+312 Find the greatest integer value of $b$ for which the expression $\frac{9x^3+4x^2+11x+7}{x^2+bx+8}$ has a domain of all real numbers.
+129899 The denomiantor cannot = 0
So...we can guarantee no reals will make the denominator = 0 if we solve this :
b^2 - 4 (8)(1) < 0
b^2 - 32 < 0
b^2 < 32 take the positive root
b < sqrt 32
b < 5.6
So 5 is the greatest integer
+876 Find the greatest integer value of b for which the expression (9x^3+4x^2+11x+7)/(x^2+bx+8) has a domain of all real numbers.
Domain is all real numbers iff x^2+bx+8 ≠ 0.
Let x^2 + bx + 8 = 0 and then we will find a discriminant that will result in no real solutions if x^2+bx+8 = 0 because the denominator ≠ 0.
A equation is unreal iff b^2 - 4ac < 0
So b^2 - 32 < 0
b^2 < 32
floor(sqrt(32)) = 5
5 is the answer I think
