Obtain all the zeroes of \(3x^4 + 6x^3 – 2x^2 – 10x – 5\), if two of its zeroes are \(\sqrt{5\over 3}\) and -\(\sqrt{5\over 3}\).

Solution :

Since two zeroes are \(\sqrt{5\over 3}\) and -\(\sqrt{5\over 3}\),

x = \(\sqrt{5\over 3}\) and x = -\(\sqrt{5\over 3}\)

\(\implies\) (x – \(\sqrt{5\over 3}\))(x + \(\sqrt{5\over 3}\)) = \(3x^2 – 5\) is a factor of the given polynomial. Now, we apply the division algorithm to the given polynomial and \(3x^2 – 5\).

polynomial image

First term of quotient is \(3x^4\over 3x^2\) = \(x^2\)

Second term of quotient is \(6x^3\over 3x^2\) = 2x

Third term of the quotient is \(3x^2\over 3x^2\) = 1

So, \(3x^4 + 6x^3 – 2x^2 – 10x – 5\) = (\(3x^2 – 5\))(\(x^2 + 2x + 1\)) + 0 = (\(3x^2 – 5\))\({(x + 1)}^2\)

Quotient = \(x^2 + 2x + 1\) = \({(x + 1)}^2\)

Zeroes of \({(x + 1)}^2\) are -1 and -1.

Hence, all its zeroes are \(\sqrt{5\over 3}\), -\(\sqrt{5\over 3}\), -1, -1.

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