$2 + \sqrt{3}$ and $2 - \sqrt{3}$ are the zeros of $p (x) = x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35$ . Find the remaining zeros of $p (x)$ .

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Solution

$P (n) = x^{4} - 6 x^{3} - 26 x^{2} + 138 - 35$
as $(2 + \sqrt{3})$ and $(2 - \sqrt{3})$ are joas of polynomial of $P (a)$
So, we divid $(R - (2 + \sqrt{3})) (x - (2 \sqrt{3}))$ then find quotient
$x^{2} - 2 x - \sqrt{3} x - 2 x - \sqrt{3} x + 1 = x^{2} - 4 x + 1$

$x^{2} + 4 x + 1) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ x^{4} - {6 x}^{3} - {26 x}^{2} + 138 x - 35 (n^{2} - 2 x - 35 x^{4} - {4 x}^{3} + x^{2}_____________- {2 x}^{3} - {27 x}^{2} + 138 x - {2 x}^{3} + {8 x}^{2} - 2 a_- 35 x^{2} + 140 x - 35 - 35 x^{2} + 140 x - 35______________0$

$x^{2} - 2 x - 35 = x^{2} - 7 x + 5 x - 35 = x (x - 7) + 5 (x - 7) = 0$
$x = - 5, 7$
$∴ 7$ and $- 5$ are zeros of polynomial

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Similar questions

x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35

2 + \sqrt{3}

2 - \sqrt{3}

(x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35)

(2 + \sqrt{3})

(2 - \sqrt{3})

x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35

2 \pm \sqrt{3}

f (x) = x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35

2 \pm \sqrt{3}

If two zeros of the polynomial f(x) = $x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35$ are $2 \pm \sqrt{3},$ find other zeros.

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