91Ó°ÊÓ

The Rational Root Theorem is a powerful tool that helps us find possible rational solutions, or roots, of polynomial equations. It works by narrowing down the potential rational roots to a finite set of fractions. Here's how it works: If we have a polynomial in the form \[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \text{...} + a_1 x + a_0 \] where the coefficients \(a_n, a_{n-1},... , a_0 \) are integers, the theorem states that every rational solution can be expressed as a fraction \( \frac{p}{q} \) with:

• \( p \) being a factor of the constant term \(a_0 \)
  
• \( q \) being a factor of the leading coefficient \(a_n \)
  

For example, in our polynomial \(x^4 - 5x^3 + 5x^2 + 5x - 6 = 0 \), the constant term is -6 and the leading coefficient is 1. Hence, the possible rational roots are:

• Factors of -6: \( \pm 1, \pm 2, \pm 3, \pm 6 \)
  
• Factors of 1: \( \pm 1 \)
  

So, the possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6 \). We then test these values by substitution into the polynomial to determine which, if any, are actual roots.

Descartes' Rule of Signs provides a way to estimate the number of positive and negative real roots of a polynomial by examining the signs of its coefficients. Here's how it works:

• To find the number of positive real roots, look at the polynomial as it is and count how many times the sign of its coefficients changes. Each sign change represents a potential positive root.
  
• To find the number of negative real roots, replace \(x\) with \(-x\) in the polynomial and then count the sign changes in the resulting polynomial.
  

For example, if we apply Descartes' Rule to our polynomial \(x^4 - 5x^3 + 5x^2 + 5x - 6 = 0 \), we count the sign changes between the terms:

1. \( x^4 \) to \( -5x^3 \) (positive to negative)
   
2. \( -5x^3 \) to \( 5x^2 \) (negative to positive)
   
3. \( 5x^2 \) to \( 5x \) (no change)
   
4. \( 5x \) to \( -6 \) (positive to negative)
   

So, we have three sign changes, indicating up to three positive roots. For negative roots, substituting \( -x \) leads to no sign changes, suggesting no negative roots.

Polynomial division is a method used to simplify polynomials by dividing them by other polynomials, especially when simplifying to find the roots. Synthetic division is a streamlined form of long division particularly useful when dividing by linear polynomials (like \(x - c\)).
Using the root \(x = 1 \) from our initial test using the Rational Root Theorem on \(P(x) = x^4 - 5x^3 + 5x^2 + 5x - 6 \), we perform synthetic division:\[x^4 - 5x^3 + 5x^2 + 5x - 6\] becomes \((x - 1)(x^3 - 4x^2 + x + 6)\).Next, we use \(x = -1 \):

• We divide \(x^3 - 4x^2 + x + 6 \) by \(x + 1\) to get:\[ (x - 1)(x + 1)(x^2 - 3x + 6)\].

Now that we have simplified it further, we are closer to finding all roots.

The Quadratic Formula is a straightforward way to find the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Consider our simplified polynomial \(x^2 - 3x + 6 = 0\): Here, \(a = 1\), \(b = -3\), and \(c = 6\). Substituting these values into the formula gives: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(6)}}{2(1)} \] \[ x = \frac{3 \pm \sqrt{9 - 24}}{2} \] \[ x = \frac{3 \pm \sqrt{-15}}{2} \] Since our result involves a negative number under the square root, we introduce the imaginary unit \(i\), so we get \(x = \frac{3 \pm \sqrt{15}i}{2} \). Now we have identified all the roots: \(x = 1, -1\), and the complex roots \( \frac{3 \pm \sqrt{15}i}{2} \).