If $p(x)=x+3,$ then $p(x)+p(-x)$ is equal to
$3$
$2x$
$0$
$6$
From the following polynomials find out which of them has $(x+1)$ as a factor
$x^{3}-5 x^{2}+2 x+8$
Find the following products:
$\left(\frac{x}{2}+2 y\right)\left(\frac{x^{2}}{4}-x y+4 y^{2}\right)$
Without actually calculating the cubes, find the value of each of the following
$(14)^{3}+(27)^{3}-(41)^{3}$
Find $p(1), p(2)$ and $p(4)$ for each of the following polynomials
$p(t)=t^{2}-6 t+8$
The following expressions are polynomials? Justify your answer:
$\frac{1}{x+1}$