Thread: zeros of polynomials

Solve:
x^4-10x^3+42x^2-82x+65=0 the product of two of the roots of this equation is 13. The answers are complex numbers.
Thanks for helping me

Originally Posted by Anand Parikh

Solve:
x^4-10x^3+42x^2-82x+65=0 the product of two of the roots of this equation is 13. The answers are complex numbers.
Thanks for helping me

Vieta's Root formulas should help here ..... Start with the product of all the roots being equal to 65 .......

Originally Posted by Anand Parikh

Solve:
x^4-10x^3+42x^2-82x+65=0 the product of two of the roots of this equation is 13. The answers are complex numbers.
Thanks for helping me

Because this is a quartic with real coefficients this factorises into a pair of real quadratics:



where are all real. Also as the product of a pair of roots is , this suggests we try and . Doing this we find:



and from there it is easy.

RonL

Hello, Anand Parikh!

Solve: .
The product of two of the roots of this equation is 13.
The roots are complex numbers.

I approached it like CaptainBlack and got very lucky . . .

Since the product of two roots is 13,
. . I guessed that the roots were: .


If the roots are: .
. . then: . .are factors of
Hence: . .is a factor of

. . But does not divide


If the roots are: .
. . then: . .are factors of
Hence: . .is a factor of

. . And we have: .