Question

How do you find the roots of ${{x}^{3}}+x+10=0$?

Answer 1

Hint: We have to factorise the polynomial in the given equation. By the hit and trial method, we can find the factor of the polynomial and with the help of the factor theorem we will get one linear and one quadratic factor of the polynomial. From the linear factor we will get a linear equation solving which we will get a solution. From the quadratic factor, we will get a quadratic equation which can be solved by using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step by step answer:
The equation given in the question is
${{x}^{3}}+x+10=0........(i)$
In order to solve the equation, we need to factorise the polynomial on the left hand side.
Considering the polynomial on the LHS, we have
$\Rightarrow p\left( x \right)={{x}^{3}}+x+10........(ii)$
Substituting $x=-1$ in the polynomial, we get
$\begin{align}
  & \Rightarrow p\left( -1 \right)={{\left( -1 \right)}^{3}}+\left( -1 \right)+10 \\
 & \Rightarrow p\left( -1 \right)=8\ne 0 \\
\end{align}$
Now, we substitute $x=-2$ to get
$\begin{align}
  & \Rightarrow p\left( -2 \right)={{\left( -2 \right)}^{3}}+\left( -2 \right)+10 \\
 & \Rightarrow p\left( -2 \right)=-8-2+10 \\
 & \Rightarrow p\left( -2 \right)=0 \\
\end{align}$
Therefore, $x=-2$ is a factor of $p\left( x \right)$. By the factor theorem, we can say that $\left( x+2 \right)$ is a factor of $p\left( x \right)$. Also, by the remainder theorem, on dividing $p\left( x \right)$ by $\left( x+2 \right)$, we will get remainder equal to zero.
Therefore, on dividing $p\left( x \right)$ by $\left( x+2 \right)$, we get
\[x+2\overset{{{x}^{2}}-2x+5}{\overline{\left){\begin{align}
  & \underline{\begin{align}
  & {{x}^{3}}+x+10 \\
 & {{x}^{3}}+2{{x}^{2}} \\
\end{align}} \\
 & \underline{\begin{align}
  & -2{{x}^{2}}+x+10 \\
 & -2{{x}^{2}}-4x \\
\end{align}} \\
 & \underline{\begin{align}
  & 5x+10 \\
 & 5x+10 \\
\end{align}} \\
 & \underline{0} \\
\end{align}}\right.}}\]
From the above, $p\left( x \right)$ can be written as
$\Rightarrow p\left( x \right)=\left( x+2 \right)\left( {{x}^{2}}-2x+5 \right).......(iii)$
From (ii) and (iii) we can say that
$\Rightarrow {{x}^{3}}+x+10=\left( x+2 \right)\left( {{x}^{2}}-2x+5 \right)$
Substituting this in the equation (i), we get
$\left( x+2 \right)\left( {{x}^{2}}-2x+5 \right)=0$
We can say from the above equation that
$x+2=0$ and ${{x}^{2}}-2x+5=0$
From the first equation, we get
$x=-2$
So one root of the given equation is $-2$.
From the second equation, we have
$\Rightarrow {{x}^{2}}-2x+5=0........(iv)$
So we have a quadratic equation. By the quadratic formula, we know that the solution can be given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From (iv) we note that $a=1,b=-2,c=5$. Substituting these above, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 5 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{4-20}}{2} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{-16}}{2} \\
 & \Rightarrow x=\dfrac{4\pm 4i}{2} \\
 & \Rightarrow x=2\pm 2i \\
\end{align}\]

Hence, the roots of the given equation are $-2,2+i,2-i$.

Note: In the hit and trial method, we can decide the sign of the root of the cubic equation from the value of the constant term. If the constant term is positive, and all the other coefficients of the equation are positive, then the root will be negative. And if the constant term is negative, while all the other coefficients are positive, then the root will be positive.