Solution :
We have : \(2 \pm \sqrt{3}\) are two zeroes of the polynomial
p(x) = \(x^4 – 6x^3 – 26x^2 + 138x – 35\)
Let x = \(2 \pm \sqrt{3}\), So, x – 2 = \(\pm \sqrt{3}\)
Squaring, we get
\(x^2 – 4x + 4\) = 3, i.e. \(x^2 – 4x + 1\) = 0
Let us divide p(x) by \(x^2 – 4x + 1\) to obtain other zeroes.
\(\therefore\) p(x) = \(x^4 – 6x^3 – 26x^2 + 138x – 35\)
= (\(x^2 – 4x + 1\))(\(x^2 – 2x – 35\))
= (\(x^2 – 4x + 1\))(\(x^2 – 7x + 5x – 35\))
= (\(x^2 – 4x + 1\))(x + 5)(x – 7)
So, (x+ 5) and (x – 7) are the other factors of p(x).
\(\therefore\) -5 and 7 are other zeroes of the given polynomial.
