Question

Find all the zeros of \[{x^3} + 6{x^2} + 11x + 6\;\]if \[(x + 1)\;\]is a factor.

Answer 1

Hint: We start with taking \[(x + 1)\;\] common from the equation \[{x^3} + 6{x^2} + 11x + 6\;\] as this is a factor of the cubic equation. Then we will try to factorize the result we get by the splitting of terms. In this way, we reach the roots of the equation which we need to find.

Complete step by step answer:

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, the zeros of \[{x^3} + 6{x^2} + 11x + 6\;\]if \[(x + 1)\;\] is a factor are, \[ - 1, - 2, - 3\]

Note: A cubic equation is an equation which can be represented in the form \[{\text{a}}{{\text{x}}^{\text{3}}}{\text{ + b}}{{\text{x}}^{\text{2}}}{\text{ + cx + d}}\], where \[{\text{a}} \ne {\text{0}}\]
As a cubic polynomial has three roots (not necessarily distinct) by the fundamental theorem of algebra, at least one root must be real.