Final answer:
The zeros of P(x) = x³ - 10x² + 42x - 72 are 3-3i, 3+3i, and 4. Since complex zeros come in pairs, and using the sum of zeros, the remaining real zero is determined to be 4.
Step-by-step explanation:
To find all other zeros of the polynomial P(x) = x³ - 10x² + 42x - 72, given that 3-3i is a zero, we need to use the fact that complex zeros of polynomials with real coefficients come in conjugate pairs. This means that if 3-3i is a zero, then its conjugate 3+3i is also a zero.
Now we have two zeros of the polynomial: 3-3i and 3+3i. To find the remaining zero, we can perform synthetic division or long division with one of these zeros to reduce the polynomial to a quadratic form, after which we can use the quadratic formula to find the last zero.
However, we can bypass the division step by recognizing that the sum of the zeros of a cubic polynomial (with leading coefficient 1) is equal to the negative of the coefficient of the x² term. So, if we let the remaining zero be r, we have 3-3i + 3+3i + r = 10. Simplifying the left side gives 6 + r = 10, therefore r is 4.
Thus, the zeros of P(x) are 3-3i, 3+3i, and 4.