# zeros of polynomials

• May 18th 2008, 03:39 AM
Anand Parikh
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(Rofl)
• May 18th 2008, 04:22 AM
mr fantastic
Quote:

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(Rofl)

Vieta's Root formulas should help here ..... Start with the product of all the roots being equal to 65 .......
• May 18th 2008, 06:40 AM
CaptainBlack
Quote:

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(Rofl)

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

$x^4-10x^3+42x^2-82x+65=(x^2+a x +b)(x^2+ c x +d)$

where $a, b, c, d$ are all real. Also as the product of a pair of roots is $13$, this suggests we try $b=13$ and $d=5$. Doing this we find:

$x^4-10x^3+42x^2-82x+65=(x^2-6x+13)(x^2-4x+5)$

and from there it is easy.

RonL
• May 18th 2008, 08:01 AM
Soroban
Hello, Anand Parikh!

Quote:

Solve: . $f(x) \:=\:x^4-10x^3+42x^2-82x+65\:=\:0$
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: . $2 \pm 3i\,\text{ or }\,3\pm2i$

If the roots are: . $2 \pm3i$
. . then: . $\left(x - [2 + 3i]\right)\,\text{ and }\,\left(x-[2-3i]\right)$ .are factors of $f(x).$
Hence: . $(x-2-3i)(x-2+3i) \:=\:x^2-4x+13$ .is a factor of $f(x).$

. . But $x^2-4x+13$ does not divide $f(x).$

If the roots are: . $3 \pm2i$
. . then: . $\left(x - [3+2i]\right),\text{ and }\, \left(x - [3-2i]\right)$ .are factors of $f(x).$
Hence: . $(x-3-2i)(x-3+2i) \:=\:x^2 -6x + 13$ .is a factor of $f(x).$

. . And we have: . $f(x)\;=\;(x^2-6x+13)(x^2-4x+5) \quad\hdots\quad \text{ta-}DAA!$