Question

How do you use the rational roots theorem to find all possible zeros of ${{x}^{4}}-5{{x}^{3}}-5{{x}^{2}}+23x+10$?

Answer 1

Hint: We factor the given equation with the help of vanishing method. In this method we find a number $a$ such that for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We assume $f\left( x \right)={{x}^{4}}-5{{x}^{3}}-5{{x}^{2}}+23x+10$ and take the value of $a$ as 5.

Complete step by step solution:
We find the value of $x=a$ for which the function \[f\left( x \right)={{x}^{4}}-5{{x}^{3}}-5{{x}^{2}}+23x+10=0\].
We take $x=a=5$.
We can see \[f\left( 5 \right)={{5}^{4}}-5\times {{5}^{3}}-5\times {{5}^{2}}+23\times 5+10=-125+115+10=0\].
So, the root of the $f\left( x \right)={{x}^{4}}-5{{x}^{3}}-5{{x}^{2}}+23x+10$ will be the function $\left( x-5 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
We can now divide the polynomial ${{x}^{4}}-5{{x}^{3}}-5{{x}^{2}}+23x+10$ by $\left( x-5 \right)$.
\[x-5\overset{{{x}^{3}}-5x-2}{\overline{\left){\begin{align}
  & {{x}^{4}}-5{{x}^{3}}-5{{x}^{2}}+23x+10 \\
 & \underline{{{x}^{4}}-5{{x}^{3}}} \\
 & -5{{x}^{2}}+23x \\
 & \underline{-5{{x}^{2}}+25x} \\
 & -2x+10 \\
 & \underline{-2x+10} \\
 & 0 \\
\end{align}}\right.}}\]
Now we have ${{x}^{4}}-5{{x}^{3}}-5{{x}^{2}}+23x+10=\left( x-5 \right)\left( {{x}^{3}}-5x-2 \right)$.
We again use the same process and take $x=a=-2$.
So, the root of the $f\left( x \right)={{x}^{3}}-5x-2$ will be the function $\left( x+2 \right)$.
\[x+2\overset{{{x}^{2}}-2x-1}{\overline{\left){\begin{align}
  & {{x}^{3}}-5x-2 \\
 & \underline{{{x}^{3}}+2{{x}^{3}}} \\
 & -2{{x}^{2}}-5x-2 \\
 & \underline{-2{{x}^{2}}-4x} \\
 & -x-2 \\
 & \underline{-x-2} \\
 & 0 \\
\end{align}}\right.}}\]
Now we have ${{x}^{4}}-5{{x}^{3}}-5{{x}^{2}}+23x+10=\left( x-5 \right)\left( x+2 \right)\left( {{x}^{2}}-2x-1 \right)$.
We now use quadratic form to find roots for $\left( {{x}^{2}}-2x-1 \right)$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{x}^{2}}-2x-1=0$. The values of a, b, c is $1,-2,-1$ respectively.
We put the values and get $x$ as \[x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times \left( -1 \right)\times 1}}{2\times 1}=\dfrac{2\pm \sqrt{8}}{2}=1\pm \sqrt{2}\]
Therefore, the zeroes of the polynomial ${{x}^{4}}-5{{x}^{3}}-5{{x}^{2}}+23x+10$ are \[\left( 1\pm \sqrt{2} \right),-2,5\].

Note: We need to remember that the highest value of the indices for the polynomial decides the number of roots for the polynomial. The number of irrational roots will always be in even number of times if the coefficients of the polynomial are rational.