Problem 15 - Precalculus : Building Concepts and Connections

Question

Write each polynomial in the form \(p(x)=d(x) q(x)+r(x),\) where \(p(x)\) is the given polynomial and \(d(x)\) is the given factor. You may use synthetic division wherever applicable. $$x^{2}+x+1 ; x+1$$

Step-by-Step Solution

Answer

The given polynomial can be expressed as \(p(x) = (x+1)(x) + 2\).

Step 1: Set up Synthetic Division

Set up synthetic division with the divisor \(x+1\) and the dividend \(x^{2}+x+1\). The numbers to be written are the coefficients of the powers of \(x\) in the dividend.

Step 2: Synthetic Division

Now, start the synthetic division process. Bring down the first number in the row (which is the coefficient of the highest power of \(x\)). Multiply this number by the root (which is -1, opposite of the number in the divisor) and write the result under the next number in the row. Add these numbers and continue the process.

Step 3: Write the Quotient and Remainder

The last number in the row is the remainder. The other numbers in the row represent the coefficients of the powers of \(x\) in the quotient polynomial \(q(x)\). In this case, the quotient \(q(x)\) will be \(x + 0\) (since the coefficients represent \(x\) and \(1\)) and the remainder \(r(x)\) will be 2.

Step 4: Write the Final Result

Combine the results of Step 3 to form the representation of the given polynomial as dictated in the exercise. Thus, we get \(p(x) = (x+1)(x) + 2\).

Key Concepts

Synthetic Division

Synthetic Division is a shortcut method for dividing polynomials, specifically designed for cases where the divisor is a linear polynomial of the form \(x - c\). This method makes the process less cumbersome and more efficient compared to long division.

• To start synthetic division, you write down the coefficients of the dividend polynomial.
• In the exercise, for \(x^2 + x + 1\), you have the coefficients \([1, 1, 1]\).
• Next, use the root derived from the divisor \(x + 1\), which is \(-1\), for the calculations.

The idea is to work across a row where you bring down the first coefficient and multiply it by the root, placing the result under the next coefficient. Continue this multiplication and addition process moving left to right. The simplicity of this method lies in its systematic approach to polynomials without variables cluttering the view.

Quotient and Remainder

When you perform division on polynomials using either synthetic or long division, the results can be expressed in the form of \(p(x) = d(x)q(x) + r(x)\). This is similar to the division process in arithmetic where you express numbers as the product of a quotient and divisor plus a remainder.

• In our example, the divisor \(d(x)\) is \(x+1\).
• The quotient \(q(x)\) after division is \(x\).
• The remainder \(r(x)\) is the constant term 2.

Just like fitting the pieces of a puzzle, each part of the equation must fall into the right place. The remainder should always have a degree less than that of the divisor, allowing everything to balance perfectly on the division scale of polynomials.

Polynomial Factorization

Polynomial factorization involves expressing a polynomial as a product of its factors. Understanding how to factor polynomials helps in breaking down complex expressions into simpler, more manageable parts for analysis or application.

• Factorization makes it easier to solve polynomial equations by setting either factor equal to zero.
• In the exercise, through synthetic division, you've factored \(x^2 + x + 1\) involving \(x + 1\) as a divisor.

This process is also essential in calculus and algebra for solving equations, simplifying expressions, and even finding limits and derivatives. The ultimate goal is to simplify and solve polynomials efficiently, and factorization forms the backbone of these techniques.