Theory of Equations

Introduction

In this blog, we will dive into the Theory of Equations, with a focus on polynomial equations, which are essential in both mathematics and real-world applications. The Theory of Equations plays a crucial role in understanding how to find the roots (solutions) of an equation and explore the relationship between these roots and the coefficients of the polynomial. Mastering this theory allows us to solve complex problems involving equations of higher degrees, like cubic and quartic equations, which are often encountered in advanced mathematics.

We will cover several key topics, including how the roots of an equation relate to its coefficients, and how symmetrical functions of the roots provide insights without solving the equation directly. We’ll also explore special types of roots, such as surd and imaginary roots, which occur in pairs. Important tools like De-Cardes’ rule of signs will be introduced to help predict the number of positive and negative real roots, and Horner’s process will simplify evaluating polynomials.

Furthermore, we’ll discuss how polynomial functions behave and how to identify where the roots lie using properties of odd and even degree equations. We’ll also look at techniques for transforming equations by changing the roots, and finally, we’ll explore methods for solving cubic and quartic equations, including trigonometric solutions and Ferrari’s method. By the end of this blog, you will have a clear understanding of how to solve and manipulate polynomial equations, equipping you with valuable tools for tackling algebraic problems.

What is an Equation?

An equation is a mathematical statement that shows two expressions are equal. It’s like a balance scale, where both sides must weigh the same. For example, \(2x + 3 = 7\) is an equation.

What is the degree of an equation?

The degree of an equation is the highest power of the variable in the equation. For example, \(2x^3 + 5x^2 – 7x + 1 =0\) is a third-degree equation because the highest power of \(x\) is 3. Here are some common types of equations based on their degree:

Linear equations: These have a degree of 1. For example, \(2x + 3 = 7\).
Quadratic equations: These have a degree of 2. For example, \(x^2 – 4 = 0\).
Cubic equations: These have a degree of 3. For example, \(x^3 + 2x^2 – 5x + 1 = 0\).
Biquadratic equations: These have a degree of 4. For example, \(x^4 + 3x^2 – 2 = 0\).

What are roots or zeros of an equation?

The roots or zeros of an equation are the values of the variable that make the equation true. In other words, they are the values that make both sides of the equation equal. For example, the roots of the equation \(x^2 – 4 = 0\) are 2 and -2, because when you substitute 2 or -2 for \(x\), both sides of the equation equal 0.

Every equation of the \(n^{th}\) degree has \(n\) roots(counting multiplicity), and no more.

Proof of the Theorem

Proof:

Denote the given equation by \(f(x)=0\), where

\[ f(x)=p_0x^n+p_1x^{n-1}+p_2x^{n-2}+\cdots+p_n \]

The equation \(f(x)=0\) has a root, real or imaginary; let this root be denoted by \( a_1\); then \(f(x)\) is divisible by \( (x-a_1)\), so that:

\[f(x)=(x-a_1)\phi_1(x),\]

where \(\phi_1(x)\) is a polynomial of degree \(n-1\).

Again, the equation \(\phi_1(x)=0\) has a root, real or imaginary; let this be denoted by \( a_2\); then \(\phi_1(x)\) is divisible by \((x-a_2)\), so that:

\[\phi_1(x)=(x-a_2)\phi_2(x),\]

where \(\phi_2(x)\) is a polynomial of degree \(n-2\).

Thus:

\[ f(x)=(x-a_1)(x-a_2)\phi_2(x).\]

Proceeding in this way, we obtain:

\[ f(x)=p_0(x-a_1)(x-a_2)\ldots(x-a_n)\]

Hence, the equation \(f(x)=0\) has \(n\) roots, since \(f(x)\) vanishes when \(x\) has any one of the values \( a_1, a_2, a_3, \ldots, a_n\).

Also, the equation cannot have more than \(n\) roots; for if \(x\) has any value different from any of the quantities \(a_1, a_2, \ldots, a_n\), all the factors on the right are different from zero, and therefore \(f(x)\) cannot vanish for the values of \(x\).

Vieta's Relations between Roots and Coefficients of an Equation

Relations between Roots and Coefficients

Consider the equation:

\[ f(x) = p_0x^n + p_1x^{n-1} + p_2x^{n-2} + \cdots + p_n = 0 \]

Let \(a_1, a_2, \dots, a_n\) be the roots of the polynomial equation. The relations between the roots and the coefficients are given as follows:

1. Sum of the roots:

\[ a_1 + a_2 + \cdots + a_n = -\frac{p_1}{p_0} \]

2. Sum of the products of roots taken two at a time:

\[ a_1a_2 + a_1a_3 + \cdots + a_{n-1}a_n = \frac{p_2}{p_0} \]

3. Sum of the products of roots taken three at a time:

\[ a_1a_2a_3 + a_1a_2a_4 + \cdots + a_{n-2}a_{n-1}a_n = -\frac{p_3}{p_0} \]

4. General formula for the sum of the products of roots taken \(k\) at a time:

\[ (-1)^k \sum_{1 \leq i_1 < \cdots < i_k \leq n} a_{i_1} \cdots a_{i_k} = (-1)^k \frac{p_k}{p_0} \]

5. Product of the roots:

\[ a_1a_2 \cdots a_n = (-1)^n \frac{p_n}{p_0} \]

Example 1: Solve the equation \( 4x^3-24x^2+23x+18=0\), having given that the roots are in Arithmetical Progression.

Proof of the Theorem Solution: The given cubic equation is:

\[4x^3 - 24x^2 + 23x + 18 = 0\]

We compare this with the general form:

\[p_0x^3 + p_1x^2 + p_2x + p_3 = 0\]

From the given equation, we have:

\[p_0 = 4, \quad p_1 = -24, \quad p_2 = 23, \quad p_3 = 18\]

Let the roots be \(a - b\), \(a\), and \(a + b\).
Using Vieta’s Relations:

1. Sum of the roots:

\[(a - b) + a + (a + b) = \frac{-p_1}{p_0}\] \[ \implies 3a = \frac{24}{4} = 6 \quad \Rightarrow \quad a = 2\]

2. Sum of the products of the roots taken two at a time:

\[(a - b)a + a(a + b) + (a - b)(a + b) = \frac{p_2}{p_0}\] \[ \implies3a^2 - b^2 = \frac{23}{4}\]

Substituting \(a = 2\):

\[12 - b^2 = \frac{23}{4} \quad \Rightarrow \quad 4b^2 = 25 \quad \Rightarrow \quad b^2 = \frac{25}{4} \quad \Rightarrow \quad b = \frac{5}{2}\]

3. Product of the roots:

\[(a - b)a(a + b) = \frac{-p_3}{p_0}\]

Substituting the values:

\[\left(-\frac{1}{2}\right)(2)\left(\frac{9}{2}\right) = -\frac{9}{2}\]

Final Answer:

The roots of the cubic equation are:

\[a - b = -\frac{1}{2}, \quad a = 2, \quad a + b = \frac{9}{2}\]

Example 2: Solve the equation \(24x^3 - 14x^2 - 63x + 45 = 0\), given that one root being double another.

Proof of the Theorem Solution:
We are given the cubic equation:

\[24x^3 - 14x^2 - 63x + 45 = 0\]

and that one root is double another.
Comparing this equation with:

\[p_0x^3 + p_1x^2 + p_2x + p_3 = 0\]

we get:

\[p_0 = 24, \quad p_1 = -14, \quad p_2 = -63, \quad p_3 = 45\]

Let the roots of the equation be \(a\), \(2a\), and \(b\).
From Vieta's relations, we know:
1. The sum of the roots:

\[ \begin{align*} a + 2a + b &= \frac{-p_1}{p_0} \\ \implies 3a + b &= \frac{-(-14)}{24} = \frac{7}{12} \tag{1} \end{align*} \]

2. The sum of the products of the roots taken two at a time:

\[ \begin{align*} (a)(2a) + (a)(b) + (2a)(b) &= \frac{p_2}{p_0} \\ \implies 2a^2 + 3ab &= \frac{-63}{24} = \frac{-21}{8} \tag{2} \end{align*} \]

3. The product of the roots:

\[ \begin{align*} a \cdot 2a \cdot b &= \frac{-p_3}{p_0} \\ \implies 2a^2b &= \frac{-45}{24} = \frac{-15}{8} \tag{3} \end{align*} \]

Now, solving equations (1) and (2), we obtain:

\[ \begin{align*} 8a^2 - 2a - 3 &= 0 \\ \therefore a &= \frac{3}{4} \quad \text{or} \quad \frac{-1}{2} \\ \implies b &= \frac{-5}{3} \quad \text{or} \quad \frac{25}{12} \end{align*} \]

Here, the values \(a = \frac{-1}{2}\) and \(b = \frac{25}{12}\) do not satisfy equation (3).

Final Answer:

Thus, the roots of the given equation are:

\[ a = \frac{3}{4}, \quad 2a = \frac{3}{2}, \quad b = \frac{-5}{3} \]

Example 3: Solve the equation \(4x^3 +16x^2 - 9x -36 = 0\), given that the sum of two roots being zero.

Solution of the Equation Solution:
We are given the cubic equation:

\[4x^3 + 16x^2 - 9x - 36 = 0\]

and that the sum of two roots is zero.
Comparing this equation with:

\[p_0x^3 + p_1x^2 + p_2x + p_3 = 0\]

we get:

\[p_0 = 4, \quad p_1 = 16, \quad p_2 = -9, \quad p_3 = -36\]

Let the roots of the equation be \(a\), \(-a\), and \(b\).
From Vieta's relations, we know:
1. The sum of the roots:

\[ \begin{align*} a - a + b &= \frac{-p_1}{p_0} \\ \implies b &= \frac{-16}{4} = -4 \tag{1} \end{align*} \]

2. The sum of the products of the roots taken two at a time:

\[ \begin{align*} (a)(-a) + (a)(b) + (-a)(b) &= \frac{p_2}{p_0} \\ \implies -a^2 + ab - ab &= \frac{-9}{4} \\ \implies -a^2 &= \frac{-9}{4} \\ \implies a^2 &= \frac{9}{4} \\ \implies a &= \frac{3}{2} \quad \text{or} \quad a = -\frac{3}{2}\tag{2} \end{align*} \]

3. The product of the roots:

\[ \begin{align*} a \cdot (-a) \cdot b &= \frac{-p_3}{p_0} \\ \implies -a^2b &= \frac{36}{4} = 9 \\ \implies a^2b &= -9 \tag{3} \end{align*} \]

Substituting the values of \( a\) and \( b\) into equation (3), we see that the equation (3) is satisfied: Thus, the roots of the given equation are:

\[ a = \frac{3}{2}, \quad -a = -\frac{3}{2}, \quad b = -4 \]

Example 4: Solve the equation \( 4x^3 + 20x^2 - 23x + 6 = 0\), given that two of the roots being same.

Solution of the Equation Solution:
We are given the cubic equation:

\[4x^3 + 20x^2 - 23x + 6 = 0\]

and that two of the roots are the same.
Comparing this equation with:

\[p_0x^3 + p_1x^2 + p_2x + p_3 = 0\]

we get:

\[p_0 = 4, \quad p_1 = 20, \quad p_2 = -23, \quad p_3 = 6\]

Let the roots of the equation be \(a\), \(a\), and \(b\).
From Vieta's relations, we know:
1. The sum of the roots:

\[ \begin{align*} a + a + b &= \frac{-p_1}{p_0} \\ \implies 2a + b &= \frac{-20}{4} = -5 \\ \implies b &=-5-2a \tag{1} \end{align*} \]

2. The sum of the products of the roots taken two at a time:

\[ \begin{align*} (a)(a) + (a)(b) + (a)(b) &= \frac{p_2}{p_0} \\ \implies a^2 + 2ab &= \frac{-23}{4} \tag{2} \end{align*} \]

3. The product of the roots:

\[ \begin{align*} a \cdot a \cdot b &= \frac{-p_3}{p_0} \\ \implies a^2b &= \frac{-6}{4} = -\frac{3}{2} \tag{3} \end{align*} \]

Now, Substituting the value of \( b\) from equations (1) into (2) we get:

\[ \begin{align*} a^2+2a(-5-2a)&=\frac{-23}4 \\ \implies -3a^2-10a&=\frac{-23}4 \\ \implies 12a^2+40a-23&=0 \\ \implies a&=\frac 12 \quad \text{or} \quad \frac{-23}6 \\ \therefore \quad b&= -6 \quad \text{or}\quad \frac 83 \end{align*} \]

Here, the values \(a = \frac{-23}{6}\) and \(b = \frac{8}{3}\) do not satisfy equation (3).
Thus, the roots of the given equation are:

\[ a = \frac{1}{2}, \quad a = \frac{1}{2}, \quad b = -6 \]

Example 5: Solve the equation \( 54x^3 -39x^2 - 26x + 16 = 0\), given that the roots being in Geometrical Progression.

Solution of the Equation Solution:
We are given the cubic equation:

\[54x^3 - 39x^2 - 26x + 16 = 0\]

and that the roots are in geometric progression.
Comparing this equation with:

\[p_0x^3 + p_1x^2 + p_2x + p_3 = 0\]

we get:

\[p_0 = 54, \quad p_1 = -39, \quad p_2 = -26, \quad p_3 = 16\]

Let the roots of the equation be \(\frac{a}{r}\), \(a\), and \(ar\).
From Vieta's relations, we know:
1. The sum of the roots:

\[ \begin{align*} \frac{a}{r} + a + ar &= \frac{-p_1}{p_0} \\ \implies a\cdot \left( \frac{1}{r} + 1 + r \right) &= \frac{39}{54} = \frac{13}{18} \tag{1} \end{align*} \]

2. The sum of the products of the roots taken two at a time:

\[ \begin{align*} \left(\frac{a}{r}\right)(a) + \left(\frac{a}{r}\right)(ar) + (a)(ar) &= \frac{p_2}{p_0} \\ \implies \frac{a^2}{r} + a^2 + a^2r &= \frac{-26}{54} = -\frac{13}{27} \tag{2} \end{align*} \]

3. The product of the roots:

\[ \begin{align*} \left(\frac{a}{r}\right)(a)(ar) &= \frac{-p_3}{p_0} \\ \implies a^3 &= \frac{-16}{54} = -\frac{8}{27}\\ \therefore \quad a&=\frac{-2}3 \tag{3} \end{align*} \]

Now, substituing the value of \( a=\frac{-2}3\) in the equation (1), we get

\[ \begin{align*} -\frac 23 \left( \frac 1r +1 + r \right) &=\frac{13}{18} \\ \implies \frac 1r +1 + r&=-\frac{13}{12} \\ \implies 12 + 12r+12r^2&=-13r \\ \implies 12r^2+25r+12&=0 \\ \therefore r&= -\frac 34 \quad \text{or} \quad -\frac 43 \end{align*} \]

Using any of the values for \( r\), we will obtain the same roots. Thus, the roots of the given equation are:

\[ ar= \frac 89, \quad a=-\frac 23,\quad ar=\frac 12 \]

Example 6: Solve the equation \(x^4 - 16x^3 + 86x^2 - 176x + 105 = 0\) using synthetic division, given two roots are 1 and 7.

Solution of the Equation

Step 1: Divide by \(x - 1\)
We will perform synthetic division of the polynomial

\(x^4 - 16x^3 + 86x^2 - 176x + 105\)

by \(x - 1\).

\[ \begin{array}{r|rrrrr} 1 & 1 & -16 & 86 & -176 & 105 \\ & & 1 & -15 & 71 & -105 \\ \hline & 1 & -15 & 71 & -105 & 0 \\ \end{array} \]

The quotient is \(x^3 - 15x^2 + 71x - 105\).

Step 2: Divide by \(x - 7\)
Now, we'll divide the quotient \(x^3 - 15x^2 + 71x - 105\) by \(x - 7\).

\[ \begin{array}{r|rrrr} 7 & 1 & -15 & 71 & -105 \\ & & 7 & -56 & 105 \\ \hline & 1 & -8 & 15 & 0 \\ \end{array} \]

The quotient is \(x^2 - 8x + 15\).

Step 3: Solve the quadratic equation
Now, solve the quadratic equation \(x^2 - 8x + 15 = 0\) using the quadratic formula:

\[ \begin{align*} x &= \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(15)}}{2(1)}\\ x &= \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm \sqrt{4}}{2}\\ x &= \frac{8 \pm 2}{2} \end{align*} \]

Therefore, the roots are \(x = 5\) and \(x = 3\).

Final Answer:
The roots of the equation

\[x^4 - 16x^3 + 86x^2 - 176x + 105=0\]

are

\[ x= 1,\quad 3, \quad 5, \quad 7 \]

In an equation with real coefficients imaginary roots occur in pairs

Proof:

Suppose that \(f(x)=0\) is an equation with real coefficients, and suppose that it has an imaginary root \( a+ib\).
We Shall show that \(a-ib\) is also a root.
The factor of \(f(x)\) corresponding to these two roots is

\[ \left(x-(a+ib)\right)\left(x-(a-ib)\right)=(x-a-ib)(x-a+ib)=(x-a)^2+b^2\]

Let \(f(x)\) be divided by \((x-a)^2+b^2\); denote the quotient by \(Q\) and the remainder,if any, by \(Rx+R'\); then

\[ f(x)=Q\left\{(x-a)^2+b^2\right\}+Rx+R'\]

In this equation put \(x=a+ib\), then \(f(x)=0\) by hypothesis; also \((x-a)^2+b^2=0\); hence
\[ R(a+ib)+R'=0\] Equating to zero the real and imaginary parts, we get
\[ Ra+R'=0;\quad Rb=0\] and \(b\) by hypothesis is not zero. \[ R=0 \quad \& \quad R'=0\] Hence \(f(x)\) is exactly divisible by \((x-a)^2+b^2\), that is, by

\[ (x-a-ib)(x-a+ib)\]

Hence \(x=a-ib\) is also a root.

In an equation with rational coefficients surd roots occur in pairs; that is; if \(a+\sqrt{b}\) is a root then \( a-\sqrt{b}\) is also a root.

Proof:

Suppose that \(f(x)=0\) is an equation with rational coefficients, and suppose that it has a surd root \(a + \sqrt{b}\).
We shall show that \(a - \sqrt{b}\) is also a root.

The factor of \(f(x)\) corresponding to these two roots is

\[ \left(x - (a + \sqrt{b})\right)\left(x - (a - \sqrt{b})\right) = (x - a - \sqrt{b})(x - a + \sqrt{b}) = (x - a)^2 - b \]

This factor \((x - a)^2 - b\) is a quadratic expression with rational coefficients because it involves only rational numbers \(a\) and \(b\) (since \(b\) is rational).

Let \(f(x)\) be divided by \((x - a)^2 - b\); denote the quotient by \(Q(x)\) and the remainder, if any, by \(Rx + R'\); then

\[ f(x) = Q(x)\left\{(x - a)^2 - b\right\} + Rx + R' \]

In this equation, substitute \(x = a + \sqrt{b}\). By hypothesis, \(f(x) = 0\), and also \((x - a)^2 - b = 0\). Therefore, we have:

\[ R(a + \sqrt{b}) + R' = 0 \]

Equating the rational and surd parts separately, we get:

\[ Ra + R' = 0 \quad \text{and} \quad R\sqrt{b} = 0 \]

Since \(\sqrt{b} \neq 0\), we conclude that \(R = 0\) and thus \(R' = 0\). This shows that \(f(x)\) is exactly divisible by \((x - a)^2 - b\), that is, by the factor:

\[ (x - a - \sqrt{b})(x - a + \sqrt{b}) \]

Hence, \(a - \sqrt{b}\) is also a root of the equation.

Synthetic Division with Bottom Divider

1	1	-16	86	-176	105
	↓	1	-15	71	-105
7	1	-15	71	-105	0
	↓	7	-56	105
	1	-8	15	0

Example 1: Solve the equation \( 6x^4-13x^3-35x^2-x+3=0\), having given that one root is \( 2-\sqrt{3}\).

Proof of the Theorem Solution: The given equation is: \[6x^4-13x^3-35x^2-x+3=0\] Since the equation has rational coefficients and one root is \(2-\sqrt{3}\), then its conjugate \(2+\sqrt{3}\) must also be a root.
The quadratic factor corresponding to these roots \(2-\sqrt{3} \quad \&\quad 2+\sqrt{3}\) is

\[ \begin{align*} \left(x - (2 - \sqrt{3})\right)\left(x - (2 + \sqrt{3})\right) &= (x - 2 + \sqrt{3})(x - 2 - \sqrt{3}) \\ &= (x - 2)^2 - 3 \\ &= x^2-4x+1 \end{align*} \]

Now dividing the biquadratic polynomial \(6x^4-13x^3-35x^2-x+3\) by the quadratic factor \(x^2-4x+1\) using long division we get;

\[ \frac{6x^4-13x^3-35x^2-x+3}{x^2-4x+1}=6x^2+11x+3 \]

Hence, the other roots are obtained from

\[ \begin{align*} 6x^2+11x+3 &=0 \\ (3x+1)(2x+3)&=0 \\ \therefore \quad x&= -\frac 13, \quad -\frac 32 \end{align*} \]

Thus the roots of the equation \(6x^4-13x^3-35x^2-x+3=0\) are

\[ x=-\frac 13, \quad -\frac 32, \quad 2-\sqrt{3}, \quad 2+\sqrt{3} \]

Example 2: Solve the equation \( 3x^4-10x^3+4x^2-x-6=0\), having given that one root is \( \frac{1+i\sqrt{3}}{2}\).

Solution of the Equation Solution: The given equation is: \[ 3x^4 - 10x^3 + 4x^2 - x - 6 = 0 \] Since the equation has rational coefficients and one root is \( \frac{1 + i \sqrt{3}}{2} \), then its conjugate \( \frac{1 - i \sqrt{3}}{2} \) must also be a root.
The quadratic factor corresponding to these roots \( \frac{1 + i \sqrt{3}}{2} \quad \& \quad \frac{1 - i \sqrt{3}}{2} \) is

\[ \begin{align*} \left( x - \frac{1 + i \sqrt{3}}{2} \right) \left( x - \frac{1 - i \sqrt{3}}{2} \right) &= \left( x - \frac{1}{2} - \frac{i \sqrt{3}}{2} \right) \left( x - \frac{1}{2} + \frac{i \sqrt{3}}{2} \right) \\ &= \left( x - \frac{1}{2} \right)^2 - \left( \frac{i\sqrt{3}}{2} \right)^2 \\ &= x^2 - x + 1 \end{align*} \]

Now dividing the quartic polynomial \( 3x^4 - 10x^3 + 4x^2 - x - 6 \) by the quadratic factor \( x^2 - x + 1 \) using long division, we get:

\[ \frac{3x^4 - 10x^3 + 4x^2 - x - 6}{x^2 - x + 1} = 3x^2 - 7x - 6 \]

Hence, the other roots are obtained from:

\[ \begin{align*} 3x^2 - 7x - 6 &= 0 \\ (3x + 2)(x - 3) &= 0 \\ \quad x &= \frac{-2}{3}, \quad 3 \end{align*} \]

Thus, the roots of the equation \( 3x^4 - 10x^3 + 4x^2 - x - 6 = 0 \) are:

\[ x = \frac{1 + i \sqrt{3}}{2}, \quad \frac{1 - i \sqrt{3}}{2}, \quad \frac{-2}{3}, \quad 3 \]

Example 3: Solve the equation \( x^4+4x^3+5x^2+2x-2=0\), having given that one root is \( -1+i\).

Solution of the Equation Solution: The given equation is: \[ x^4 + 4x^3 + 5x^2 + 2x - 2 = 0 \] Since the equation has rational coefficients and one root is \( -1 + i \), then its conjugate \( -1 - i \) must also be a root.
The quadratic factor corresponding to these roots \( -1 + i \quad \& \quad -1 - i \) is

\[ \begin{align*} \left( x - (-1 + i) \right) \left( x - (-1 - i) \right) &= (x + 1 - i)(x + 1 + i) \\ &= (x + 1)^2 - i^2 \\ &= x^2 + 2x + 2 \end{align*} \]

Now dividing the quartic polynomial \( x^4 + 4x^3 + 5x^2 + 2x - 2 \) by the quadratic factor \( x^2 + 2x + 2 \) using long division, we get:

\[ \frac{x^4 + 4x^3 + 5x^2 + 2x - 2}{x^2 + 2x + 2} = x^2 + 2x - 1 \]

The other roots are obtained from:

\[ \begin{align*} x^2 + 2x - 1 &= 0 \\ x &= -1 \pm \sqrt{2} \end{align*} \]

Thus, the roots of the equation \( x^4 + 4x^3 + 5x^2 + 2x - 2 = 0 \) are:

\[ x = -1 + i, \quad -1 - i, \quad -1 + \sqrt{2}, \quad -1 - \sqrt{2} \]

To find the value of \( f(x+h)\), where \( f(x)\) is a polynomial, using Horner's process

Solution of the Equation Let

\[ f(x)=p_0x^n+p_1x^{n-1}+p_2x^{n-2}+\cdots+p_{n-1}x+p_n \]

Put \(x=y+h\) , and suppose that \(f(x)\) then becomes;

\[ \begin{align*} f(x)&=q_0y^n+q_1y^{n-1}+q_2y^{n-2}+\cdots+q_{n-1}y+q_n \\[1.5mm] \therefore \quad f(x)&=q_0(x-h)^n+q_1(x-h)^{n-1}+q_2(x-h)^{n-2}+\cdots+q_{n-1}(x-h)+q_n \end{align*} \]

Therefore \(q_n\) is the remainder found by dividing \(f(x)\) by \(x-h\) ; also the quotient arising from the division is

\[ q_0(x-h)^{n-1}+q_1(x-h)^{n-2}+q_2(x-h)^{n-3}+\cdots+q_{n-2}(x-h)+q_{n-1} \]

Similarly, \(q_{n-1}\) is the remainder found by dividing the last expression by \(x-h\) ; and the quotient arising from the division is

\[ q_0(x-h)^{n-2}+q_1(x-h)^{n-3}+\cdots+q_{n-3}(x-h)+q_{n-2} \]

and so on.
Thus \(q_n, q_{n-1}, q_{n-2},\ldots,q_0\) may be found by dividing \(f(x)\) successively by \(x-h\)
The last quotient is \(q_0\) , and is obviously equal to \(p_0\)

\[ \begin{align*} \therefore \quad f(x+h)=q_0x^n+q_1x^{n-1}+q_2x^{n-2}+\cdots+q_{n-1}x+q_n \end{align*} \]

Example 1: Find the result of changing \( x\) into \( x+3\) in the expression \(2x^4-x^3-2x^2+5x-1\).

Proof of the Theorem Solution: The given equation is: \[f(x)=2x^4-x^3-2x^2+5x-1\] Here we divide \( f(x)\) successively by \( x-3\)

\[ \begin{array}{r|rrrrr} 3 & 2 & -1 & -2 & 5 & -1 \\ & ↓ & 6 & 15 & 39 & 132 \\ \hline 3 & 2 & 5 & 13 & 44 & 131 &=q_4 \\ & ↓ & 6 & 33 & 138 \\ \hline 3 & 2 & 11 & 46 & 182 & =q_3\\ & ↓ & 6 & 51 \\ \hline 3 & 2 & 17 & 97 & =q_2 \\ & ↓ & 6 \\ \hline & 2 & 23 & =q_1 \end{array} \]

Thus,

\[ \begin{align*} f(x+3)&=q_0x^4+q_1x^3+q_2x^2+q_3x+q_4 \\[1.5mm] \therefore \quad f(x+3)&=2x^4+23x^3+97x^2+182x+131 \end{align*} \]

Example 2: If \( f(x) = x^4 + 10x^3 + 39x^2 + 76x + 65 \) , find the value of \( f(x-4)\) .

Solution for f(x-4) Solution: The given equation is: \[ f(x) = x^4 + 10x^3 + 39x^2 + 76x + 65 \] Here we divide \( f(x) \) successively by \( x + 4 \)

\[ \begin{array}{r|rrrrr} -4 & 1 & 10 & 39 & 76 & 65 \\ & \downarrow & -4 & -24 & -60 & -64 \\ \hline & 1 & 6 & 15 & 16 & 1 &= q_4 \\ & \downarrow & -4 & -8 & -28 \\ \hline & 1 & 2 & 7 & -12 &= q_3 \\ & \downarrow & -4 & 8 \\ \hline & 1 & -2 & 15 &= q_2 \\ & \downarrow & -4 \\ \hline & 1 & -6 &= q_1 \\ \end{array} \]

Thus,

\[ \begin{align*} f(x-4) &= q_0 x^4 + q_1 x^3 + q_2 x^2 + q_3 x + q_4 \\[1.5mm] \therefore \quad f(x-4) &= x^4 - 6x^3 +15x^2 - 12x + 1 \end{align*} \]