Problem 28 - Algebra 2

Question

Use synthetic substitution to show that \(x-8\) is a factor of \(x^{3}-4 x^{2}-29 x-24 .\) Then find any remaining factors.

Step-by-Step Solution

Answer

The factors are \((x - 8)(x + 1)(x + 3)\).

Step 1: Set Up Synthetic Division

First, list the coefficients of the polynomial \(x^3 - 4x^2 - 29x - 24\), which are [1, -4, -29, -24]. Since we are using synthetic substitution for \(x - 8\), use \(8\) as the divisor.

Step 2: Perform Synthetic Division

Write 8 on the left and the coefficients [1, -4, -29, -24] on the right. Begin by bringing down the 1. Multiply it by 8, resulting in 8, and write it under -4. Add to get 4. Repeat the process: multiply 4 by 8 to get 32, add to get 3, multiply 3 by 8 to get 24, and add to get 0.

Step 3: Analyze the Remainder

The final number, 0, is the remainder. Because the remainder is 0, this confirms \(x - 8\) is a factor. The result of the division is \(x^2 + 4x + 3\).

Step 4: Factor the Quotient

Factor the quadratic \(x^2 + 4x + 3\). Look for two numbers that multiply to 3 and add to 4. These numbers are 1 and 3. Therefore, the quadratic factors as \((x + 1)(x + 3)\).

Step 5: Write the Complete Factorization

Combine all factors: since \(x - 8\) is a factor and \(x^2 + 4x + 3\) factors as \((x + 1)(x + 3)\), the complete factorization of \(x^3 - 4x^2 - 29x - 24\) is \((x - 8)(x + 1)(x + 3)\).

Key Concepts

Polynomial Factorization

Polynomial factorization is the process of expressing a polynomial as a product of its simpler factors. This is similar to integer factorization, where we break down a number into the product of its prime factors.
For polynomials, these factors are typically polynomials of lower degrees. Factoring helps simplify expressions and solve polynomial equations more easily.
For example, when we factor the polynomial \( x^3 - 4x^2 - 29x - 24 \), we express it as a product of lower-degree polynomials.

• To start with factorization, it's helpful to use techniques like finding common factors or applying special factorization formulas such as difference of squares, sum/difference of cubes, or completing the square.
• Advanced methods involve synthetic division or methods like the quadratic formula for finding roots and further factorization.

Recognizing when a polynomial can be further factored or when it has been completely factored is a key skill in algebra.

Synthetic Division

Synthetic division is a simplified version of polynomial division that is particularly useful when dividing a polynomial by a linear factor, like \(x - a\). It streamlines the division process by focusing only on coefficients.
To perform synthetic division, follow these steps:

• Write the coefficients of the polynomial in descending order of degree.
• Use the root from the binomial (e.g., the \(a\) in \(x-a\)) to the left of the setup.
• Bring down the leading coefficient as is.
• Multiply the root by this leading coefficient and add it to the next coefficient, repeating as you move across all terms.

This method is quick and efficient for handling polynomials and helps in determining factors, as seen when \(x-8\) results in a zero remainder with \( x^3 - 4x^2 - 29x - 24 \).
Synthetic division provides both the quotient and remainder, aiding in confirming factors and simplifying complex polynomial expressions.

Quadratic Factoring

Quadratic factoring deals with finding two binomials whose product is a given quadratic polynomial. This is often used after reducing a higher-degree polynomial to a quadratic form using techniques like synthetic division.
The typical form for a quadratic polynomial is \( ax^2 + bx + c \). To factor it, you'll look for two numbers that multiply to \( ac \) (the product of the coefficient of \(x^2\) and the constant term) and sum to \(b\).

• These numbers are used to split the middle term and make factoring by grouping possible.
• Alternatively, if the quadratic is simple, direct factoring by inspecting possibilities for terms that add and multiply correctly may suffice.

In the exercise example, the quadratic \(x^2 + 4x + 3\) is factored as \((x + 1)(x + 3)\), completing the factorization of the original cubic polynomial. Mastering quadratic factoring is essential for simplifying equations and solving polynomial expressions in algebra.