If two zeroes of the polynomial p(x) = x4 - 6x3 -26x2 +138x -35 are 2 +root 3, find the - Maths - Polynomials

Question

If two zeroes of the polynomial p(x) = x4 -
6x3 -26x2 +138x -35 are 2 +root 3,
find the other zeroes

Santosh Kumar · Accepted Answer

As $x = 2 \pm \sqrt{3}$ are the zeroes of p(x), so $x - (2 \pm \sqrt{3})$ are the factors of p(x).

$Now, {x - (2 + \sqrt{3})} {x - (2 - \sqrt{3})} = {(x - 2) - \sqrt{3}} {(x - 2) + \sqrt{3}} = (x - 2)^{2} - (\sqrt{3})^{2} {= x}^{2} - 4 x + 1$

Dividing p(x) by x² – 4x + 1

$∴ p (x) = {x - (2 + \sqrt{3})} {x - (2 - \sqrt{3})} (x^{2} - 2 x - 35) As, x^{2} - 2 x - 35 = (x + 5) (x - 7)$

Hence, other two zeroes of p(x) are – 5 and 7.

$∴ p (x) = {x - (2 + \sqrt{3})} {x - (2 - \sqrt{3})} (x^{2} - 2 x - 35) As, x^{2} - 2 x - 35 = (x + 5) (x - 7)$ Hence, other two zeroes of p(x) are â€“ 5 and 7

If two zeroes of the polynomial p(x) = x4 - 6x3 -26x2 +138x -35 are 2 +root 3, find the other zeroes.

If two zeroes of the polynomial p(x) = x⁴ - 6x³ -26x² +138x -35 are 2 +root 3, find the other zeroes.