Question

Solve the 4 degree polynomial equation \( x^4 − 2x^3 + 4x^2+ 6x − 21 = 0 \), given that the sum of two its roots is zero.

Answer 1

Let the roots be \( \alpha, \beta, \gamma, \delta \) such that \( \alpha + \beta = 0 \)

Also     \( \alpha+ \beta+ \gamma+ \delta=2 \)

It's mean    \( \gamma+ \delta=2 \)

Thus the quadratic factor corresponding to \( \alpha , \beta \) is of the form \( x^2 + 0x + p \)

and that corresponding to \( \gamma, \delta \) is of the form of \( x^2 − 2x + q. \)

So the eqation can also written as

\( x^4− 2x^3 + 4x^2 + 6x − 21 = (x^2 + p)(x^2 − 2x + q).......(i) \)

Equating coefficients of \( x^2 \)

 and x from both sides of (i), we get

\( 4 = p + q \)

\( 6 =- 2p \)

\( p = −3 \)

\( q = 7. \)

Hence the given equation is equivalent to

 \( (x^2 - 3)(x^2 - 2x + 7) = 0 \)

\( The\ roots\ are\ x = ±\sqrt3,\ 1 ± i \sqrt6. \)

\( The\ roots\ are\ x = ±\sqrt3,\ 1 ± i \sqrt6. \)