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.)

Mike

use grouping to factor ...

x^3 + x^2 + 16x + 16 > 0
x^2(x+1) + 16(x+1) > 0
(x+1)(x^2 + 16) > 0

only one critical value , x = -1

so y = x^3 + x^2 + 16x + 16
is a cubic which is above the x-axis for all values of x, such that x > -1

I will let you write that in interval notation

To solve the polynomial inequality x^3 + x^2 + 16x + 16 > 0, we can begin by finding the critical points of the polynomial. We can do this by finding the roots of the polynomial. However, the polynomial does not factor nicely, so let's solve it using another method.

First, let's find the critical points by setting the polynomial equal to zero:

x^3 + x^2 + 16x + 16 = 0

There are no rational roots for this polynomial, so we'll use a different approach. We can use synthetic division or long division to find the quotient and remainder when dividing the polynomial by a binomial factor. By using long division or synthetic division, we find that (x + 2) is a factor of the polynomial.

Now we can rewrite the polynomial as:

(x + 2)(x^2 - x + 8) > 0

To solve this inequality, we need to consider three cases:

1. When (x + 2) > 0 and (x^2 - x + 8) > 0
2. When (x + 2) < 0 and (x^2 - x + 8) < 0
3. When (x + 2) < 0 and (x^2 - x + 8) > 0

Let's solve each case individually:

1. When (x + 2) > 0 and (x^2 - x + 8) > 0:
From the first inequality, we have x > -2.
From the second inequality, we can solve the quadratic equation (x^2 - x + 8) = 0 by completing the square or using the quadratic formula. However, this quadratic equation has no real roots, which means it is always positive. Therefore, we don't need to consider this inequality further.
Hence, the solution to this case is x > -2.

2. When (x + 2) < 0 and (x^2 - x + 8) < 0:
From the first inequality, we have x < -2.
From the second inequality, we have (x^2 - x + 8) < 0. However, this inequality has no real solutions, since the quadratic equation has no real roots. Therefore, we don't need to consider this inequality further.
Hence, there are no solutions for this case.

3. When (x + 2) < 0 and (x^2 - x + 8) > 0:
From the first inequality, we have x < -2.
From the second inequality, we have (x^2 - x + 8) > 0. This quadratic equation has no real roots, but by using the discriminant, we can determine that the parabola opens upward and never crosses the x-axis. Therefore, it is always positive.
Hence, the solution to this case is x < -2.

Combining all the cases, the solution set for the inequality x^3 + x^2 + 16x + 16 > 0 is:
x < -2

In interval notation, the solution set is (-∞, -2).

To solve the polynomial inequality x^3 + x^2 + 16x + 16 > 0, we can follow these steps:

Step 1: Factorize the polynomial, if possible. Unfortunately, in this case, the polynomial cannot be easily factorized.

Step 2: Identify the critical points. These are the points where the polynomial is equal to zero. To find them, set the polynomial equal to zero and solve for x:
x^3 + x^2 + 16x + 16 = 0

Step 3: Plot the critical points on a number line. In this case, we don't have any critical points, as the polynomial cannot be factored and equals zero.

Step 4: Choose test points. Select a few values on each side of the critical points and plug them into the polynomial inequality to determine the sign.

Step 5: Determine the sign of the polynomial for each interval. Based on the results from using test points, we can determine if the polynomial is positive or negative in each interval.

Now, let's apply these steps to find the solution set and express it in interval notation:

Step 1: Since the polynomial cannot be factored easily, we move on to the next step.

Step 2: Set the polynomial equal to zero and solve for x:
x^3 + x^2 + 16x + 16 = 0

Unfortunately, this equation cannot be solved algebraically. Therefore, we move on to the next step.

Step 3: As mentioned earlier, we don't have any critical points in this case, so there are no numbers to plot on the number line.

Step 4: Choose test points. We can select values for x, such as -10, -1, 0, 1, and 10, to represent different intervals.

Step 5: Determine the sign of the polynomial for each interval:
When we substitute -10 into the inequality, we get a negative value: (-10)^3 + (-10)^2 + 16(-10) + 16 = -804, meaning the polynomial is negative in this interval.
When we substitute -1 into the inequality, we get a positive value: (-1)^3 + (-1)^2 + 16(-1) + 16 = 10, meaning the polynomial is positive in this interval.
When we substitute 0 into the inequality, we get a positive value: 0^3 + 0^2 + 16(0) + 16 = 16, meaning the polynomial is positive in this interval.
When we substitute 1 into the inequality, we get a positive value: 1^3 + 1^2 + 16(1) + 16 = 34, meaning the polynomial is positive in this interval.
When we substitute 10 into the inequality, we get a positive value: 10^3 + 10^2 + 16(10) + 16 = 1356, meaning the polynomial is positive in this interval.

Based on the signs determined in Step 5, we can conclude that the polynomial is positive for all x. Therefore, the solution set to the inequality x^3 + x^2 + 16x + 16 > 0 is all real numbers.

In interval notation, the solution set is (-∞, ∞).