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Our given equation is, \[{x^3} + 6{x^2} + 11x + 6\;\]

Since \[(x + 1)\;\]is a factor of the given cubic equation, it will completely divide the given equation.

Now, we will factorize the given equation

\[\left( {{x^3} + 6{x^2} + 11x + 6} \right)\]

On expansion of terms we get,

\[{x^3} + {x^2} + 5{x^2} + 5x + 6x + 6\]

On taking factors common we get,

\[{x^2}(x + 1) + 5x(x + 1) + 6(x + 1)\]

\[ \Rightarrow \left( {{x^2} + 5x + 6} \right)\left( {x + 1} \right)\]

On expansion we get,

\[({x^2} + 3x + 2x + 6)(x + 1)\]

On taking factors common we get,

\[[x(x + 3) + 2(x + 3)](x + 1)\]

\[ = \left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 1} \right)\]

∴ now, the zeros of \[\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\;\]would be,

For, \[(x + 1) = 0\] we get, \[x = - 1\]

As,

\[x + 1 = 0 \Rightarrow x = - 1\]

For, \[(x + 2) = 0\] we get, \[x = - 2\]

As,

\[x + 2 = 0 \Rightarrow x = - 2\]

For, \[(x + 3) = 0\] we get, \[x = - 3\]

As,

\[x + 3 = 0 \Rightarrow x = - 3\]

Therefore,

As a cubic polynomial has three roots (not necessarily distinct) by the fundamental theorem of algebra, at least one root must be real.