Number of discontinuous values We have to find the number of values of $x$ at which the function $$ f(x) = \frac{2x^5-8x^2+11}{x^4+4x^3+8x^2+8x+4}$$ is discontinuous. 
I thought that since both numerator and denominator are polynomials, and so are continuous, that the function would be continuous except when the denominator is zero.
But I am unable to find points where the denominator is zero. 

Note $(x+1)^4=x^4+4x^3+6x^2+4x+1$. 
Then, $$x^4+4x^3+8x^2+8x+4=(x+1)^4+2x^2+4x+3=(x+1)^4+2(x+1)^2+1$$
Thus we conclude that it is never discontinuous because  $$x^4+4x^3+8x^2+8x+4=(x^2+2x+2)^2>0$$
