Problem 14 - Algebra for College Students

Question

Use synthetic division to determine the quotient and remainder for each problem. $$ \left(x^{3}-6 x^{2}+5 x+14\right) \div(x-4) $$

Step-by-Step Solution

Answer

Quotient: \(x^2 - 2x - 3\), Remainder: 2.

Step 1: Setup Synthetic Division

First, write down the coefficients of the dividend polynomial \(x^3 - 6x^2 + 5x + 14\), which are \(1, -6, 5, 14\). Write the divisor \(x - 4\) as \(x - c\) where \(c = 4\). Set up the synthetic division by writing 4 to the left and the coefficients in a row.

Step 2: Bring Down the Leading Coefficient

Bring down the leading coefficient \(1\) to the bottom row. This will be the start of your quotient.

Step 3: Multiply and Add Iteratively

Multiply the number you've just brought down (\(1\)) by \(4\) and write the result under the next coefficient (\(-6\)). Then add the number you've just written to the \(-6\) above it. This gives the next number in the bottom row. Repeat this process for every coefficient. - Multiply \(1\) by \(4\) to get \(4\), add to \(-6\) to get \(-2\).- Multiply \(-2\) by \(4\) to get \(-8\), add to \(5\) to get \(-3\).- Multiply \(-3\) by \(4\) to get \(-12\), add to \(14\) to get \(2\).

Step 4: Interpret the Bottom Row

The numbers in the bottom row represent the coefficients of the quotient polynomial and the remainder. The quotient is \(x^2 - 2x - 3\) and the remainder is \(2\).

Key Concepts

Polynomials

Polynomials are mathematical expressions made up of variables and coefficients, organized in terms of powers or exponents of the variable. The general form of a polynomial is expressed as: \[ a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \]where each term consists of a coefficient (such as \(a_n, a_{n-1}, \) etc.) and a variable raised to an exponent (like \(x^n, x^{n-1}, \) etc.).
- The highest power of the variable is called the degree of the polynomial. In this exercise, the polynomial is of degree 3: \(x^3 - 6x^2 + 5x + 14\).
- Coefficients tell you how much of each power of the variable is present. Here, they are 1, -6, 5, and 14.
Understanding polynomials is crucial as they are foundational in algebra and calculus, and they frequently appear in equations that model real-world situations.

Quotient and Remainder

In mathematics, when dividing polynomials, the result can be broken down into a quotient and a remainder. This process is quite similar to long division with numbers.
- The **quotient** is the result of the division. It's what you get when you divide the polynomial completely except for the remainder part.
- The **remainder** is what's left after the division is complete. In polynomial division, this remainder will have a degree lower than the divisor polynomial.
For the given exercise, when dividing \(x^3 - 6x^2 + 5x + 14\) by \(x - 4\), we end up with a quotient of \(x^2 - 2x - 3\) and a remainder of 2. When you put it all together, the original polynomial can be expressed as:
\[ \text{Dividend} = (\text{Divisor}) \times (\text{Quotient}) + \text{Remainder} \]
So, in this case:
\[ x^3 - 6x^2 + 5x + 14 = (x - 4)(x^2 - 2x - 3) + 2 \]
This identity verifies the division process and is useful for checking your work.

Step-by-Step Solution

Learning how to perform synthetic division step-by-step can simplify complex polynomial division tasks. Here is how it's done:
1. **Set up Synthetic Division**: Write down the coefficients of the polynomial you are dividing. For \(x^3 - 6x^2 + 5x + 14\), the coefficients are 1, -6, 5, 14. The divisor \(x - 4\) gives \(c = 4\).
2. **Bring Down the Leading Coefficient**: Start by bringing down the leading coefficient, which is 1, to the bottom row.
3. **Multiply and Add**: Multiply the "brought down" number by \(c\) (4) and add it to the next coefficient. Repeat for each coefficient:
- Multiply 1 by 4, add to -6 to get -2.
- Multiply -2 by 4, add to 5 to get -3.
- Multiply -3 by 4, add to 14 to get 2.
4. **Result Interpretation**: The bottom row gives the coefficients of the quotient, and the last number is the remainder. Thus, the quotient is \(x^2 - 2x - 3\), and the remainder is \(2\).
This method is valued for its simplicity and efficiency, especially when dealing with polynomials of higher degrees.