$\left(x+3\right)\left(3x-4\right)$
$x\left(3x^2+4x+2\right)\ge\:3\left(x^3+x^2-2x\right)-16$
$\left(-1\right)\:+\:\left(-1\right)\:\cdot\:\left(-2\right)$
$\lim_{x\to1}\left(2x^2+x+4\right)$
$\left(10.7\cdot\:9.3\right)$
$\frac{3x^{-1}}{2x^{-4}}$
$5a+2b-3a+2b$
Get a preview of step-by-step solutions.
Earn solution credits, which you can redeem for complete step-by-step solutions.
Save your favorite problems.
Become premium to access unlimited solutions, download solutions, discounts and more!