Problem 75 - A Graphical Approach to College Algebra

Question

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=-6 x^{3}-13 x^{2}+14 x-3 ; \quad k=-3$$

Step-by-Step Solution

Answer

\(P(x) = -6(x + 3)(x)(6x + 31)\).

Step 1: Verify k as a Zero

To verify that \(k = -3\) is a zero of \(P(x)\), substitute \(-3\) into \(P(x)\): \[P(-3) = -6(-3)^3 - 13(-3)^2 + 14(-3) - 3.\] Calculate: \(-6(-27) - 13(9) - 42 - 3 = 162 - 117 - 42 - 3 = 0.\) Since \(P(-3) = 0\), \(k = -3\) is indeed a zero.

Step 2: Perform Synthetic Division

Use synthetic division on \(P(x)\) with \(k = -3\):- Coefficients: \([-6, -13, 14, -3]\).- Bring down \(-6\).- Multiply \(-3\) by \(-6\) and add to \(-13\) to get \(-31\).- Multiply \(-3\) by \(-31\) and add to \(14\) to get \(107\).- Multiply \(-3\) by \(107\) and add to \(-3\) to get \(0\).The quotient is \(-6x^2 - 31x + 107\) and the remainder is \(0\), confirming \(x + 3\) as a factor.

Step 3: Factor the Quadratic

Factor \(-6x^2 - 31x + 107\). First, check if it can be factored using simple techniques or find its roots using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] where \(a = -6, b = -31, c = 107\). Calculate \[b^2 - 4ac = (-31)^2 - 4(-6)(107) = 961,\] leading to roots \[x = \frac{31 \pm 31}{-12},\] which gives \(x = 0\) or \(x = \frac{62}{-12} = -\frac{31}{6}.\)Thus, \(-6x^2 - 31x + 107 = -6(x - 0)(x + \frac{31}{6})\).

Step 4: Write the Factorized Form

Combine all factors: \[P(x) = -6(x + 3)(x)(x + \frac{31}{6}).\]Convert to linear factors for simplicity: \[P(x) = -6(x + 3)(x)\left(6x + 31\right).\] Take \(-1\) common out of \(6x + 31\) to simplify.

Key Concepts

Synthetic Division

Synthetic Division is a handy shortcut for dividing polynomials when you know one of the roots. In essence, it simplifies the long division process, which can save you time and effort.

To use synthetic division, you only need the coefficients of the polynomial. Here's a step-by-step process using our example:

• List the coefficients of the polynomial \([-6, -13, 14, -3]\).
• Write the value of the known root, which is \(-3\) here, outside the division box.
• Bring down the leading coefficient, \(-6\), to start.
• Multiply \(-3\) by \(-6\) to get \(18\), and add this to the next coefficient, \(-13\), resulting in \(-31\).
• Repeat this process: multiply \(-3\) by the new result \(-31\) and add the product to the next coefficient to continue the process.
• The final result should have no remainder if the divisor is indeed a factor.

This method confirms whether the divisor is a factor. If the remainder is zero, it means that \(x + 3\) is indeed a factor. This simplifies the larger polynomial into manageable quadratic factors.

Quadratic Formula

The Quadratic Formula is a well-known tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It provides a straightforward means to find the roots when factoring directly isn't feasible.

The formula is written as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula stems from completing the square on the general quadratic equation. In our example:

• Coefficients are \(a = -6\), \(b = -31\), \(c = 107\).
• First, calculate the discriminant, \(b^2 - 4ac\).
• Use the discriminant to check the number of real roots. A positive discriminant indicates two real roots.
• Plug values into the formula for precise roots.

This method is powerful because it guarantees finding the roots of any quadratic equation, no matter the complexity.

Roots of Polynomial

Roots of a polynomial indicate the values for which the polynomial equals zero. These values are where the graph of the polynomial crosses the x-axis. Finding all roots is crucial because they determine the polynomial's behavior and can assist in factorization.

To find roots:

• Start with the given polynomial and substitute suspected roots to verify they make the polynomial zero.
• If confirmed, use techniques like synthetic division to reduce the polynomial's degree and find other roots.
• Repeated roots indicate the polynomial touches the x-axis rather than crossing it.
• The number of roots corresponds to the polynomial's degree.

Identifying roots facilitates the factorization process, as each root corresponds to a linear factor of the polynomial.

Linear Factors

Linear Factors represent the simplest form of a polynomial, broken down into its root-based components. Each linear factor corresponds to a root, expressed in the form \((x - r)\) where \(r\) is a root.

For the polynomial to be expressed in linear factors:

• Start by finding all roots of the polynomial.
• Translate each root into a linear factor. For example, if \(r = -3\), the factor is \((x + 3)\).
• Combine all factors to express the polynomial's factorized form. If a root appears multiple times, it is included repeatedly.
• Simplify to ensure all coefficients and terms are accurate.

Converting to linear factors simplifies the polynomial's expression and can help in further analysis, such as graphing or simplifying larger expressions.