Problem 59 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The factorization of \(P(x)\) is \((2x+1)^2 (6x-1)\).

Step 1: Identify the Constant and Leading Coefficient

The polynomial is given as \(P(x) = 12x^3 + 20x^2 - x - 6\). The constant term (the term without \(x\)) is \(-6\), and the leading coefficient (the coefficient of the highest power of \(x\)) is 12.

Step 2: List the Possible Rational Zeros

According to the Rational Zeros Theorem, the possible rational zeros are the factors of the constant term divided by the factors of the leading coefficient. The factors of \(-6\) are \(\pm 1, \pm 2, \pm 3, \pm 6\), and the factors of 12 are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\). So, the possible rational zeros are \(\pm 1, \pm 2, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}\).

Step 3: Test the Possible Rational Zeros

Using synthetic division or direct substitution, check which of the possible rational zeros actually make \(P(x) = 0\). For instance, testing \(x = 1\) shows it is not a zero: \[P(1) = 12(1)^3 + 20(1)^2 - (1) - 6 = 12 + 20 - 1 -6 = 25\].Continue testing other values. When \(x = -\frac{1}{2}\), \[P\left(-\frac{1}{2}\right) = 12\left(-\frac{1}{2}\right)^3 + 20\left(-\frac{1}{2}\right)^2 -\left(-\frac{1}{2}\right) - 6 = 0,\]indicating \(-\frac{1}{2}\) is a zero.

Step 4: Perform Polynomial Division

Knowing \(x + \frac{1}{2}\) is a factor, perform synthetic division or long division of \(12x^3 + 20x^2 - x - 6\) by \(2x + 1\) (rewrite as this factor).Performing synthetic division:\[\begin{array}{c|cccc} -\frac{1}{2} & 12 & 20 & -1 & -6 \ & & -6 & -7 & 4 \\hline & 12 & 14 & -8 & -2 \\end{array}\]The quotient from the division is \(12x^2 + 14x - 2\), and it simplifies to \(0\), confirming \(x + \frac{1}{2}\) is a factor.

Step 5: Factor the Quotient

Next, factor the quadratic quotient \(12x^2 + 14x - 2\) by checking for factorizable pairs or using the quadratic formula.Trying factorization:\[12x^2 + 14x - 2 = (6x - 1)(2x + 1)\]This factors further since testing roots and factor pairs confirms this factorization.

Step 6: Write the Complete Factorization

Combine all factors together. The complete factorization of \(P(x)\) is:\[P(x) = (2x + 1)(6x - 1)(x + \frac{1}{2})\].Simplifying it, this becomes:\[P(x) = (2x + 1)^2 (6x - 1).\]Final check confirms these factors generate back the original polynomial.

Key Concepts

Polynomial Factorization

Polynomial factorization is the process of breaking down a polynomial into simpler, multiplicative factors. These factors are polynomials of lower degrees that, when multiplied, will give back the original polynomial. Factorization is crucial because it simplifies complex expressions and can solve polynomial equations by finding their roots.

• The first step in factorization involves identifying the polynomial's degree and leading coefficient, which helps in determining potential rational zeros.
• Using the Rational Zeros Theorem, one can list all possible rational zeros, providing an approach to test which values, if any, make the polynomial equate to zero.
• Once a zero is found, it's factorized out of the polynomial, reducing the residual polynomial into a simpler degree.

Factorization often comes down to identifying patterns such as the sum or difference of cubes or quadratics that can be factored further, simplifying the solution process.

Synthetic Division

Synthetic division is a simplified form of polynomial division, used specifically when dividing by a linear factor of the form \(x - c\). It's a method that allows for faster computation and is especially useful for finding polynomial zeros. In synthetic division, only the coefficients of the polynomial are used in the calculation. Here's a brief overview:

• Write down the coefficients of the polynomial in descending order. If any terms are missing, use a zero for its coefficient.
• Place the zero of the divisor, \(c\), outside the bracket.
• Bring down the leading coefficient, multiply by \(c\), add the result to the next coefficient, and continue the process.
• The last value obtained is the remainder; if zero, the division is exact, confirming a factor.

Synthetic division provides an efficient way to test potential zeros quickly, as shown in the example where testing \(-\frac{1}{2}\) confirmed it as a zero.

Polynomial Division

Polynomial division is a broader method that includes both long division and synthetic division to divide one polynomial by another. It works similarly to numerical long division but with polynomial terms. Long division helps when synthetic division is not applicable, particularly when the divisor is not a linear factor like \(2x + 1\) instead of \(x + \frac{1}{2}\). The process is as follows:

• Divide the highest degree term of the dividend by the highest degree term of the divisor, writing the result.
• Multiply the entire divisor by this result and subtract it from the dividend to form a new polynomial.
• Repeat the process until the remainder's degree is less than the divisor's or the remainder is zero.

This thorough method ensures that every step is clear, leading to accurate factorization or verification of zeros. It showed that \(2x + 1\) divides into the polynomial cleanly.

Quadratic Factorization

Quadratic factorization involves breaking down a quadratic expression (a polynomial of degree 2) into simpler binomial factors. This factorization is essential because it helps in solving quadratic equations and finding polynomial roots. To factor a quadratic like \(12x^2 + 14x - 2\), several methods can be used:

• Factoring by grouping: This involves rearranging the terms and grouping them in pairs, making it easier to spot common factors.
• The quadratic formula: For any quadratic \(ax^2 + bx + c\), the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) can give the roots, which directly translate to factors.
• Trial and error or mental math: Sometimes, factors can quickly be guessed if the polynomial components are simple, as \((6x - 1)(2x + 1)\) factors of the quadratic portion here.

This factorization technique simplifies complex algebraic expressions into much more manageable parts and is key in solving higher-degree polynomials, as seen in our given solution.