Problem 1 - College Algebra

Question

Divide using long division. State the quotient, \(q(x),\) and the remainder, \(r(x)\). $$ \left(x^{2}+8 x+15\right) \div(x+5) $$

Step-by-Step Solution

Answer

The quotient, \(q(x),\) is \(x+3\), and the remainder, \(r(x),\) is 0.

Step 1: Divide the leading terms

Begin the long division process by dividing the leading term of the dividend \(x^{2}\) by the leading term of the divisor \(x\), which gives your first term of the quotient \(q(x)\), \(x\).

Step 2: Multiply and subtract

Multiply the divisor \(x+5\) by the first term of the quotient \(x\), resulting in \(x^{2}+5x\). Subtract this from the dividend \(x^{2}+8x+15\) to get a new dividend of \(3x+15\).

Step 3: Repeat the process

Repeat Step 1 with the new dividend. Divide the leading term of the new dividend \(3x\) by the leading term of the divisor \(x\) to get \(3\) as the next term of the quotient. The quotient \(q(x)\) is now \(x+3\).

Step 4: Complete the division

Multiply the divisor \(x+5\) by \(3\) to get \(3x+15\), and subtract this from the new dividend \(3x+15\) to get 0. Therefore, the division is complete with no remainder, so \(r(x) = 0\).

Key Concepts

Dividing Polynomials

Polynomial division, similar to numerical long division, is a method used to divide one polynomial by another. It is essential to understand that polynomials are expressions that consist of variables raised to whole number exponents and their coefficients. When you divide polynomials, the goal is to determine how many times the divisor can 'fit' into the dividend, resulting in the quotient.

The division process starts by comparing the leading terms of both the dividend and the divisor. Just like in our example, where we divided the leading term of the dividend, which is an expression like \(x^{2}+8x+15\), by the leading term of the divisor \(x+5\). It's an iterative process where subtraction and multiplication are also used to gradually reduce the dividend until no further division is possible.

In the provided exercise, dividing \(x^{2}\) by \(x\) gave us the first term of the quotient. This step is crucial as it sets the foundation for the rest of the polynomial long division process.

Algebraic Division

Algebraic division is a more general term that encompasses dividing polynomials but can also refer to division involving simple algebraic terms or more complex algebraic expressions. The fundamentals are the same: divide, multiply, subtract, bring down the next term, and repeat. Accuracy at each step is vital to ensure the validity of the quotient and remainder.

In our exercise example, to find \(x\) as the first term of the quotient, we executed an algebraic division where we divided \(x^{2}\) by \(x\). This concept applies to various forms of algebraic expressions, and mastering the process of dividing polynomials is essential for tackling more complex algebraic division scenarios.

Quotient and Remainder

In polynomial long division, as well as in regular arithmetic division, determining the quotient and remainder is the aim of the division process. The quotient \(q(x)\) is the result of the division before considering the remainder, and represents the number of times the divisor fits into the dividend. The remainder \(r(x)\) is what is left over after the division is complete.

If the remainder is zero, as noted in our exercise where we ended up with a remainder of \(r(x) = 0\), it means the divisor is a factor of the dividend. The concept of quotient and remainder is inherent to many areas of mathematics, including Euclidean division in number theory, and is pivotal in understanding the division of polynomials.

Synthetic Division

Synthetic division is a shortcut method for dividing polynomials when the divisor is a linear binomial of the form \(x - c\). It's a simpler, more efficient algorithm compared to the traditional long division method, requiring less written work and involving only the coefficients. However, the traditional method used in our exercise is more versatile, whereas synthetic division's primary limitation lies in its inability to handle divisors that are not first-degree polynomials.

Although synthetic division was not used in the provided example, understanding this method can be advantageous for students when they encounter divisors suitable for its application. It's particularly useful for quickly determining whether a given term is a factor of the polynomial and for factoring and solving polynomial equations.