Proving bc=d for Cubic Function with r=-s

[Aki]

Hey I need help with this proof:

Consider the cubic function: x^3 + bx^2 + cx + d = 0 . If the two solutions of the cubic function are not equal, ie r != s, but r=-s, then prove that bc=d.

Thanks

 

[Gib Z]

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.

 

[Aki]

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.

-------------
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.

Very helpful. Thanks a lot Gib Z