Question

Given that $\left( {x - \sqrt 5 } \right)$ is a factor of the cubic polynomial (${x^3} - 3\sqrt 5 {x^2} + 13x - 3\sqrt 5 $), find all its other zeroes.

Answer 1

Hint: In this particular question use the concept that in a cubic polynomial if one of the root is given then other two roots of the polynomial together makes a quadratic equation so assume general quadratic equation and multiply with the given factor and then compare its terms with the given cubic polynomial so use these concepts to reach the solution of the question.

Complete step by step answer:
Given cubic polynomial is
${x^3} - 3\sqrt 5 {x^2} + 13x - 3\sqrt 5 $
Given factor of the cubic polynomial is $\left( {x - \sqrt 5 } \right)$
So one of the zero or root of the cubic polynomial is = $\sqrt 5 $
As we all know that are at most three roots or zeros in a cubic polynomial.
In which one of the roots or zero is given which is $\sqrt 5 $.
Now together other two roots of the cubic polynomial make a quadratic equation, so when this quadratic equation multiplied by the given root it is given the original cubic polynomial.
So the quadratic equation be $a{x^2} + bx + c$, so the multiplication of this quadratic equation with the given root i.e. $\left( {x - \sqrt 5 } \right)$ it will give us the resultant cubic polynomial.
Where, a, b and c are some constants.
Therefore,
$ \Rightarrow \left( {a{x^2} + bx + c} \right)\left( {x - \sqrt 5 } \right) = {x^3} - 3\sqrt 5 {x^2} + 13x - 3\sqrt 5 $
Now simplify this we have,
$ \Rightarrow \left( {a{x^3} - a\sqrt 5 {x^2} + b{x^2} - bx\sqrt 5 + cx - c\sqrt 5 } \right) = {x^3} - 3\sqrt 5 {x^2} + 13x - 3\sqrt 5 $
$ \Rightarrow \left( {a{x^3} - \left( {a\sqrt 5 - b} \right){x^2} - \left( {b\sqrt 5 - c} \right)x - c\sqrt 5 } \right) = {x^3} - 3\sqrt 5 {x^2} + 13x - 3\sqrt 5 $
Now on comparing we have,
$ \Rightarrow a = 1$................. (1)
$ \Rightarrow a\sqrt 5 - b = 3\sqrt 5 $.............. (2)
$ \Rightarrow b\sqrt 5 - c = - 13$................ (3)
$ \Rightarrow c\sqrt 5 = 3\sqrt 5 $................ (4)
Now from equation (4) we have,
c = 3
Now substitute the value of c in equation (3) we have,
$ \Rightarrow b\sqrt 5 - 3 = - 13$
$ \Rightarrow b\sqrt 5 = - 10$
$ \Rightarrow b = - \dfrac{{10}}{{\sqrt 5 }}$
$ \Rightarrow b = - 2\sqrt 5 $
So the quadratic equation becomes,
$ \Rightarrow {x^2} - 2\sqrt 5 x + 3 = 0$
Now apply quadratic formula we have,
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow x = \dfrac{{2\sqrt 5 \pm \sqrt {{{\left( {2\sqrt 5 } \right)}^2} - 4\left( 1 \right)\left( 3 \right)} }}{{2\left( 1 \right)}}$
Now simplify we have,
\[ \Rightarrow x = \dfrac{{2\sqrt 5 \pm \sqrt {20 - 12} }}{2}\]
\[ \Rightarrow x = \dfrac{{2\sqrt 5 \pm \sqrt 8 }}{2} = \dfrac{{2\sqrt 5 \pm 2\sqrt 2 }}{2} = \sqrt 5 \pm \sqrt 2 \]
$ \Rightarrow x = \left( {\sqrt 5 + \sqrt 2 } \right),\left( {\sqrt 5 - \sqrt 2 } \right)$
So this is the required other two zeroes of the cubic polynomial.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the quadratic formula to solve the complex quadratic equation which is stated above, so first find the quadratic equation as above, then use this formula we will get the required other zeroes of the cubic polynomial.