$f\left(x\right)\:=\:-2x^2\:-4x\:-3\:$
$4j+4+3j$
$\left(x\:-\:5\:\right)\left(x+6\right)$
$\frac{\tan\left(x\right)+1}{\sec\left(x\right)+\csc\left(x\right)}$
$4x^3-5x^2+3x^3+2x^2-8x$
$\left(3x^3+2x^2-x\right):\left(x+2\right)$
$-\left(1-3+5-7\right)+4-12+13-\left(2+15-4+3\right)$
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