Finding Real Roots of Polynomial Functions

Generating Rational Zero Lists

• Systematic Construction of Candidate Roots:

  • To construct a complete list of candidate rational zeros using the Rational Zero Theorem, systematically iterate through potential numerators and denominators.
  • Hold the numerator constant and pair it with every available denominator.
  • Move to the next numerator and repeat the process across all available denominators.
  • Example with numerators $1, 2, 3, 6$ and denominators $1, 2$:
  • With numerator $1$: denominator $1$ gives $\pm \frac{1}{1}$; denominator $2$ gives $\pm \frac{1}{2}$.
  • With numerator $2$: denominator $1$ gives $\pm \frac{2}{1}$; denominator $2$ gives $\pm \frac{2}{2}$.
  • With numerator $3$: denominator $1$ gives $\pm \frac{3}{1}$; denominator $2$ gives $\pm \frac{3}{2}$.
  • With numerator $6$: denominator $1$ gives $\pm \frac{6}{1}$; denominator $2$ gives $\pm \frac{6}{2}$.

• Interpretation of Plus-or-Minus Notation:

  • The $\pm$ symbol serves as a compact representation for two separate numbers: the positive value and the negative value.
  • Example: $\pm \frac{6}{2}$ represents both $+\frac{6}{2}$ and $-\frac{6}{2}$.
  • Using $\pm$ prevents the need to duplicate candidate lists into separate positive and negative groups, though each sign must be tested as an independent root.

• Simplification and Removal of Duplicate Values:

  • Simplification prevents testing redundant numerical values during synthetic division.
  • List simplification steps:
  • $\pm \frac{1}{1} = \pm 1$
  • $\pm \frac{1}{2} = \pm \frac{1}{2}$
  • $\pm \frac{2}{1} = \pm 2$
  • $\pm \frac{2}{2} = \pm 1$ (Duplicate of $\pm 1$; omitted)
  • $\pm \frac{3}{1} = \pm 3$
  • $\pm \frac{3}{2} = \pm \frac{3}{2}$
  • $\pm \frac{6}{1} = \pm 6$
  • $\pm \frac{6}{2} = \pm 3$ (Duplicate of $\pm 3$; omitted)
  • Final simplified candidate root list without repetitions:
  • $\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 6$

Analyzing Polynomial Functions via Descartes' Rule of Signs

• Target Polynomial Function Characteristics:

  • Polynomial function: $f(x) = 2x^3 + 11x^2 - 7x - 6$
  • The terms share no greatest common factor (GCF) that can be factored out to reduce the expression directly to a quadratic form.
  • Finding the roots requires using Descartes' Rule of Signs, the Rational Zero Theorem, and synthetic division in combination.

• Application of Descartes' Rule of Signs for Positive Real Roots:

  • Descartes' Rule of Signs determines the maximum number of positive real roots by counting the sign changes between consecutive nonzero coefficients in $f(x)$.
  • Evaluating sign changes in $f(x) = +2x^3 + 11x^2 - 7x - 6$:
  • Term 1 ($+2x^3$) to Term 2 ($+11x^2$): No sign change ($+$ to $+$).
  • Term 2 ($+11x^2$) to Term 3 ($-7x$): First sign change ($+$ to $-$).
  • Term 3 ($-7x$) to Term 4 ($-6$): No sign change ($-$ to $-$).
  • Total sign changes in $f(x) = 1$
  • Possible positive real roots must equal the number of sign changes or be less than it by an even integer.
  • Subtracting $2$ from $1$ yields a negative value ($1 - 2 = -1$), which is not mathematically valid for a count of roots.
  • Conclusion: The polynomial has exactly $1$ positive real root.

Determining Possible Negative Real Roots

• Evaluating $f(-x)$ for Negative Real Roots:

  • Substitute $-x$ into the original function $f(x) = 2x^3 + 11x^2 - 7x - 6$:
  • $f(-x) = 2(-x)^3 + 11(-x)^2 - 7(-x) - 6$
  • Exponent evaluation rules:
  • Raising $-x$ to an odd power preserves the negative sign: $(-x)^3 = -x^3$. Thus, $2(-x)^3 = -2x^3$.
  • Raising $-x$ to an even power eliminates the negative sign: $(-x)^2 = x^2$. Thus, $11(-x)^2 = 11x^2$.
  • Linear evaluation: $-7(-x) = +7x$
  • Constant term remains unchanged: $-6$
  • Fully simplified expression:
  • $f(-x) = -2x^3 + 11x^2 + 7x - 6$

• Counting Sign Changes in $f(-x)$:

  • Evaluating sign changes in $f(-x) = -2x^3 + 11x^2 + 7x - 6$:
  • Term 1 ($-2x^3$) to Term 2 ($+11x^2$): First sign change ($-$ to $+$).
  • Term 2 ($+11x^2$) to Term 3 ($+7x$): No sign change ($+$ to $+$).
  • Term 3 ($+7x$) to Term 4 ($-6$): Second sign change ($+$ to $-$).
  • Total sign changes in $f(-x) = 2$
  • Subtracting even integers ($2$) gives two possibilities: $2 - 2 = 0$
  • Conclusion: The polynomial has $2$ or $0$ negative real roots.

Root Testing with Synthetic Division

• Testing Strategy:

  • Candidate roots from the rational zero list are tested using synthetic division (or long division).
  • A tested value $c$ is a true root if and only if the remainder of the division equals $0$
  • A remainder of $0$ proves that $(x - c)$ is a factor of $f(x)$.

• Execution of Root Test 1 ($x = 1$):

  • Tested candidate root: $x = 1$
  • Associated factor: $(x - 1)$
  • Coefficients of $f(x) = 2x^3 + 11x^2 - 7x - 6$ in standard form: $2, 11, -7, -6$
  • Synthetic division steps:
  1. Bring down the leading coefficient: $2$
  2. Multiply the root by the coefficient ($1 \times 2 = 2$) and add to the second coefficient ($11 + 2 = 13$).
  3. Multiply the root by the result ($1 \times 13 = 13$) and add to the third coefficient ($-7 + 13 = 6$).
  4. Multiply the root by the result ($1 \times 6 = 6$) and add to the constant term ($-6 + 6 = 0$).
  • Remainder result: $0$
  • Conclusion: $x = 1$ is a confirmed real root, and $(x - 1)$ is a factor of $f(x)$.

Updating the Rational Zero Candidates

• Impact of Descartes' Rule of Signs on Remaining Candidates:

  • Descartes' Rule of Signs established that $f(x)$ contains exactly $1$ positive real root.
  • Because $x = 1$ was confirmed as a positive real root, no other positive real roots can exist for $f(x)$.
  • All remaining positive values in the rational zero candidate list are eliminated.

• Downsized Candidate List for Subsequent Testing:

  • The remaining candidate rational roots are strictly negative:
  • $-1, -\frac{1}{2}, -2, -3, -\frac{3}{2}, -6$