Question

hey..branlics
HERE IS UR QUE..



If the zeros of the polynomial
${x}^{4} - {6x}^{3} - {26x}^{2} + {138}^{x} - 35 \: are \: 2 + \sqrt{3} \: and \: 2 - \sqrt{3}$
FIND OTHER

NO SPAM..!!
THANK U...☺☺

Answer 1

HEY..!! HERE IS UR ANSWER.
______________________

INTRODUCTION....

Factorising 6
6 = 2 × 3
2 is factor of 6
3 is also factor of 6
so, 2×3 is also factor of 6

WE USE SAME IN OUR QUE..

$p(x) = {x}^{4} - {6x}^{3} - {26x}^{2} + 138x - 35 \\ given \: roots \: 2 + \sqrt{3} \: and \: 2 - \sqrt{3}$
$x = 2 + \sqrt{3} \: - \: is \: also \: factor \\ x - (2 + \sqrt{3} ) \: - is \: also \: factor\\ x - 2 - \sqrt{3} \: - \: is \: also \: factor$

$and \: x = 2 - \sqrt{3} \: - \: is \: also \: factor \\ \: \: \: \: \: \: \: \: x - (2 - \sqrt{3}) \: - \: is \: also \: factor\\ \: \: \: \: \: \: \: x - 2 + \sqrt{3} \: - \: is \: also \: factor$
$x - 2 - \sqrt{3} \times x - 2 + \sqrt{3} - \: is \: also \: factor \\ ((x - 2) - \sqrt{3} ) \times ((x - 2) + \sqrt{3} ) - \: is \: also \: factor$
${(x - 2)}^{2} - ({ \sqrt{3} })^{2}$

${ x}^{2} + {2}^{2} - 4x - 3 \\ {x}^{2} + 4 - 4x - 3 \\ {x}^{2} - 4x + 1$
THEREFORE,
${x}^{2} - 4x + 1 \: is \: a \: factor \: of \: p(x)$
NOW DIVIDING,
$p(x) \: by \: ( {x}^{2} - 4x + 1)$
WE CAN FIND OUT OTHER FACTOR...
${x}^{2} - 2x - 35$
NOW,
$we \: find \: zeroesof \: {x}^{2} - 2x - 35$
BY MIDDLE TERM SPILITTING METHOD...
${x}^{2} - 7x + 5x - 35 = 0 \\ x(x - 7) + 5(x - 7) = 0 \\ (x - 7) \: (x + 5) = 0 \\ x = 7 \: and \: x = - 5$
THEREFORE,
THE ZEROES OF P(X) ARE
$2 + \sqrt{3} and \: 2 - \sqrt{3} \: and \: - 5 \: and \: 7$
#SR

HOPE THIS HELPS...☺☺