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#