Questions: Find a formula for the polynomial P(x) with - degree 3 - real coefficients - zeros at x=-2-2i and x=1 - y-intercept at (0,8) P(x)=

Transcript text: Find a formula for the polynomial $P(x)$ with - degree 3 - real coefficients - zeros at $x=-2-2 i$ and $x=1$ - $y$-intercept at $(0,8)$ \[ P(x)= \]

Solution

Solution Steps

Step 1: Identify the Zeros

The polynomial $ P(x) $ is of degree 3 with zeros at $ x = -2-2i $, $ x = -2+2i $ (the complex conjugate), and $ x = 1 $.

Step 2: Construct the Polynomial

The polynomial can be expressed in factored form as: \[ P(x) = a(x - (-2-2i))(x - (-2+2i))(x - 1) \] This simplifies to: \[ P(x) = a((x + 2 + 2i)(x + 2 - 2i))(x - 1) \]

Step 3: Simplify the Complex Factors

Using the difference of squares, we have: \[ (x + 2 + 2i)(x + 2 - 2i) = (x + 2)^2 + 4 \] Expanding $(x + 2)^2$: \[ (x + 2)^2 = x^2 + 4x + 4 \] Thus: \[ (x + 2)^2 + 4 = x^2 + 4x + 8 \]

Step 4: Expand the Polynomial

Now, substituting back, we get: \[ P(x) = a(x^2 + 4x + 8)(x - 1) \] Expanding this expression: \[ P(x) = a(x^3 + 3x^2 + 4x - 8) \]

Step 5: Determine the Leading Coefficient

Given the $ y $-intercept at $ (0, 8) $, we find: \[ P(0) = a(-8) = 8 \implies a = -1 \]

Step 6: Write the Final Polynomial

Substituting $ a $ back into the polynomial gives: \[ P(x) = -(x^3 + 3x^2 + 4x - 8) \]

Step 7: Factor the Polynomial

The polynomial can be factored as: \[ P(x) = -\left(x - 1\right)\left(x^{2} + 4x + 8\right) \]

This is the complete solution for the polynomial $ P(x) $.