Problem 22 - Precalculus with Calculus Previews

Question

Find all real zeros of the given polynomial function \(f\). Then factor \(f(x)\) using only real numbers. $$ f(x)=6 x^{3}+23 x^{2}+3 x-14 $$

Step-by-Step Solution

Answer

The real zeros are 1, -\frac{4}{3}, and -4. The factorization is \((x - 1)(3x + 4)(2x + 8)\).

Step 1: Use the Rational Root Theorem

The Rational Root Theorem suggests potential rational zeros are all combinations of the factors of the constant term (-14) divided by factors of the leading coefficient (6). Therefore, the possible rational roots are \(\pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}, \pm \frac{1}{6}, \pm \frac{7}{6}\).

Step 2: Test Rational Roots

Check these potential roots by plugging them into the polynomial and finding which ones yield a remainder of zero when divided into the polynomial using synthetic division or direct substitution. After testing, we find that \( x = 1 \) is a root.

Step 3: Perform Synthetic Division

Perform synthetic division with \( x = 1 \) on \( f(x) \). Dividing \( 6x^3 + 23x^2 + 3x - 14 \) by \( x - 1 \) gives a quotient of \( 6x^2 + 29x + 32 \) with a remainder of 0, confirming \( x = 1 \) is a root.

Step 4: Factor the Quadratic

Factor the quadratic polynomial \( 6x^2 + 29x + 32 \). Use either the AC method or quadratic formula. By applying the quadratic formula, \( x = \frac{-29 \pm \sqrt{(29)^2-4 \cdot 6 \cdot 32}}{2 \cdot 6} \), and solving, find the quadratic factors to be \( (3x + 4)(2x + 8) \).

Step 5: Write the Complete Factorization

With roots \( x = 1, \ x = -\frac{4}{3}, \text{ and } x = -4 \), the complete factorization of the polynomial \( f(x) \) over the real numbers is \((x - 1)(3x + 4)(2x + 8)\).

Key Concepts

Polynomial Function

A polynomial function is an expression made of variables and coefficients that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. For example, the polynomial function given in the exercise is

• \( f(x) = 6x^3 + 23x^2 + 3x - 14 \).

This specific form is called a cubic polynomial because the highest exponent of the variable \( x \) is 3. Cubic polynomials typically have up to three real roots or zeros, and the degree of the polynomial (here, 3) indicates the maximum number of possible real roots.

When analyzing a polynomial function, we are interested in finding its roots. Roots are simply the values of \( x \) that make the function equal to zero. This means solving the equation \( f(x) = 0 \). By finding the roots, we can also express the polynomial as a product of simpler factors.

Synthetic Division

Synthetic division is a simplified method of dividing a polynomial by a linear factor of the form \( x - c \). It is particularly useful when applying the Rational Root Theorem to find potential roots of a polynomial. In synthetic division, you only need to write down the coefficients of the polynomial and the suspected root. Here’s a brief overview of the process:

• Write down the coefficients of \( f(x) = 6x^3 + 23x^2 + 3x - 14 \).
• Choose \( c = 1 \), since \( x = 1 \) is a root.
• Perform synthetic division, using these coefficients: \( 6, 23, 3, -14 \).

During this method, you repeatedly multiply the divisor (the suspected root) by the difference of the coefficients from the previous step, adding it to the next coefficient. If the final result has a remainder of 0, it confirms that your suspected root is indeed a root of the polynomial. This process helps in breaking down the polynomial into a simpler form, thereby identifying other possible roots.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of a quadratic equation, which is an equation of the form \( ax^2 + bx + c = 0 \). The formula is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the exercise, after using synthetic division, the polynomial \( 6x^2 + 29x + 32 \) is left to be solved. This equation is a quadratic, so the quadratic formula is perfectly suited to solve for its roots.

Applying the formula, we correctly identified the roots:

• \( x = -\frac{4}{3} \)
• \( x = -4 \)

These roots, like any solution to a quadratic equation, occur through algebraic manipulation and calculation involving the coefficients and constants of the quadratic equation.

Polynomial Factorization

Polynomial factorization is the process of breaking down a polynomial into a product of simpler polynomials. The goal is to express the original polynomial function as a product of factors and identify its roots. For the exercise's polynomial \( f(x) = 6x^3 + 23x^2 + 3x - 14 \), we discovered that its real roots are \( x = 1 \), \( x = -\frac{4}{3} \), and \( x = -4 \). Through factorization, the original polynomial can be expressed as:

• \((x - 1)(3x + 4)(2x + 8)\)

Each factor corresponds to a root, and multiplying these factors together will recreate the original polynomial. This factorized form is particularly helpful as it shows both the roots of the polynomial and simplifies solving problems related to the polynomial function. Factorization is crucial in mathematics because it facilitates easier computation and understanding of polynomial behavior by breaking it into manageable parts.