Question

complex numbers & higher order polynomials quick check
one of the solutions to the equation $x^3 + x^2 - 16x - 16 = 0$ is $-4$.
test the solution to find the full solution set.
which of the following statements correctly describes the solution set?
(1 point)
the equation has all non-real solutions
the equation has two real solutions and one non-real solution
the equation has one real solution and two non-real solutions
the equation has all real solutions

Explanation:

Step1: Factor the polynomial

Given the equation \(x^3 + x^2 - 16x - 16 = 0\) and a root \(x = -4\), we can factor the polynomial by dividing it by \((x + 4)\) (since if \(x = -4\) is a root, \((x + 4)\) is a factor). Using polynomial long division or synthetic division:

Using synthetic division with root \(-4\):

$$
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$$

So the polynomial factors as \((x + 4)(x^2 - 3x - 4)\).

Step2: Factor the quadratic

Now factor the quadratic \(x^2 - 3x - 4\). We need two numbers that multiply to \(-4\) and add to \(-3\). Those numbers are \(-4\) and \(1\). So, \(x^2 - 3x - 4 = (x - 4)(x + 1)\).

Step3: Find all roots

Putting it all together, the factored form of the polynomial is \((x + 4)(x - 4)(x + 1) = 0\). Setting each factor equal to zero gives the roots:

• \(x + 4 = 0 \implies x = -4\)
• \(x - 4 = 0 \implies x = 4\)
• \(x + 1 = 0 \implies x = -1\)

All three roots (\(-4\), \(4\), and \(-1\)) are real numbers.

Answer:

The equation has all real solutions.