Question

What are the zeros in ${x^3} + 3{x^2} - 3x - 5$ ?

Answer 1

Hint: Given equation is a cubic equation. So there should be definitely three roots for the equation above. Finding the zeros is nothing but finding the values of x that would satisfy the equation above. So, we will first find one of the zeroes of the given equation through the hit and trial method. Then, we will factorise the equation to get the other quadratic factor. Then, we will apply the quadratic formula to get the remaining two zeroes from the quadratic equation.

Complete step by step answer:
Given equation is ${x^3} + 3{x^2} - 3x - 5$. Now, we will use the hit and trial method to get the first zero. So, we try to put x as $ - 1$. So, we get,
$ \Rightarrow {\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} - 3\left( { - 1} \right) - 5$
Opening the brackets,
$ \Rightarrow \left( { - 1} \right) + 3\left( 1 \right) - 3\left( { - 1} \right) - 5$
$ \Rightarrow - 1 + 3 + 3 - 5 = 0$
So, $ - 1$ is a zero of the polynomial ${x^3} + 3{x^2} - 3x - 5$.

Now, according to the factor theorem, if $ - 1$ is a zero, then $x - \left( { - 1} \right)$ is a factor of the polynomial. So, we get $x + 1$ as a factor of ${x^3} + 3{x^2} - 3x - 5$.Now, we divide ${x^3} + 3{x^2} - 3x - 5$ by $x + 1$ to get the other quadratic factor. So, we get,
\[x+1\overset{{{x}^{2}}+2x-5}{\overline{\left){\begin{align}
  & {{x}^{3}}+{3{x}^{2}}-3x-5 \\
 & \underline{{{x}^{3}}+{{x}^{2}}} \\
 & 2{{x}^{2}}-3x-5 \\
 & \underline{2{{x}^{2}}+2x} \\
 & -5x-5 \\
 & \underline{-5x-5} \\
 & 0 \\
\end{align}}\right.}}\]
So, we get ${x^2} + 2x - 5$ as a quotient. So, ${x^2} + 2x - 5$ is the other quadratic factor.

Now, we get, ${x^3} + 3{x^2} - 3x - 5 = \left( {x + 1} \right)\left( {{x^2} + 2x - 5} \right)$. So, we need to find the zeros of the polynomial $\left( {{x^2} + 2x - 5} \right)$. So, we use quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to get the other two zeros from $\left( {{x^2} + 2x - 5} \right)$, where $a = 1$, $b = 2$ and $c = - 5$. Hence, we get,
$x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4\left( 1 \right)\left( { - 5} \right)} }}{{2\left( 1 \right)}}$

Simplifying the calculations, we get,
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 + 20} }}{{2\left( 1 \right)}}$
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {24} }}{2}$
Cancelling common factors from numerator and denominator, we get,
$ \Rightarrow x = \dfrac{{ - 2 \pm 2\sqrt 6 }}{2}$
$ \therefore x = - 1 \pm \sqrt 6 $

So, we get the zeroes of the polynomial ${x^3} + 3{x^2} - 3x - 5$ as: $ - 1 - \sqrt 6 $, $\sqrt 6 - 1$ and $\left( { - 1} \right)$.

Note: The polynomial ${x^3} + 3{x^2} - 3x - 5$ has the highest degree of the variable x as three. So, it is called a cubic polynomial. If one root of a quadratic equation with rational coefficients is irrational, then the other root will also be the conjugate pair of the first irrational root. For example: If one root is $a + \sqrt b $, then the root will be $a - \sqrt b $.