Problem 36 - Algebra 2

Question

Open-Ended Write a polynomial division that has a quotient of \(x+3\) and a remainder of 2 .

Step-by-Step Solution

Answer

Given a quotient of \(x+3\) and a remainder of 2, a possible divisor is \(x\) and dividend is \(x^2+3x+2\). Therefore, the division is \(\frac{x^2 + 3x + 2}{x}\).

Step 1: Select a Divisor

To begin, select a polynomial that will be used as the divisor. Since the quotient will be \(x+3\), it is necessary for the divisor to be of one degree lower than the dividend. Therefore, choose a linear polynomial, e.g., \(x\) or \(x-1\). For this example, let's select \(x\) as the divisor.

Step 2: Create the Dividend

Construct the dividend by multiplying the divisor \(x\) by the given quotient \(x+3\). This will give the initial part of the dividend. \[ (x)(x+3)=x^2+3x \] Then add the remainder to the product. Since the remainder is 2, the complete dividend becomes: \[ x^2 + 3x + 2 \]

Step 3: Write the Division Equation

With the dividend \(x^2 + 3x + 2\) and the divisor \(x\), you can set up the polynomial division: \[ \frac{x^2 + 3x + 2}{x} \] This equation represents a polynomial division that has a quotient of \(x+3\) and a remainder of 2.

Key Concepts

Dividend and Divisor

Understanding the roles of dividend and divisor is crucial for comprehending the process of polynomial division. In this context, the dividend is the polynomial that is being divided, while the divisor is the polynomial by which the dividend is divided. Think of it like a pizza (the dividend) being cut into slices (the divisor).
In the provided exercise, the dividend is constructed based on the desired quotient, which is the result of the division after the divisor is divided into the dividend, and a specific remainder. By reversing the division process, one multiplies the quotient, which is \(x+3\), by the divisor, chosen as \(x\), to get the first part of the dividend, resulting in \(x^2+3x\). To complete the dividend, the remainder, which is 2, is added, yielding the polynomial \(x^2 + 3x + 2\). This polynomial, when divided by our divisor \(x\), will indeed provide the desired quotient and remainder.

• Dividend: The polynomial being divided (\(x^2 + 3x + 2\))
• Divisor: The polynomial by which you are dividing (\(x\))

Remainder Theorem

The Remainder Theorem is an essential concept in polynomial algebra, which states that when a polynomial \(f(x)\) is divided by a linear divisor of the form \(x-c\), the remainder of the division is the value of \(f(c)\). This theorem can be particularly useful for quickly evaluating polynomials at specific points and for confirming the remainder from polynomial division without needing to perform long division or synthetic division.
In our exercise, we see the Remainder Theorem in reverse: we’re given the remainder (2) and are asked to construct a polynomial that, when divided by \(x\), leaves this particular remainder. Hence, the final piece of dividend, \(x^2 + 3x + 2\), when evaluated at \(x=0\) using the Remainder Theorem, will yield the remainder 2. This clever back-engineering use of the Remainder Theorem helps in creating the problem scenario outlined in the initial exercise.

Synthetic Division

Synthetic division is a simplified form of polynomial division that is faster and requires less writing than traditional long division. It's particularly effective when dividing a polynomial by a linear divisor of the form \(x-c\). In synthetic division, you only deal with the coefficients of the dividend and the value of \(c\) from the divisor, discarding variables and exponent notations.
To perform synthetic division, one would write down the coefficients of the dividend, bring down the leading coefficient, multiply it by \(c\) from the divisor, place this value under the next coefficient, sum them up, and continue the process until all coefficients have been addressed. The final line of numbers represents the coefficients of the quotient, and any remainder if one exists.

Why Synthetic Division Wasn't Used Here

For the given exercise, synthetic division wasn't used because the exercise itself was about constructing a division situation based on a given quotient and a remainder. However, if one wished to confirm that \(x^2 + 3x + 2\) divided by \(x\) indeed yields a quotient of \(x+3\) and a remainder of 2, synthetic division would offer a quick and efficient verification method.