The equation ${x^5} - 209x + 56 = 0$ has two roots whose product is unity. These two roots are
- $2 + \sqrt 3 ,2 - \sqrt 3 $
- $4 + \sqrt {15} ,4 - \sqrt {15} $
- $8 + \sqrt {63} ,8 - \sqrt {63} $
- $6 + \sqrt {35} ,6 - \sqrt {35} $
If the product of two roots is 1, let the roots be a and 1/a. Using Vieta's formulas for a quintic x^5 - 209x + 56 = 0, the product of all roots is -56. The roots are of the form (a, 1/a, r1, r2, r3). Testing the options, 2+sqrt(3) and 2-sqrt(3) have a product of 4-3=1. Checking if they satisfy the equation confirms they are roots.
We can factor the given polynomial by assuming a quadratic factor of the form $x^2 - px + 1$, since the product of the two roots is unity. Let the polynomial be expressed as $(x^2 - px + 1)(x^3 + px^2 + qx - 56) = x^5 - 209x + 56$. By expanding and matching the coefficients of $x^4$, we find $p - p = 0$, which is consistent. Matching the coefficients of $x$ yields the equation $q - 56p = -209$, and matching the constant terms gives $-56 = 56$ if the constant isn't handled carefully, but testing the options directly shows that if the roots are $2 + \sqrt{3}$ and $2 - \sqrt{3}$, their sum is 4 and their product is 1. This makes the quadratic factor $x^2 - 4x + 1$, which when multiplied by $x^3 + 4x^2 + 15x + 56$ perfectly reconstructs $x^5 - 209x + 56$.