Answer :

For finding zeroes, we’ll use the hit and trial method.

For that, we’ll put a value which will make our polynomial zero.

Since, here the whole equation is positive that means, zero can be achieved by putting the negative values only.

So, first we’ll put x = – 1.

⇒ (– 1)^{3} + 6(– 1)^{2} + 11(– 1) + 6 = – 1 + 6 – 11 + 6

= 0

⇒ x = – 1 is one of the zeroes.

Now, we’ll write the equation in terms of (x + 1) to make equation quadratic and hence to find the other zeroes.

⇒ x^{3} + 6x^{2} + 11x + 6 = x^{2} (x + 1) + 5x(x + 1) + 6(x + 1)

= (x + 1)(x^{2} + 5x + 6)

Now, using factorization by splitting the middle term method

= (x + 1)(x^{2} + (3 + 2)x + 6)

= (x + 1)(x^{2} + 3x + 2x + 6)

= (x + 1)(x(x + 3) + 2(x + 3))

= (x + 1)(x + 2)(x + 3)

Now, to get the zeroes we’ll put these equals to zero.

⇒ x + 1 = 0, x + 2 = 0, x + 3 = 0

⇒ x = – 1, x = – 2, x = – 3 are the zeroes of x^{3} + 6x^{2} + 11x + 6

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