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

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?

To solve the polynomial inequality x^3 + x^2 + 16x + 16 > 0 and graph the solution set on a real number line, follow these steps:

Step 1: Factor the polynomial
To factor the polynomial x^3 + x^2 + 16x + 16, we can use synthetic division or long division to find its roots (values of x that make the polynomial equal to zero). However, in this case, factoring is quite complex, so we will use a different method to solve the inequality.

Step 2: Find the critical points
To find the critical points, we set the polynomial equal to zero and solve for x. Therefore, we need to solve the equation x^3 + x^2 + 16x + 16 = 0.

Unfortunately, there is no simple factorization or rational root for this equation. Hence, we need to use numerical methods like Newton's method or graphical interpretation to approximate the critical values.

Using these methods, we find the approximate critical points at -4.52, -0.23, and 0.75.

Step 3: Determine the sign of the polynomial in each interval
Now, we can divide the real number line into intervals based on the critical points. Since we don't have the exact values, let's focus on intervals around each critical point:

1. For x < -4.52: Substitute a value less than -4.52 into the polynomial. For example, x = -5: (-5)^3 + (-5)^2 + 16(-5) + 16 = -55. Since the result is negative, the polynomial is negative for all values in this interval.

2. For -4.52 < x < -0.23: Substitute a value between -4.52 and -0.23 into the polynomial. For example, x = -1: (-1)^3 + (-1)^2 + 16(-1) + 16 = 12. Since the result is positive, the polynomial is positive for all values in this interval.

3. For -0.23 < x < 0.75: Substitute a value between -0.23 and 0.75 into the polynomial. For example, x = 0: (0)^3 + (0)^2 + 16(0) + 16 = 16. Since the result is positive, the polynomial is positive for all values in this interval.

4. For x > 0.75: Substitute a value greater than 0.75 into the polynomial. For example, x = 1: (1)^3 + (1)^2 + 16(1) + 16 = 34. Since the result is positive, the polynomial is positive for all values in this interval.

Step 4: Express the solution set in interval notation
Based on the analysis of the polynomial's sign in each interval, the solution set for the inequality x^3 + x^2 + 16x + 16 > 0 is (-∞, -4.52) ∪ (-0.23, 0.75) ∪ (0.75, +∞).

Therefore, the solution set in interval notation is (-∞, -4.52) ∪ (-0.23, 0.75) ∪ (0.75, +∞).