Questions: A polynomial P is given. P(x)=x^4+8x^2+16 (a) Find all zeros of P, real and complex. (Enter your answers as a comma-separated list. Enter all answers including repetitions) x= (b) Factor P completely. P(x)=

Solution Steps

(a) To find the zeros of the polynomial \( P(x) = x^4 + 8x^2 + 16 \), we can perform a substitution \( u = x^2 \), transforming the polynomial into a quadratic equation in terms of \( u \). Solve the quadratic equation for \( u \), and then solve for \( x \) by taking the square root of the solutions for \( u \).

(b) Once the zeros are found, we can express the polynomial as a product of its linear factors.

To find the zeros of the polynomial \( P(x) = x^4 + 8x^2 + 16 \), we first substitute \( u = x^2 \). This transforms the polynomial into:

\[ P(u) = u^2 + 8u + 16 \]

Next, we solve the quadratic equation \( u^2 + 8u + 16 = 0 \) using the quadratic formula:

\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{-8 \pm \sqrt{0}}{2} = -4 \]

Since \( u = x^2 \), we have:

\[ x^2 = -4 \]

Taking the square root gives us the zeros:

\[ x = \pm 2i \]

Thus, the zeros of \( P \) are:

\[ x = -2i, \quad x = 2i \]

Next, we factor the polynomial \( P(x) \) completely. Since we found that the only roots are \( x = -2i \) and \( x = 2i \), we can express \( P(x) \) as:

\[ P(x) = (x - 2i)(x + 2i)(x - 2i)(x + 2i) = ((x^2 + 4))^2 \]

Final Answer