The discussion focuses on proving that for the cubic function defined as x³ + bx² + cx + d = 0, if the roots are r and -s with r ≠ s, then it can be established that bc = d. The proof utilizes the complex conjugate root theorem, which asserts that if two roots are real (s and -s), the third root must also be real. By applying the factor theorem, the polynomial can be expressed as (x - k)(x² - s²), allowing for the coefficients to be equated to derive the relationship between b, c, and d.
PREREQUISITESMathematics students, educators, and anyone interested in polynomial equations and their properties, particularly those studying cubic functions and root relationships.
Gib Z said:I'll give you help on this one, but please, next time post these problems in the Homework Help section. Its not just for our sake, you'll get a faster reply there too.
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If all coefficients are real, by the complex conjugate root theorem you know that since two of the roots are real, s and -s, then the third root must be real as well. So now use the factor theorem and write the polynomial as a product of its factors,
x^3+bx^2+cx+d = (x-k)(x^2-s^2).
Equate coefficients and see what you get.