$\lim_{x\to-4}\left(\frac{x^2+3x-4}{x-3}\right)$
$c\left(d\:-\:4\right)$
$\frac{\left(2\left(z+i\right)\right)^n}{n^2}$
$4x\left(x-2\right)+4x=8\left(x-9\right)$
$3x^2\left(2x-2\right)$
$49q^8+16b^4-56b^2q^4$
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