Questions: Factor completely using complex-numbers. y=x^4-5x^2-36. A. y=(x-3)^2(x+2i)(x-2i) B. y=(x+3)(x-3)(x-2i)(x+2i) C. y=(x-3)(x+3)(x-2i)^2 D. y=(x^2-2i)(x-3)(x+3)

Transcript text: 6. Factor completely using complex-numbers. $y=x^{4}-5 x^{2}-36$. A. $y=(x-3)^{2}(x+2 i)(x-2 i)$ B. $y=(x+3)(x-3)(x-2 i)(x+2 i)$ C. $y=(x-3)(x+3)(x-2 i)^{2}$ D. $y=\left(x^{2}-2 i\right)(x-3)(x+3)$

Solution

Solution Steps

To solve the problem of factoring the polynomial $ y = x^4 - 5x^2 - 36 $ completely using complex numbers, we can start by recognizing it as a quadratic in terms of $ x^2 $. We can substitute $ z = x^2 $, turning the equation into a quadratic $ z^2 - 5z - 36 = 0 $. We then solve for $ z $ using the quadratic formula. Once we have the roots for $ z $, we substitute back $ x^2 = z $ and solve for $ x $, which may involve complex numbers.

Step 1: Recognize the Polynomial as a Quadratic in Terms of $ x^2 $

The given polynomial is $ y = x^4 - 5x^2 - 36 $. We can rewrite this as a quadratic in terms of $ z = x^2 $, resulting in the equation $ z^2 - 5z - 36 = 0 $.

Step 2: Solve the Quadratic Equation for $ z $

Using the quadratic formula $ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = -5 $, and $ c = -36 $, we find the discriminant:

\[ \text{discriminant} = \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-36)} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

Thus, the roots for $ z $ are:

\[ z_1 = \frac{5 + 13}{2} = 9, \quad z_2 = \frac{5 - 13}{2} = -4 \]

Step 3: Solve for $ x $ Using the Roots of $ z $

Substitute back $ x^2 = z $ to find $ x $:

For $ z_1 = 9 $: \[ x^2 = 9 \implies x = \pm 3 \]

For $ z_2 = -4 $: \[ x^2 = -4 \implies x = \pm 2i \]