Write a polynomial with the following given zeros (roots): \( -2,-3,4-3 i \)

We are given the zeros –2, –3, and 4 – 3i. Since we want a polynomial with real coefficients, we must include the complex conjugate of 4 – 3i, which is 4 + 3i, as a zero. Step 1. Write the linear factors corresponding to each zero:   Factor 1: (x + 2)   Factor 2: (x + 3)   Factor 3: (x – (4 – 3i))  = (x – 4 + 3i)   Factor 4: (x – (4 + 3i))  = (x – 4 – 3i) Step 2. Multiply the complex conjugate factors first:   (x – 4 + 3i)(x – 4 – 3i) This product is of the form (a + b)(a – b) = a² – b², where a = x – 4 and b = 3i. Thus,   = (x – 4)² – (3i)²   = (x – 4)² – (–9)    [since (3i)² = –9]   = (x – 4)² + 9 Expanding (x – 4)²:   = x² – 8x + 16 + 9   = x² – 8x + 25 Step 3. Now multiply with the remaining factors (x + 2) and (x + 3). First, combine them:   (x + 2)(x + 3) = x² + 5x + 6 Step 4. Multiply the quadratic factors:   (x² + 5x + 6)(x² – 8x + 25)   Performing the multiplication:    x²·x² = x⁴    x²·(–8x) = –8x³    x²·25 = 25x²    5x·x² = 5x³    5x·(–8x) = –40x²    5x·25 = 125x    6·x² = 6x²    6·(–8x) = –48x    6·25 = 150 Now, combine like terms:   x⁴ term: x⁴   x³ terms: (–8x³ + 5x³) = –3x³   x² terms: (25x² – 40x² + 6x²) = –9x²   x terms: (125x – 48x) = 77x   Constant: 150 Thus, the polynomial is:   P(x) = x⁴ – 3x³ – 9x² + 77x + 150 This is a monic polynomial with the given zeros.