What is the solution to x^3 +4x^2 > x + 4

B) −4<x<−1  or  x>1

Inequality Form: −4<x<−1  or  x>1

Interval Notation: (−4,−1)  ∪  (1,∞)

Step-by-step explanation:

Solve: x^3 + 4x^2 > x + 4

Subtract x from both sides of the inequality.

x^3 + 4x^2 − x > 4

Convert the inequality to an equation.

x^3 + 4x^2 − x = 4

Move 4 to the left side of the equation by subtracting it from both sides.

x^3 + 4x^

2 − x − 4 = 0

Factor the left side of the equation

(x + 4)(x + 1)(x − 1) = 0

If any individual factor on the left side of the equation is equal to , the entire expression will be equal to 0

x + 4 = 0

x + 1 = 0

x − 1 = 0

Set x + 4 equal to 0 and solve for x

x = −4

Set x + 1 equal to 0 and solve for x

x = −1

Set x + -1 equal to 0 and solve for x

x = 1

The final solution is all the values that make (x + 4)(x + 1)(x − 1) = 0  true

x = −4, −1, 1

Use each root to create test intervals.

x < −4

−4 < x < −1

−1 < x < 1

x > 1

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the

inequality.

x < −4  ←False

−4 < x < −1  ←True

−1 < x < 1  ←False

x > 1  ←True

The solution consists of all of the true intervals.

−4 < x < −1 x > 1

I believe the answer would be C) 4 < x < –1 or x > 1

Step-by-step explanation:

If you use desmos, you can see that the solution are located between -4 and -1 and that the solutions also include any number greater than 1.

Have a great day! You got this :)

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step-by-step explanation: