Problem 59 - A Graphical Approach to College Algebra

Question

Use the rational zeros theorem to factor \(P(x)\). $$P(x)=12 x^{3}+20 x^{2}-x-6$$

Step-by-Step Solution

Answer

The factored form of \( P(x) \) is \((x - \frac{1}{2})(2x + 3)(6x + 4)\).

Step 1: Define the Rational Zeros Theorem

The Rational Zeros Theorem states that if a polynomial has a rational zero \( \frac{p}{q} \), where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient, then all possible rational zeros are given by the set of values \( \frac{p}{q} \).

Step 2: Identify factors of constant and leading coefficients

For the polynomial \( P(x) = 12x^3 + 20x^2 - x - 6 \), the constant term is -6 and the leading coefficient is 12. We need to find all factors of -6 (±1, ±2, ±3, ±6) and 12 (±1, ±2, ±3, ±4, ±6, ±12).

Step 3: List all possible rational zeros

The possible rational zeros are given by the combinations of factors of the constant over factors of the leading coefficient: ±\( \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{12}, 2, 3, 6 \). Simplifying this gives the list: ±1, ±\( \frac{1}{2} \), ±\( \frac{1}{3} \), ±\( \frac{1}{4} \), ±\( \frac{1}{6} \), ±2, ±3, ±6.

Step 4: Test possible zeros using substitution

Substitute each potential zero into the polynomial \( P(x) \) to check if it results in 0 (indicating a root). For example, substitute \( x = 1 \) into \( P(x) \): \[ P(1) = 12(1)^3 + 20(1)^2 - 1 - 6 = 25 \], which is not zero. Continue this process for \( x = -1, x = 2, x = -2, x = \frac{1}{2}, x = -\frac{1}{2}, \) and others until zero is found.

Step 5: Find a valid rational zero

Testing \( x = \frac{1}{2} \)...\[ P\left(\frac{1}{2}\right) = 12\left(\frac{1}{2}\right)^3 + 20\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) - 6 = 0 \].Hence, \( \frac{1}{2} \) is a root.

Step 6: Perform polynomial division

Divide \( P(x) \) by \( x - \frac{1}{2} \) to find the other factors. Properly carry out synthetic or long division to divide \( 12x^3 + 20x^2 - x - 6 \) by \( 2x - 1 \), as multiplying by reciprocal to make integer division easier. The quotient is \( 12x^2 + 26x + 12 \).

Step 7: Factor the quadratic polynomial

Factor \( 12x^2 + 26x + 12 \) by using methods like the quadratic formula or factoring by grouping. The factored form is \((2x + 3)(6x + 4)\).

Step 8: Write the complete factorization of P(x)

Combine all factors, including the initial factor, to express the complete factorization of \( P(x) \). Thus, the factored form of \( P(x) \) is: \( (x - \frac{1}{2})(2x + 3)(6x + 4) \).

Key Concepts

Polynomial Factorization

Factoring a polynomial is like breaking down a complex puzzle into simpler pieces. It involves expressing the polynomial as a product of its factors, which are polynomials of lower degree. This process helps in solving polynomial equations, simplifying expressions, and finding polynomial roots.

To factor a polynomial, we look for numbers or expressions that multiply together to give the original polynomial. In the exercise, we started with the cubic polynomial \( P(x) = 12x^3 + 20x^2 - x - 6 \).

• Firstly, by using the Rational Zeros Theorem, we identified a rational root, \( x = \frac{1}{2} \).
• We then used polynomial division to divide \( P(x) \) by \( x - \frac{1}{2} \), which gave us a quadratic polynomial.
• The final step was to factor the resulting quadratic polynomial into simpler binomial factors. This resulted in the product \( (x - \frac{1}{2})(2x + 3)(6x + 4) \).

The factorization process is crucial in algebra as it allows us to solve polynomial equations by setting each factor to zero and solving for the variable.

Rational Roots

Rational roots are potential solutions to a polynomial equation that can be expressed as fractions of integers. According to the Rational Zeros Theorem, these roots are formed by factors of the constant term divided by factors of the leading coefficient. In our exercise, the polynomial is \( P(x) = 12x^3 + 20x^2 - x - 6 \).

The whole idea is to list all possible candidate roots that could be solutions of the polynomial function.

• The constant term in our polynomial is \(-6\), with factors \(±1, ±2, ±3, ±6\).
• The leading coefficient is \(12\), with factors \(±1, ±2, ±3, ±4, ±6, ±12\).
• You then form all possible fractions of these factors, which gave us things like \(±\frac{1}{1}, ±\frac{1}{2}, ±1, ±\frac{1}{4}, ±3, ±\frac{3}{2}, ±6\), etc.

We then had to test each of these rational numbers in the polynomial equation to identify which one, if any, result in zero. Once a valid root like \( x = \frac{1}{2} \) is found, it validates the potential rational roots given by the theorem.

Synthetic Division

Synthetic division is a streamlined method of dividing a polynomial by a linear factor of the form \( x - r \), which helps to quickly determine whether a number is a root of the polynomial and facilitates further factorization. Let's see how it played into our exercise with \( P(x) = 12x^3 + 20x^2 - x - 6 \).

After identifying \( x = \frac{1}{2} \) as a root, we could use synthetic division to divide \( P(x) \) by \( x - \frac{1}{2} \) to simplify it. Here's how it works:

• Write \( \frac{1}{2} \) on the left and the coefficients of the polynomial on the right: \( (12, 20, -1, -6) \).
• Bring down the first coefficient, \( 12 \).
• Multiply \( 12 \) by \( \frac{1}{2} \) and add this product to the next coefficient \(20\), which gives 26.
• Repeat the multiplication and addition process, and if the last number is 0, it confirms \( \frac{1}{2} \) is a root and helps factor further.

This procedure gives us the reduced polynomial \( 12x^2 + 26x + 12 \), which we factor further into \( (2x + 3)(6x + 4) \). Synthetic division is preferred for its simplicity and speed, especially for polynomials with higher degrees.