One of the solutions to the equation x3+x2−16x−16=0 is −4.%0D%0A%0D%0ATest the solution to find the full solution set.%0D%0A%0D%0AWhich of the following statements correctly describes the solution set?%0D%0A%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0AThe equation has all non-real solutions.%0D%0AThe equation has all non-real solutions.%0D%0A%0D%0AThe equation has two real solutions and one non-real solution.%0D%0AThe equation has two real solutions and one non-real solution.%0D%0A%0D%0AThe equation has all real solutions.%0D%0AThe equation has all real solutions.%0D%0A%0D%0AThe equation has one real solution and two non-real solutions.

1 answer

To determine the full solution set for the equation \(x^3 + x^2 - 16x - 16 = 0\), we start by verifying the given root \(-4\).

Step 1: Verify if \(-4\) is a root.

Substituting \(x = -4\) into the equation: \[ (-4)^3 + (-4)^2 - 16(-4) - 16 = -64 + 16 + 64 - 16 = 0. \] Since the expression equals \(0\), \(-4\) is indeed a root.

Step 2: Factor the polynomial.

Since \(-4\) is a root, we can factor the polynomial using synthetic division or polynomial long division. Let's perform synthetic division with \(-4\):

\[ \begin{array}{r|rrrr} -4 & 1 & 1 & -16 & -16 \ & & -4 & 12 & 16 \ \hline & 1 & -3 & -4 & 0 \ \end{array} \]

The result shows that: \[ x^3 + x^2 - 16x - 16 = (x + 4)(x^2 - 3x - 4). \]

Step 3: Factor the quadratic.

Now, we need to factor the quadratic \(x^2 - 3x - 4\). We can either use the quadratic formula or factor it directly: \[ x^2 - 3x - 4 = (x - 4)(x + 1). \]

Step 4: Write the full factorization.

The complete factorization of the original polynomial is: \[ (x + 4)(x - 4)(x + 1) = 0. \]

Step 5: Find the roots.

Setting each factor to zero gives the roots:

  1. \(x + 4 = 0 \Rightarrow x = -4\)
  2. \(x - 4 = 0 \Rightarrow x = 4\)
  3. \(x + 1 = 0 \Rightarrow x = -1\)

The solutions to the equation are: \[ x = -4, \quad x = 4, \quad x = -1. \]

Conclusion:

Since all solutions are real numbers, the statement that accurately describes the solution set is:

The equation has all real solutions.