Problem 17 - Algebra 2

Question

Find all of the rational zeros of each function. \(p(x)=x^{3}-3 x-2\)

Step-by-Step Solution

Answer

The rational zeros of the function are \( x = -1 \) and \( x = 2 \).

Step 1: Identify Possible Rational Zeros

Use the Rational Root Theorem, which states that any rational solution \( \frac{p}{q} \) of the polynomial \( a_n x^n + a_{n-1} x^{n-1} + ... + a_0 \) is such that \( p \) is a factor of the constant term \( a_0 \) and \( q \) is a factor of the leading coefficient \( a_n \). Here, the constant term is \(-2\) and the leading coefficient is \(1\), so the possible rational zeros are factors of \(-2\), which are \( \pm 1, \pm 2\).

Step 2: Test Each Possible Rational Zero

Now test each of the possible rational zeros by substituting them into the polynomial \( p(x) \). First, try \( x = 1 \). Calculate \( p(1) = 1^3 - 3(1) - 2 = 1 - 3 - 2 = -4 \). Since the result is not zero, \( x = 1 \) is not a root.

Step 3: Continue Testing Possible Zeros

Continue testing: try \( x = -1 \). Substitute into the polynomial: \( p(-1) = (-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0 \). Since the result is zero, \( x = -1 \) is a root of the polynomial.

Step 4: Divide the Polynomial by \( x + 1 \)

Now that we know \( x = -1 \) is a root, use synthetic division to divide \( p(x) = x^3 - 3x - 2 \) by \( x + 1 \). Perform synthetic division: write the coefficients \( 1, 0, -3, -2 \). Bring down the first coefficient (1), multiply by \(-1\) and add to the next coefficient (0), continue the process to find the quotient \( x^2 - 1x - 2 \).

Step 5: Factor the Quotient

The quotient from the division is \( x^2 - x - 2 \). Factor this quadratic equation. Look for factors of \(-2\) that add up to \(-1\). The factors are \(-2\) and \( 1 \), so it factors to \( (x - 2)(x + 1) \).

Step 6: List All Rational Zeros

The original polynomial \( p(x) = x^3 - 3x - 2 \) is factored as \((x+1)(x-2)(x+1)\). The rational zeros are the values of \( x \) that make each factor zero; thus \( x = -1 \) and \( x = 2 \).

Key Concepts

Synthetic Division

Synthetic division is a simplified method of dividing a polynomial by a linear factor of the form \( x - c \). It's especially useful for quickly finding a polynomial's quotient and remainder. Here’s a step-by-step guide to applying synthetic division:

• Identify the root \( c \) in the linear divisor \( x - c \).
• Write the coefficients of each term in the polynomial. For instance, if you have \( p(x) = x^3 - 3x - 2 \), the coefficients are \( 1, 0, -3, -2 \).
• Bring down the leading coefficient to the bottom row.
• Multiply this value by the root \( c \) and write in the next column, beneath the row of coefficients.
• Add the column values to get a new number for the bottom row, and repeat the process for all columns.

This results in the polynomial's quotient and a remainder. For instance, dividing \( x^3 - 3x - 2 \) by \( x+1 \) results in \( x^2 - x - 2 \). This simplifies the task of finding further roots and factors. Synthetic division is a handy tool for polynomial factoring and helps streamline the equation-solving process.

Polynomial Factoring

Polynomial factoring involves breaking down a polynomial expression into a product of simpler polynomials. Understanding this concept is crucial for determining a polynomial's roots or zeros. Here's how you factor polynomials:

• First, identify any common factors among all terms of the polynomial. This is your greatest common factor (GCF).
• Next, for a quadratic polynomial like \( x^2 - x - 2 \), list the factor pairs of the constant term (in this case, \(-2\)).
• Choose the pair that adds up to the linear coefficient (here, \(-1\)). The appropriate pair for this example is \(-2\) and \(1\), which factor the expression into \( (x - 2)(x + 1) \).
• If dealing with higher-order polynomials, continue applying factoring techniques or synthetic division until fully simplified.

Factorization transforms polynomial problems into simpler tasks, allowing for straightforward determination of zeros or solutions to equations. This method is foundational in algebra and aids in solving polynomial equations effectively.

Quadratic Equation

A quadratic equation is a second-degree polynomial typically expressed as \( ax^2 + bx + c = 0 \). Solving these equations is pivotal for finding roots where the polynomial equals zero. There are several ways to tackle quadratic equations:

• Factoring: As illustrated earlier, this involves expressing the quadratic in its factorized form, such as \( (x - m)(x - n) = 0 \). The solutions are then \( x = m \) and \( x = n \).
• Quadratic Formula: In cases where factoring is complex, use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This equation always gives solutions, provided the discriminant \( b^2 - 4ac \) is non-negative.
• Completing the Square: Another method involves rewriting the equation in the form \( (x - p)^2 = q \), from which the roots can be extracted as \( x = p \pm \sqrt{q} \).

Having multiple strategies for solving quadratic equations ensures that any polynomial of degree two can be tackled with flexibility and confidence, regardless of its complexity.