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+4x+4<0
The solution set is=
x^3+x^2+4x+4<0
(x^2(x+1) + 4(x+1) < 0
(x+1)(x^2+4) < 0
well, x^2 + 4 is always positive, so the only way the whole thing < 0 if
x+1 < 0
x < -1
check:
good ol' Wolfram webpage
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E3%2Bx%5E2%2B4x%2B4
notice where the curve is below the x-axis ?
Thank you!!!
To solve the polynomial inequality x^3 + x^2 + 4x + 4 < 0, we need to find the values of x that make the inequality true.
Step 1: Factor the polynomial
x^3 + x^2 + 4x + 4
Unfortunately, this polynomial cannot be easily factored. So, we'll proceed using a different method.
Step 2: Find the critical points
To find the critical points, we set the polynomial equal to zero and solve for x.
x^3 + x^2 + 4x + 4 = 0
Since we cannot factor this equation, we can use numerical methods or a graphing calculator to find the approximate solutions. Using a graphing calculator, we find that the solutions are approximately x ≈ -2.239 and x ≈ -0.881.
Step 3: Test intervals
Now, we need to test intervals to determine the sign of the polynomial in each interval.
For simplicity, let's choose the intervals:
Interval 1: (-∞, -2.239)
Interval 2: (-2.239, -0.881)
Interval 3: (-0.881, ∞)
Step 4: Determine the sign of the polynomial in each interval
Evaluate the polynomial at a point in each interval to determine the sign.
Interval 1: Choose x = -3
Evaluate: f(-3) = (-3)^3 + (-3)^2 + 4(-3) + 4 = -17
The polynomial is negative in this interval.
Interval 2: Choose x = -1
Evaluate: f(-1) = (-1)^3 + (-1)^2 + 4(-1) + 4 = 1
The polynomial is positive in this interval.
Interval 3: Choose x = 0
Evaluate: f(0) = (0)^3 + (0)^2 + 4(0) + 4 = 4
The polynomial is positive in this interval.
Step 5: Determine the solution set
Since we're looking for when the polynomial is less than zero, we only need to consider the intervals where the polynomial is negative.
Interval 1: (-∞, -2.239), the polynomial is negative.
Step 6: Express the solution set in interval notation
The solution set in interval notation is:
(-∞, -2.239)
Step 7: Graph the solution set on a real number line
On a real number line, plot an open circle at -2.239 and shade to the left.
---------------●--->
-3 -2.239
This represents the solution set: (-∞, -2.239).
To solve the polynomial inequality x^3+x^2+4x+4<0, we first need to find the critical points, which are the values of x where the inequality changes.
Step 1: Set the inequality to zero:
x^3+x^2+4x+4 = 0
Step 2: Factorize the equation if possible:
(x+1)(x^2+4) = 0
The polynomial x^2+4 does not have any real roots, so the only root is x = -1.
Step 3: Divide the real number line into intervals using the critical point (-1) as a reference. Choose a test point from each interval and substitute it into the original inequality.
For x < -1, let's test x = -2:
(-2)^3 + (-2)^2 + 4(-2) + 4 = -8 + 4 - 8 + 4 = -8 < 0
For -1 < x, let's test x = 0:
0^3 + 0^2 + 4(0) + 4 = 4 > 0
Step 4: Analyze the signs of the test points to determine the solution set.
From the test points, we can observe that the inequality is true for x < -1 and false for -1 < x. Therefore, the solution set is x < -1.
Step 5: Express the solution set in interval notation.
The solution set, expressed in interval notation, is (-∞, -1).