Answer :

p(x) = 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

Since the two zeroes are

is a factor of 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

Therefore, we divide the given polynomial by

We know,

Dividend = (Divisor × quotient) + remainder

3 x^{4} + 6 x^{3} – 2x^{2} – 10 x – 5

3 x^{4} + 6 x^{3} – 2 x^{2} – 10 x – 5

As (a + b)^{2} = a^{2} + b^{2} + 2ab

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

Therefore, its zero is given by x + 1 = 0, x = −1

**Hence, the zeroes of the given polynomial are** **and – 1, –1.**

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