it helps if you can find the roots of the function. In this case, you have
y=(x+1)(x^2+16)
So, you know the graph crosses the x-axis only at x = -1.
Since the 2nd factor is always positive, y>0 where (x+1) > 0.
Think you can take it from there?
Solve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation
x^3+x^2+16x+16>0
What is the solution set?
(Type your answer in interval notation.)
qaz
1 answer