Problem 10 - Algebra and Trigonometry

Question

\(9-14\) . Two polynomials \(P\) and \(D\) are given. Use either synthetic or long division to divide \(P(x)\) by \(D(x),\) and express the quotient \(P(x) / D(x)\) in the form $$ \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)} $$ $$ P(x)=x^{3}+6 x+5, \quad D(x)=x-4 $$

Step-by-Step Solution

Answer

The quotient is \(x^2 + 4x + 22\) and the remainder is \(93\).

Step 1: Set up for Synthetic Division

Since the divisor is of the form \( x-c \), where \( c = 4 \), we can use synthetic division. Write down the coefficients of \( P(x) = x^3 + 0x^2 + 6x + 5 \) as \( [1, 0, 6, 5] \). Place the value \( 4 \) in a box on the left.

Step 2: Begin Synthetic Division

Bring down the leading coefficient, which is \( 1 \), below the line. Multiply \( 4 \) by \( 1 \), and write the result \( 4 \) under the next coefficient.

Step 3: Add and Repeat

Add \( 0 + 4 \) to get \( 4 \). Multiply \( 4 \) by \( 4 \) to get \( 16 \) and place it under the next coefficient (6). Add \( 6 + 16 \) to get \( 22 \). Multiply \( 4 \) by \( 22 \) to get \( 88 \) and place it under the last coefficient (5).

Step 4: Finish Synthetic Division

Add \( 5 + 88 \) to obtain \( 93 \). The numbers below the line \( [1, 4, 22, 93] \) represent the coefficients of the quotient polynomial and the remainder.

Step 5: Write Quotient and Remainder

The quotient \( Q(x) \) is \( x^2 + 4x + 22 \) and the remainder \( R(x) \) is \( 93 \). Thus, \( \frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)} = x^2 + 4x + 22 + \frac{93}{x-4} \).

Key Concepts

Synthetic Division

Synthetic Division is a simplified method for dividing polynomials, notably faster and more efficient than long division. It is especially useful when the divisor is a linear polynomial of the form \(x-c\). This method reduces complex algebraic operations into a few clear steps, making it favorable for students.

To utilize synthetic division, follow these steps:

• Identify the divisor. Here, it's \(x-4\), so \(c=4\).
• List the coefficients of the polynomial \(P(x) = x^3 + 6x + 5\). Don't forget to include missing degrees with a zero place holder. Thus, our coefficients are \([1, 0, 6, 5]\).
• Place \(c\) in a 'box' style setup on the left and draw a horizontal line to separate computations.

Synthetic Division is conducted as follows:

• Bring the first coefficient down below the line.
• Multiply this number by \(c\) and write below the next coefficient.
• Add the numbers in that column, bring the result down below the line, and repeat.
• Continue this process for all coefficients.

At the end, the numbers below the line are the coefficients of the quotient and the last number is the remainder. This quick process offers a straightforward alternative to long division in polynomial division problems.

Long Division

Long Division of polynomials mirrors the long division of numbers you likely learned in elementary school. However, it involves variables and their degrees which can make it slightly more complex. This method is beneficial when the divisor is not simple or when checking results obtained from synthetic division.

Here’s a step-by-step breakdown of polynomial long division:

• Write the dividend (the polynomial being divided) and the divisor (the polynomial by which you are dividing) in standard form, with terms ordered by decrements of degree.
• Divide the leading term of the dividend by the leading term of the divisor to obtain the first term of the quotient.
• Multiply the entire divisor by this term and subtract the result from the dividend.
• Bring down the next term from the original dividend and repeat the process until all terms are used or a remainder remains.

The result is a quotient polynomial plus a remainder. This method, while perhaps lengthier than synthetic division, is versatile and helps confirm the accuracy of other division approaches.

Remainder Theorem

The Remainder Theorem is a useful tool that connects polynomial division to evaluating polynomials at specific points. According to this theorem, when a polynomial \(P(x)\) is divided by a linear divisor \((x-c)\), the remainder of the division corresponds to \(P(c)\). This is particularly practical because it allows for a quick verification of remainders without completing the full division process.

Here's how you can use the Remainder Theorem:

• Identify the divisor's form \(x-c\) and extract \(c\).
• Substitute \(c\) into the polynomial \(P(x)\).
• Evaluate \(P(c)\) to find the remainder directly.

For instance, in the exercise \(P(x) = x^3 + 6x + 5\) and \(D(x) = x-4\), the remainder from synthetic division is \(93\). By substituting \(4\) into \(P(x)\), \(P(4) = 4^3 + 6(4) + 5 = 93\), confirming that the remainder is indeed \(93\). This theorem simplifies the remainder-checking process markedly, offering a fast way to affirm results obtained from either synthetic or long division.