Question

Solve the given polynomial equation for x:
$6{{x}^{6}}-25{{x}^{5}}+31{{x}^{4}}-31{{x}^{2}}+25x-6=0$

Answer 1

- Hint: Club the coefficient of 6 in the one bracket. Similarly, club the coefficient of 25 and 31 in the brackets. After solving these bracket terms, you will find $\left( x-1 \right)\left( x+1 \right)$ as a common term in the given expression. Then solve the remaining expression which is a polynomial function with degree 4. The remaining expression is solved by dividing the expression by ${{x}^{2}}$ and then solving further.

Complete step-by-step solution -

The equation given in the question as follows:
$6{{x}^{6}}-25{{x}^{5}}+31{{x}^{4}}-31{{x}^{2}}+25x-6=0$
Clubbing the coefficient of 6 in one bracket, coefficient of 25 in the another bracket and coefficient of 31 in the other bracket we get,
$6\left( {{x}^{6}}-1 \right)-25x\left( {{x}^{4}}-1 \right)+31{{x}^{2}}\left( {{x}^{2}}-1 \right)=0$
In the above equation $\left( {{x}^{6}}-1 \right)$ is written as $\left( {{\left( {{x}^{2}} \right)}^{3}}-{{1}^{3}} \right)$ and $\left( {{x}^{4}}-1 \right)$as$\left( {{\left( {{x}^{2}} \right)}^{2}}-{{1}^{2}} \right)$.
$\begin{aligned} & 6\left( {{\left( {{x}^{2}} \right)}^{3}}-1 \right)-25x\left( {{\left( {{x}^{2}} \right)}^{2}}-1 \right)+31{{x}^{2}}\left( {{x}^{2}}-1 \right)=0 \\\ & \Rightarrow 6\left( {{x}^{2}}-1 \right)\left( {{x}^{4}}+1+{{x}^{2}} \right)-25x\left( {{x}^{2}}-1 \right)+31{{x}^{2}}\left( {{x}^{2}}-1 \right)=0 \\\ & \Rightarrow \left( {{x}^{2}}-1 \right)\left( 6\left( {{x}^{4}}+1+{{x}^{2}} \right)-25x+31{{x}^{2}} \right)=0 \\\ & \Rightarrow \left( {{x}^{2}}-1 \right)\left( 6{{x}^{4}}+6+6{{x}^{2}}-25x+31{{x}^{2}} \right)=0 \\\ & \Rightarrow \left( {{x}^{2}}-1 \right)\left( 6{{x}^{4}}+6+37{{x}^{2}}-25x \right)=0 \\\ \end{aligned}$
From the above equation, solving two brackets individually we get:
$\begin{aligned} & {{x}^{2}}-1=0 \\\ & \Rightarrow \left( x-1 \right)\left( x+1 \right)=0 \\\ \end{aligned}$
From the above equation, we have two solutions of x; 1 & -1.
And equating the expression in the other bracket to 0 we get,
$6{{x}^{4}}+37{{x}^{2}}-25{{x}^{3}}-25x+6=0$
Dividing the above equation by ${{x}^{2}}$ we get,
$\begin{aligned} & \dfrac{6{{x}^{4}}}{{{x}^{2}}}+\dfrac{37{{x}^{2}}}{{{x}^{2}}}-\dfrac{25{{x}^{3}}}{{{x}^{2}}}-\dfrac{25x}{{{x}^{2}}}+\dfrac{6}{{{x}^{2}}}=0 \\\ & \Rightarrow 6{{x}^{2}}+37-25x-25\left( \dfrac{1}{x} \right)+\dfrac{6}{{{x}^{2}}}=0 \\\ & \Rightarrow 6\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)-25\left( x+\dfrac{1}{x} \right)+37=0 \\\ \end{aligned}$
In the above equation, we can write ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}={{\left( x+\dfrac{1}{x} \right)}^{2}}-2$.
$6\left( {{\left( x+\dfrac{1}{x} \right)}^{2}}-2 \right)-25\left( x+\dfrac{1}{x} \right)+37=0$
Let us assume $x+\dfrac{1}{x}=t$ in the above equation.
$\begin{aligned} & 6\left( {{t}^{2}}-2 \right)-25t+37=0 \\\ & \Rightarrow 6{{t}^{2}}-12-25t+37=0 \\\ & \Rightarrow 6{{t}^{2}}-25t+25=0 \\\ \end{aligned}$
Now, solving the quadratic equation in t we get:
$\begin{aligned} & 6{{t}^{2}}-15t-10t+25=0 \\\ & \Rightarrow 3t\left( 2t-5 \right)-5\left( 2t-5 \right)=0 \\\ & \Rightarrow \left( 3t-5 \right)\left( 2t-5 \right)=0 \\\ \end{aligned}$
From the above equation, we have two values of t i.e. $t=\dfrac{5}{3},\dfrac{5}{2}$.
As we have assumed that $x+\dfrac{1}{x}=t$ so equating $x+\dfrac{1}{x}$ to $\dfrac{5}{3}\And \dfrac{5}{2}$ we get,
First of all we are taking $x+\dfrac{1}{x}=\dfrac{5}{3}$.
$\begin{aligned} & x+\dfrac{1}{x}=\dfrac{5}{3} \\\ & \Rightarrow {{x}^{2}}+1=\dfrac{5}{3}x \\\ & \Rightarrow 3{{x}^{2}}+3=5x \\\ & \Rightarrow 3{{x}^{2}}-5x+3=0 \\\ \end{aligned}$
Now, we are going to solve the quadratic in x.
$\begin{aligned} & x=\dfrac{5\pm \sqrt{25-4.3.3}}{6} \\\ & \Rightarrow x=\dfrac{5\pm \sqrt{25-36}}{6} \\\ & \Rightarrow x=\dfrac{5\pm \sqrt{-11}}{6} \\\ & \Rightarrow x=\dfrac{5\pm i\sqrt{11}}{6} \\\ \end{aligned}$
We are taking $x+\dfrac{1}{x}=\dfrac{5}{2}$ and then solving this equation we get,
$\begin{aligned} & x+\dfrac{1}{x}=\dfrac{5}{2} \\\ & \Rightarrow 2{{x}^{2}}+2=5x \\\ & \Rightarrow 2{{x}^{2}}-5x+2=0 \\\ & \Rightarrow 2{{x}^{2}}-4x-x+2=0 \\\ & \Rightarrow 2x\left( x-2 \right)-1\left( x-2 \right)=0 \\\ & \Rightarrow \left( 2x-1 \right)\left( x-2 \right)=0 \\\ \end{aligned}$
Solving above equation will give the values of x as follows:
$\begin{aligned} & 2x-1=0;x-2=0 \\\ & \Rightarrow x=\dfrac{1}{2},2 \\\ \end{aligned}$
Hence, all the solutions of the given equation are $x=1,-1,2,\dfrac{1}{2},\dfrac{5\pm i\sqrt{11}}{6}$.

Note: Some points should be kept in mind while solving such polynomial question that:
Degree of the polynomial will determine the number of plausible solutions of the given polynomial equation. As in this question, the degree of the polynomial is 6 so 6 solutions are possible.
Finding solutions of the quadratic equation is simple but here the given equation has a degree of 6 so through hit and trial we can find some solutions like if you put 1 in the place of x, you will get 0 then divide the equation $6{{x}^{6}}-25{{x}^{5}}+31{{x}^{4}}-31{{x}^{2}}+25x-6=0$ by $\left( x-1 \right)$. The quotient of this division is$6{{x}^{5}}-19{{x}^{4}}+12{{x}^{3}}+12{{x}^{2}}-19x+6=0$ then by hit and trial you will find that $x=-1$ will satisfy this new equation then divide this new equation by $\left( x+1 \right)$.
The quotient of this new division is $6{{x}^{4}}+37{{x}^{2}}-25{{x}^{3}}-25x+6=0$. Now, it is solved with the same method as we have used in the solution part of the question.
From this hit and trial method we have found that $x=\pm 1$ is the solution of the given equation$6{{x}^{6}}-25{{x}^{5}}+31{{x}^{4}}-31{{x}^{2}}+25x-6=0$.