3)]/cancel(x^2-3)$

$x+4$

How do you divide (x^3 + 4x^2 - 3x -12) (x^3 + 4x^2 - 3x -12) by x^2 - 3x^2 - 3?