Find a degree 3 polynomial with real coefficients having zeros 5 and \( 2 i \) and a lead coefficient of 1 . Write \( P \) in expanded form. \( P(x)=\square \)

To find a degree 3 polynomial with real coefficients having zeros 5 and \( 2i \) and a lead coefficient of 1, we can use the fact that complex zeros come in conjugate pairs. Since the zeros are 5 and \( 2i \), the conjugate of \( 2i \) is \( -2i \). The zeros of the polynomial are 5, \( 2i \), and \( -2i \). The polynomial can be expressed as: \[ P(x) = (x - 5)(x - 2i)(x + 2i) \] Expanding this expression will give us the expanded form of the polynomial. Let's calculate it. Simplify the expression by following steps: - step0: Calculate: \(\left(x-5\right)\left(x-2i\right)\left(x+2i\right)\) - step1: Multiply the terms: \(\left(x^{2}+\left(-5-2i\right)x+10i\right)\left(x+2i\right)\) - step2: Apply the distributive property: \(x^{2}\times x+x^{2}\times 2i+\left(-5-2i\right)x\times x+\left(-5-2i\right)x\times 2i+10ix+10i\times 2i\) - step3: Multiply the terms: \(x^{3}+2ix^{2}+\left(-5-2i\right)x^{2}+\left(4-10i\right)x+10ix-20\) - step4: Subtract the terms: \(x^{3}-5x^{2}+4x-20\) The expanded form of the degree 3 polynomial with real coefficients having zeros 5 and \( 2i \) and a lead coefficient of 1 is: \[ P(x) = x^{3} - 5x^{2} + 4x - 20 \]