Final answer:
The nth-degree polynomial with real coefficients, a leading coefficient of 1, and zeros-3, 5, and 4 + 3i is P(x) = x^4 - 10x^3 + 5x^2 + 170x - 375.
Step-by-step explanation:
To find an nth-degree polynomial with real coefficients, a leading coefficient of 1, and given zeros, we can use the fact that each zero corresponds to a factor of the polynomial. Given the zeros -3, 5, and 4 + 3, we must first recognize that real coefficients imply that complex zeros come in conjugate pairs. Since we have a complex zero 4 + 3i, its conjugate 4 - 3i must also be a zero.
Now we can construct the factors of the polynomial:
- (x + 3) corresponding to the zero -3
- (x - 5) corresponding to the zero 5
- (x - (4 + 3i)) corresponding to the zero 4 + 3i
- (x - (4 - 3i)) corresponding to the zero 4 - 3i
Multiplying these factors together, we get the following:
- For the complex factors: (x - 4 - 3i)(x - 4 + 3i) = (x - 4)^2 - (3i)^2 = x^2 - 8x + 16 + 9 = x^2 - 8x + 25
- For the real factors: (x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15
The nth-degree polynomial, where n = 4, is the product of these quadratic polynomials:
P(x) = (x^2 - 2x - 15)(x^2 - 8x + 25)
When expanded, this equals:
P(x) = x^4 - 10x^3 + 5x^2 + 170x - 375