Reciprocal Equations
Reciprocal equations have a distinct characteristic that if a number is a root, its reciprocal is also a root. For a polynomial equation of the form
the reciprocal roots property implies that if \( \alpha \) is a root, then \( \frac{1}{\alpha} \) must also satisfy the equation. This leads us to consider the polynomial
where the coefficients are in reverse order compared to the original polynomial. This new polynomial will have roots that are the reciprocals of the original polynomial's roots.
Since both equations [1] and [2] have the roots, their coefficients are proportional, that leads to:
Let the above proportionality be equal to \( \lambda \).
Then observe that the first ratio \(\frac{a_0}{a_n}=\lambda\) and the last ratio \(\frac{a_n}{a_0}=\lambda\).
Multiply both ratios:\(\frac{a_0}{a_n}.\frac{a_n}{a_0}={\lambda}^2\) \(\implies \lambda = \pm 1\)
and hence, \( \lambda \) is either \( 1 \) or \( 1 \). The nature of \( \lambda \) dictates two types of reciprocal equations:

Type I: When \( \lambda = 1 \), the coefficients are symmetric with respect to the midpoint of the sequence: \( a_r = a_{nr} \) for \( r = 0, 1, 2, \ldots, n \). That is, \(a_0=a_n, a_1=a_{n1}, a_2 = a_{n2}, ...\)
For example: \(2x^43x^3+40x^23x+2=0\) and \(3x^52x^47x^37x^22x+3=0\) are examples of reciprocal equations of type I.

Type II: When \( \lambda = 1 \), the coefficients are symmetric but alternate in sign: \( a_r = a_{nr} \) for \( r = 0, 1, 2, \ldots, n \). That is, \(a_0=a_n, a_1=a_{n1}, a_2 =  a_{n2}, ...\)
For example: \(2x^43x^3+3x2=0\) and \(3x^52x^4+7x^37x^2+2x3=0\) are examples of reciprocal equations of type II
Type I reciprocal equations of even degree are the most important. As other reciprocal equations Type I odd degree, Type II odd and even degree can be converted to Type I even degree. We will learn in the next section how we can solve these equations.
Important Property
I. A reciprocal polynomial equation \( p(x) = 0 \) of Type I has the property that its polynomial function \( p(x) \) satisfies the condition:
for every \( x \neq 0 \), where \( n \) is the degree of the polynomial. This means that the polynomial remains unchanged when each \( x \) term is replaced by its reciprocal and then multiplied by \( x^n \).
For example:
To verify whether the polynomial \( p(x) = x^6 + 7x^5 + 7x + 1 \) is a Type I reciprocal polynomial, we need to check if it satisfies the condition:
Since the degree \( n \) of the polynomial is 6, let's substitute \( x \) with \( \frac{1}{x} \) and multiply by \( x^6 \):
Now, let's simplify the expression:
This is the original polynomial \( p(x) \) rewritten, which shows that \( p(x) \) satisfies the condition for a Type I reciprocal polynomial. Hence, the polynomial \( p(x) \) satisfies the equation \( x^6 p\left(\frac{1}{x}\right) = p(x) \).
II. For a Type II reciprocal polynomial equation \( p(x) = 0 \), the defining characteristic is that its polynomial function \( p(x) \) satisfies the condition:
for all \( x \neq 0 \), where \( n \) is the degree of the polynomial. This means that the polynomial changes sign but otherwise remains unchanged when each \( x \) term is replaced by its reciprocal and then multiplied by \( x^n \).
For example:
Let \( p(x) = 2x^5 + 4x^4  5x^3 + 5x^2  4x  2 = 0 \). It is clearly a type II polynomial.
Replace \( x \) with \( \frac{1}{x} \) and multiply by \( x^5 \):
Simplify the equation:
Comparing this with the original polynomial \( p(x) \), we see that:
Solving Type I even degree Reciprocal Equation
To solve a Type I reciprocal equation of even degree \(2m\), with the general form
we can simplify the solving process as follows:

Divide the entire equation by \( x^m \) to normalize it:
\[ a_0x^m + a_1x^{m1} + \ldots + a_1x^{m+1} + a_0x^{m} = 0. \] 
Regroup the terms by pairing the \( x \) terms and their reciprocal power terms:
\[ a_0(x^m + \frac{1}{x^m}) + a_1(x^{m1} + \frac{1}{x^{m1}}) + \ldots + a_{m1}(x + \frac{1}{x}) = 0. \] 
Introduce a new variable \( z \) where \( z = x + \frac{1}{x} \). Notice that \( z \) will generate all the necessary powers when raised to higher powers itself:
\[ z^2 = x^2 + 2 + \frac{1}{x^2} \implies x^2 + \frac{1}{x^2} = z^2  2,\]\[ z^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \implies x^3 + \frac{1}{x^3} = z^33z\]\[ \vdots \]\[ z^m = x^m + \ldots + \frac{1}{x^m}. \] 
The original equation can now be rewritten in terms of \( z \) to get a new polynomial equation of degree \( m \), which is half the original degree:
\[ b_0z^m + b_1z^{m1} + \ldots + b_{m1}z + b_m = 0, \]with each \( b_i \) being a combination of the original coefficients \( a_i \).

Solve this new polynomial equation for \( z \) using an appropriate method.

Once the values for \( z \) are found, use them to construct quadratic equations to solve for \( x \) since \( z = x + \frac{1}{x} \) implies \( x^2  zx + 1 = 0 \).
Example
Solve:
Solution:
We first notice the symmetry in the coefficients, indicating that it is indeed a Type I equation. The next steps involve transforming the equation and solving it:

Divide the entire equation by \( x^2 \):
\[ 6x^2  35x + 62  \frac{35}{x} + \frac{6}{x^2} = 0. \] 
Regroup terms to form symmetric pairs around \( x^2 \) and \( x^{2} \):
\[ 6\left(x^2 + \frac{1}{x^2}\right)  35\left(x + \frac{1}{x}\right) + 62 = 0. \] 
Let \( z = x + \frac{1}{x} \), hence \( z^2 = x^2 + 2 + \frac{1}{x^2} \), and \( x^2 + \frac{1}{x^2} = z^2  2 \).

Substitute \( z \) into the equation:
\[ 6(z^2  2)  35z + 62 = 0, \]\[ 6z^2  35z + 50 = 0. \] 
This is now a quadratic equation in \( z \). Solve for \( z \):
Let's solve \( 6z^2  35z + 50 = 0 \) using the quadratic formula:
\[ z = \frac{(35) \pm \sqrt{(35)^2  4 \cdot 6 \cdot 50}}{2 \cdot 6}. \]This yields two possible solutions for \( z \):
\[ z_1 = \frac{35 + 5}{12} = \frac{40}{12} = \frac{10}{3} \]\[ z_2 = \frac{35  5}{12} = \frac{30}{12} = \frac{5}{2} \]
Thus, the solutions for \( z \) are \( z_1 = \frac{10}{3} \) and \( z_2 = \frac{5}{2} \).
Having found \( z \) values \( \frac{10}{3} \) and \( \frac{5}{2} \), we can now find \( x \) by solving the quadratic equations derived from the substitution \( z = x + \frac{1}{x} \). This gives us two quadratic equations to solve for \( x \):
For \( z_1 = \frac{10}{3} \):
For \( z_2 = \frac{5}{2} \):
This results in two more solutions for \( x \):
In summary, the solutions for the original quartic equation \( 6x^4  35x^3 + 62x^2  35x + 6 = 0 \) are \( x_1 = 3 \), \( x_2 = \frac{1}{3} \), \( x_3 = 2 \), and \( x_4 = \frac{1}{2} \).
Solving other Reciprocal Equations
Type I odd degree Reciprocal Equations
Theorem: Conversion of Type I OddDegree Reciprocal Equations to Type I EvenDegree
Let \( p(x) \) be a Type I reciprocal polynomial of odd degree \( 2m+1 \), given by
where the coefficients are symmetric with respect to the center, fulfilling the condition \( a_i = a_{2m+1i} \) for all \( i \). Then \( p(x) \) can be factored as \( p(x) = (x + 1)q(x) \), where \( q(x) \) is a Type I reciprocal polynomial of even degree \( 2m \). In other words, a Type I odddegree reciprocal equation can be converted to a Type I evendegree equation by dividing it with \(x + 1\).
Proof:
To prove that a Type I odddegree reciprocal equation can be converted to a Type I evendegree equation by dividing it with \(x + 1\), let's begin with the general form of a Type I odddegree reciprocal equation:
Notice that the coefficients are symmetrical and the equation is of odd degree \(2m+1\).
We group terms symmetrically from the highest and lowest powers towards the center:
By inspection, we can see that each grouped term contains a factor of \(x + 1\). So we can express \(p(x)\) as:
where \(q(x)\) is a polynomial of degree \(2m\), which makes it an evendegree polynomial.
To show that \(q(x)\) is also a Type I equation, we can consider the transformation:
Since \(p(x)\) is a Type I equation, we have \(p(\frac{1}{x}) = p(x)/x^{2m+1}\). Now, let's apply this to the equation \(q(x)\):
We know that \(p(x) = (x+1)q(x)\), so when we substitute \(p(x)\) back into the equation, we get:
This demonstrates that \(q(x)\) satisfies the property of a Type I equation since it remains unchanged under the transformation \( x \rightarrow \frac{1}{x} \) after being multiplied by \(x^{2m}\). Hence, \(q(x)\) is a Type I evendegree equation.
Example
Problem:
Solve \(4x^3  13x^2  13x + 4 = 0\).
Solution:
Since the equation is a Type I odd degree reciprocal equation, we factor \(x + 1\) from the polynomial:
Now, we find the roots of the quadratic equation \(4x^2  17x + 4 = 0\) by applying the quadratic formula or factoring. The roots are \( x = 1/4 \) and \( x = 4 \).
Therefore, the roots of the original cubic equation are \( x = 1, x = 1/4, \) and \( x = 4 \).
Type II odd degree Reciprocal Equation
Theorem: Factorization of Type II Reciprocal Polynomials of Odd Degree
Let \( p(x) \) be a Type II reciprocal polynomial of odd degree \( 2m+1 \), defined by
where the coefficients satisfy \( a_i = a_{2m+1i} \) for all \( i \). Then \( p(x) \) can be factorized as \( p(x) = (x  1)q(x) \), where \( q(x) \) is a Type I reciprocal polynomial of even degree \( 2m \).
Proof:
To prove that Type II reciprocal equations of odd degree can be factorized into a product of a Type I even degree equation and \( x1 \), let's start with a Type II reciprocal polynomial of degree \( 2m+1 \):
where the coefficients satisfy the relationship \( a_i = a_{2m+1i} \) for \( i = 0, \ldots, 2m \).
First, we show that \( p(1) = 0 \). Substituting \( x = 1 \) into \( p(x) \), we have:
Since \(a_0=a_{2m+1}, a_1=a_{2m}, ..., a_m=a_{m+1}\), we have:
We can use also prove that \(p(1)=0\) by using fact that,
Substitute \( x = 1 \):
This implies \( p(1) = 0 \), which means, by remainder theorem, \( x  1 \) is a factor of \( p(x) \). So we can write \( p(x) \) as:
where \( q(x) \) is a polynomial of degree \( 2m \).
To prove that \( q(x) \) is a Type I reciprocal polynomial, we show that it satisfies the identity \( x^{2m} q\left(\frac{1}{x}\right) = q(x) \). Using the fact that \( p(x) \) is Type II and the relationship between \( p(x) \) and \( q(x) \), we get:
Dividing both sides by \( x1 \) (assuming \( x \neq 1 \)) gives us:
which after considering the Type II property \( x^{2m} q\left(\frac{1}{x}\right) = q(x) \) implies \( q(x) \) must be of Type I since the sign change is no longer present.
Thus, \( q(x) \) is a Type I even degree polynomial, and the original Type II odd degree polynomial \( p(x) \) can indeed be factorized into \( (x1)q(x) \).
Type II Even degree Reciprocal equation
Theorem: Factorization of Type II Reciprocal Polynomials of Even Degree
Let \( p(x) \) be a Type II reciprocal polynomial of even degree \( 2m \), given by
where the coefficients satisfy \( a_i = a_{2mi} \) for \( i = 0, \ldots, m1 \) and \( a_i = a_{2mi} \) for \( i = m, \ldots, 2m \). Then \( p(x) \) can be factorized as \( p(x) = (x^2  1)q(x) \), where \( q(x) \) is a Type I reciprocal polynomial of degree \( 2m2 \).
Proof:

Since, \( p(x) \) is a Type II reciprocal polynomial of even degree \( 2m \) and it satisfies the identity
\[ x^{2m}p\left(\frac{1}{x}\right) = p(x), \]then setting \( x = 1 \) gives us
\[ 1^{2m}p\left(\frac{1}{1}\right) = p(1), \]\[ p(1) = p(1), \]which implies \( p(1) = 0 \), indicating that \( x = 1 \) is a root of \( p(x) \).
Similarly, setting \( x = 1 \) gives us
\[ (1)^{2m}p\left(\frac{1}{1}\right) = p(1), \]\[ p(1) = p(1), \]which implies \( p(1) = 0 \), indicating that \( x = 1 \) is also a root of \( p(x) \).
Since both \( x = 1 \) and \( x = 1 \) are roots, \( x  1 \) and \( x + 1 \) are factors of the polynomial \( p(x) \). The product \( (x  1)(x + 1) = x^2  1 \) is therefore a factor of \( p(x) \), and we can express \( p(x) \) as
\[ p(x) = (x^2  1)q(x), \]where \( q(x) \) is the remaining polynomial of degree \( 2m  2 \), which must be a Type I reciprocal polynomial because the Type II property is no longer in effect once the \( x^2  1 \) factor is removed.

Now, let's demonstrate that \( q(x) \) is a Type I reciprocal polynomial. By the Type II property, we have:
\[ x^{2m} p\left(\frac{1}{x}\right) = p(x), \]which we apply after factoring out \( x^2  1 \):
\[ (x^2  1)x^{2m2} q\left(\frac{1}{x}\right) = (x^2  1)q(x). \] 
Dividing both sides by \( x^2  1 \) (and noting that \( x \neq \pm1 \)), we obtain:
\[ x^{2m2} q\left(\frac{1}{x}\right) = q(x), \]
confirming that \( q(x) \) satisfies the Type I reciprocal property.
Therefore, the polynomial \( p(x) \) can be factorized into \( (x^2  1)q(x) \), with \( q(x) \) being a Type I reciprocal polynomial of degree \( 2m2 \).`