Surely. To find the monic cubic polynomial with the roots 3, 1, and -5, we follow these steps:
1. We start by knowing that a polynomial is defined by its zeros. Given a polynomial p(x), if a given real number 'a' satisfies the equation p(a) = 0, then 'a' is called the zero or root of the given polynomial.
2. The zero-product property states that a product of factors is zero if and only if at least one of the factors is zero. Thus, if 3, 1, and -5 are the zeros of our polynomial, we can express it in a factored form: p(x) = k(x - 3)(x - 1)(x + 5). Here, 'k' is a constant coefficient.
3. To make the polynomial 'monic' (i.e., making the coefficient of the highest power term equal to 1), we set k equal to 1. This allows us to write our polynomial as: p(x) = (x - 3)(x - 1)(x + 5).
4. We now expand this product to express our polynomial in standard form: p(x)=x^3 + x^2 - 17x + 15.
So, the cubic polynomial with zeros at 3, 1, and -5 is x^3 + x^2 - 17x + 15.