Problem 20 - Algebra and Trigonometry

Question

Find all rational zeros of the polynomial. $$ P(x)=x^{3}-2 x^{2}-2 x-3 $$

Step-by-Step Solution

Answer

The rational zero of the polynomial is \( x = 3 \).

Step 1: Apply Rational Root Theorem

The Rational Root Theorem suggests that any rational root of the polynomial \( P(x) = x^3 - 2x^2 - 2x - 3 \) is a factor of the constant term divided by a factor of the leading coefficient. The constant term here is \(-3\), and the leading coefficient is \(1\). Therefore, the possible rational roots are \( \pm 1, \pm 3 \).

Step 2: Test Possible Roots via Evaluation

We need to evaluate \( P(x) \) at each of the potential rational roots \( \pm 1, \pm 3 \) to see if any of them equate to zero. Start with \( x = 1 \): \[ P(1) = 1^3 - 2(1)^2 - 2(1) - 3 = 1 - 2 - 2 - 3 = -6 \] Since \( P(1) eq 0 \), \( x = 1 \) is not a root. Next, try \( x = -1 \): \[ P(-1) = (-1)^3 - 2(-1)^2 - 2(-1) - 3 = -1 - 2 + 2 - 3 = -4 \] Since \( P(-1) eq 0 \), \( x = -1 \) is not a root. Next, try \( x = 3 \): \[ P(3) = 3^3 - 2(3)^2 - 2(3) - 3 = 27 - 18 - 6 - 3 = 0 \] Thus, \( x = 3 \) is a rational root.

Step 3: Polynomial Division

Since we found that \( x = 3 \) is a root, we can perform synthetic division of \( P(x) \) by \( x - 3 \). Performing synthetic division: Dividing coefficients \(1, -2, -2, -3\) by \(3\) using synthetic division results in the quotient polynomial with coefficients \(1, 1, 1\), meaning the quotient is \( x^2 + x + 1 \).

Step 4: Find Remaining Roots

Now, we need to find the roots of the quotient polynomial \( x^2 + x + 1 \). Detecting rational roots, if any, requires checking if the quadratic can be factored nicely, but it cannot. Solve this quadratic using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1, b = 1, c = 1 \).This leads to: \[ x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm i\sqrt{3}}{2} \] These solutions are non-rational numbers.

Key Concepts

Synthetic Division

Synthetic division is a simplified method of dividing polynomials, particularly useful when dividing by a linear factor of the form \(x - c\). This technique is quicker and more efficient than long polynomial division and less prone to calculation errors.

• To start synthetic division, write down all the coefficients of the polynomial being divided, in order. If a term is missing, use a 0 for that coefficient.
• Next, write the root you are dividing by, which is the number \(c\) from \(x-c\), to the left.
• Bring down the leading coefficient to the bottom row.
• Then multiply this number by \(c\) and write that under the next coefficient. Add them together and continue this process across all the coefficients.
• The bottom row, except for the last term, represents the coefficients of the quotient.
• The final term in the bottom row is the remainder. If it is zero, then \(x-c\) is a factor of the polynomial.

This method showed us that \(x-3\) was a factor of our polynomial, leaving us with the quadratic \(x^2 + x + 1\). It's a helpful tool to simplify polynomials and identify possible roots.

Polynomial Division

Polynomial division is a more general method for dividing one polynomial by another. It is similar to long division with numbers and applies to polynomials of all types.

• Write the dividend (the polynomial you are dividing) and the divisor (the polynomial by which you are dividing).
• Align terms of similar degree, setting up for subtraction, like in numerical long division.
• Divide the first term of the dividend by the first term of the divisor, obtaining the first term of the quotient.
• Multiply the whole divisor by this term and subtract from the original polynomial.
• Repeat the process with the new polynomial formed after subtraction.

This process continues until you reach a degree smaller than that of the divisor.
Polynomial division was key in confirming that \(x = 3\) was indeed a rational root through synthetic division. The remaining polynomial we derived, \(x^2 + x + 1\), was crucial for finding the other roots.

Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of any quadratic equation of the form \(ax^2 + bx + c=0\). This formula provides a systematic approach to find even the most complex roots, including those involving imaginary numbers.

• The general format of the quadratic formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
• The term under the square root, \(b^2 - 4ac\), is called the discriminate. It determines the nature of roots:
• If the discriminant is positive, two distinct real roots exist.
• If it is zero, there is exactly one real root.
• If negative, the roots become complex or imaginary numbers.
• Using the quadratic formula effectively found the roots of \(x^2 + x + 1\) as \(\frac{-1 \pm i\sqrt{3}}{2}\), demonstrating they are non-rational complex numbers.

This calculation confirmed that no additional rational roots existed for our original polynomial. The quadratic formula provided the comprehensive solution when simpler factoring was not possible.