Question

How do you find all the real and complex roots of ${{x}^{3}}-2{{x}^{2}}+10x+136=0?$

Answer 1

Hint: We will first find all the real roots of the given polynomial equation using the rational root theorem. Then we will get a factor of the given polynomial. Using this factor, we can convert this cubic equation to a quadratic equation. Then, we will use the quadratic formula to find the solution of the obtained quadratic equation.

Complete step-by-step solution:
Let us consider the given cubic equation ${{x}^{3}}-2{{x}^{2}}+10x+136=0.$
We need to find all the real and complex roots of the given equation.
Let us use the rational root theorem to find the possible real roots of the equation.
So, the possible roots are $\pm 1,\pm 2,\pm 4,\pm 8,\pm 17,\pm 34,\pm 68,\pm 136.$
By inspection, we can see that $x=-4$ is a root of the given cubic equation.
So, we can say that $x+4$ is a factor of the polynomial.
Let us convert the cubic polynomial to a quadratic equation.
We can do this by synthetic division.
Let us write the coefficients of the given polynomial in descending powers and the obtained root in the left.
We will get $-4\left| \!{\underline {
   \begin{matrix}
   1 & -2 & 10 & 136 \\
\end{matrix} \\
  \begin{matrix}
   {} & {} & {} & {} \\
\end{matrix} \\
}} \right. $
Now, we will bring the leading coefficient $1$ to the bottom row, that is below the underline and we will multiply it with $4.$ Then we will write the product right beneath the next coefficient.
We will get
$
-4\left| \!{\underline {
\begin{matrix}
1 & -2 & 10 & 136 \\
\end{matrix} \\
\begin{matrix}
{} \;\;\;\;-4 & {} & {} \\
\end{matrix} \\
}} \right. \\
\,\,\,\,\,\,\,\,\ 1
$
We are going to add the second coefficient to the obtained number below it, the value will be written in the bottom row. Then we will multiply it with the divisor and we write the product beneath the third coefficient.
We will get
$
-4\left| \!{\underline {
\begin{matrix}
1 & -2 & 10 & 136 \\
\end{matrix} \\
\begin{matrix}
\;\;\;\;\;\; -4 & 24 & {} \\
\end{matrix} \\
}} \right. \\
\,\,\,\,\,\,\,\,\,1\,\,\,-6
$
We will repeat the procedure we have done in the previous step.
We will get
$
-4\left| \!{\underline {
\begin{matrix}
1 & -2 & 10 & 136 \\
\end{matrix} \\
\begin{matrix}
{} \;\;\;\;-4 \;\;\;\;24 \;\; -136 \\
\end{matrix} \\
}} \right. \\
\,\,\,\,\,\,\,\,\,1\,\,\,-6\,\,\,\,\,\,\, 34\,\,\,\,\,\,\,\,\,\,\,\, 0
$
We have obtained the coefficients of the corresponding quadratic equation.
So, the quadratic equation is ${{x}^{2}}-6x+34=0.$
We will use the quadratic formula to find the solution.
We will get $x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 1\times 34}}{2}$
This will give us $x=\dfrac{6\pm \sqrt{36-136}}{2}=\dfrac{6\pm \sqrt{-100}}{2}=\dfrac{6\pm \sqrt{-1\times 100}}{2}=\dfrac{6\pm 10\sqrt{-1}}{2}=\dfrac{6\pm 10i}{2}.$
We will get the complex roots $x=3\pm 5i.$
Hence the roots are $x=-4,3\pm 5i.$

Note: The rational root theorem states that for a polynomial equation ${{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+...+{{a}_{1}}x+{{a}_{0}}=0,$
If the rational number $\dfrac{r}{s}$ with $\gcd \left( r,s \right)=1$ is a root then $r$ divides ${{a}_{0}}$ and $s$ divides ${{a}_{n}}.$