If two zeroes of the polynomial \(x^4 – 6x^3 – 26x^2 + 138x – 35\) are \(2 \pm \sqrt{3}\), find other zeroes.

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.

polynomial image

\(\therefore\)  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.

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