$-8\left(-12\right)\left(-3\right)$
$-2x^2+2x+3$
$\int_1^{\infty}\left(\frac{8}{x^2+1}\right)dx$
$\lim_{x\to0}\left(-3x+1\right)\left(x^2+1\right)$
$\sqrt{2}-\:\sqrt{3}\:x\:\sqrt{2}$
$40^4+8a$
$\frac{\left(9x^3-16x+8\right)}{\left(3x^2+3x-4\right)}$
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