40. Write a polynomial of least degree with the given zeroes. a. \( -2,1+\sqrt{7} \)

To find a polynomial of the least degree with the given zeros, we'll assume that the polynomial has **rational coefficients**. This typically means that if a polynomial has an irrational zero like \(1 + \sqrt{7}\), its conjugate \(1 - \sqrt{7}\) must also be a zero to ensure that all coefficients remain rational. ### Given Zeros: - \(-2\) - \(1 + \sqrt{7}\) ### Additional Zero (Conjugate): - \(1 - \sqrt{7}\) Now, the zeros of the polynomial are: \[ -2, \quad 1 + \sqrt{7}, \quad 1 - \sqrt{7} \] ### Constructing the Polynomial: The polynomial can be constructed by multiplying factors corresponding to each zero: \[ P(x) = (x + 2)(x - (1 + \sqrt{7}))(x - (1 - \sqrt{7})) \] First, simplify the product of the last two factors using the difference of squares: \[ (x - (1 + \sqrt{7}))(x - (1 - \sqrt{7})) = (x - 1)^2 - (\sqrt{7})^2 = x^2 - 2x + 1 - 7 = x^2 - 2x - 6 \] Now, multiply this result by the first factor: \[ P(x) = (x + 2)(x^2 - 2x - 6) \] Expanding this product: \[ P(x) = x(x^2 - 2x - 6) + 2(x^2 - 2x - 6) = x^3 - 2x^2 - 6x + 2x^2 - 4x - 12 \] \[ P(x) = x^3 - 10x - 12 \] ### Final Polynomial: \[ P(x) = x^3 - 10x - 12 \] **Answer:** One such polynomial is x³ minus ten x minus twelve. In symbols, P(x) = x³ − 10 x − 12