# 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 $$3x^2 – 5$$.

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.