solve the polynomial inequality and graph the solution set on a real num ber line. express the solution set in interval notation

x^3+x^2+64x+64>0

please show work

Look for roots of

x^3+x^2+64x+64 = 0
among x = +or- 1,2,4,8 and 16
(That is called using the rational rooot theorem)

One of them is x= -1. Therefore (x+1) is a factor of the polynomial.
Now divide x^3+ x^2+ 64x+ 64 by x+1 for the other roots. Between the x values where f(x) = 0, the function is either + or -. You will have to figure that out.

If you can't, try reading
http://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.516644.html

Given the statement below, translate to an inequality. Use the variable n.

The cost of repairs will be at most $240.

To solve the polynomial inequality x^3 + x^2 + 64x + 64 > 0, we can follow these steps:

Step 1: Factor the polynomial, if possible. In this case, the polynomial cannot be easily factored, so we need to find the roots numerically or use a graphical method.

Step 2: Find the critical points. The critical points are the values of x for which the polynomial is equal to zero.

To find the critical points, we set the polynomial equal to zero:

x^3 + x^2 + 64x + 64 = 0

Unfortunately, the equation is not easily solvable. We will have to use numerical methods or graphing to find the roots. By analyzing the graph, we can see that there are no real roots.

Step 3: Determine the sign of the polynomial in each interval separated by the critical points.

Since there are no critical points, we can choose any test values to check the sign of the polynomial in each interval. Let's choose x = -1 and x = 1.

For x = -1:
(-1)^3 + (-1)^2 + 64(-1) + 64 = -2

For x = 1:
1^3 + 1^2 + 64(1) + 64 = 130

The polynomial is negative for x = -1 and positive for x = 1.

Step 4: Write the solution set in interval notation.

Since the polynomial is positive at x = 1, we know that the solution set is on the right side of x = 1. We can express this as (1, +∞).

Therefore, the solution set to the inequality x^3 + x^2 + 64x + 64 > 0 is (1, +∞) in interval notation.

Graphical representation on the real number line:
```
----------------------0----------------------1-------------------> x
(1, +∞)
```

To solve the polynomial inequality x^3 + x^2 + 64x + 64 > 0, we can follow these steps:

Step 1: Factorize the polynomial, if possible.
Unfortunately, in this case, we cannot easily factorize the polynomial, so we need to use alternative methods.

Step 2: Find the critical points.
To find the critical points, we set the inequality equal to zero and solve for x:
x^3 + x^2 + 64x + 64 = 0

However, this equation cannot be easily solved algebraically. Therefore, we need to use numerical methods such as graphing or a calculator to find the zeros of the equation. By graphing or using a calculator, we find that one of the zeros is x = -4.

Step 3: Determine the sign of the polynomial in each interval.
To determine whether the polynomial is positive or negative in specific intervals, we evaluate the polynomial at test points in those intervals. The test points are chosen to represent the intervals.

Testing in the interval ( -∞, -4 ):
Let's choose x = -5 as the test point:
f(-5) = (-5)^3 + (-5)^2 + 64(-5) + 64
= -125 + 25 - 320 + 64
= -356

Since the result is negative, we can conclude that the polynomial is negative in the interval (-∞, -4).

Testing in the interval ( -4, +∞ ):
Let's choose x = 0 as the test point:
f(0) = (0)^3 + (0)^2 + 64(0) + 64
= 64

Since the result is positive, we can conclude that the polynomial is positive in the interval (-4, +∞).

Step 4: Write the solution in interval notation.
Based on the sign analysis, we have found that the polynomial is negative for x < -4 and positive for x > -4. So the solution to the inequality x^3 + x^2 + 64x + 64 > 0 can be expressed in interval notation as:

(-∞, -4) U (0, +∞)

Now, to graph the solution on a real number line, plot an open circle at -4 and shade the regions to the left and right of it.