Problem 16 - 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

\(x^2 + x + 1 = (x - 1)(x + 2) + 3\)

Step 1: Carrying out Synthetic Division Step 1

Firstly, carry the first coefficient down and multiply it with the zero of the factor. So, carry down the 1 and multiply by 1 to give 1. This result should be placed underneath the second coefficient and lines should be drawn around this computation for clarity.

Step 2: Carrying out Synthetic Division Step 2

Secondly, add the second coefficient and the result from step 1. So, 1 (second coefficient) + 1 (result from step 1) = 2. Following this, multiply this result by the zero of the factor and place this result underneath the third coefficient. So, 2 (result from the addition) * 1 (zero of the factor) = 2.

Step 3: Carrying out Synthetic Division Step 3

Lastly, add the third coefficient and the result from step 2. So, 1 (third coefficient) + 2 (result from step 2) = 3.

Step 4: Writing Final Result

The quotient and remainder are read off from the final results. The quotient is the polynomial whose coefficients are the numbers on the bottom row before the final number, read from left to right as descending powers of \(x\). Any final number is the remainder, so the quotient is \(x + 2\) and the remainder is 3. Therefore, \(x^2 + x + 1 = (x - 1)(x + 2) + 3\).

Key Concepts

Synthetic Division

Synthetic division is a simplified method of dividing polynomials, particularly useful when dividing by a linear factor in the form of \(x - c\). This process is more efficient than polynomial long division and involves fewer written steps. Here's how it works:

• Identify the divisor, \(x - c\), and note the root \(c\).
• Write down the coefficients of the dividend polynomial in descending order of the power of \(x\).
• Bring down the leading coefficient to the bottom row initially.
• Multiply this number by \(c\), and write the result under the next coefficient.
• Add these two numbers and continue the process until you've worked through all the coefficients.
• The final row provides the coefficients for the quotient polynomial, and the last number is the remainder.

For the exercise \(x^2 + x + 1\) with \(x - 1\), the calculations show that the quotient is \(x + 2\) and the remainder is 3. Thus, the expression becomes \((x - 1)(x + 2) + 3\).

Remainder Theorem

The remainder theorem is a handy tool that connects polynomial division with evaluating polynomials. It states that if a polynomial \(p(x)\) is divided by a linear factor \(x - c\), the remainder of this division is \(p(c)\). In other words, by substituting the value of \(c\) into the polynomial, you directly find what the remainder is without needing a full division.
Using the exercise's example:\ For the polynomial \(x^2 + x + 1\) divided by \(x - 1\), you simply evaluate \(p(1)\):

• Replace \(x\) with 1: \(1^2 + 1 + 1\)
• Calculate to find \(3\)

And indeed, the remainder is 3, matching what we found using synthetic division. This theorem makes it especially easy to test whether a divisor is a factor of a given polynomial: if the remainder is zero, \(x - c\) is a factor.

Polynomial Long Division

Polynomial long division is a process similar to numerical long division but involves polynomials. It is a versatile method that works for all polynomial divisions, unlike synthetic division which is limited to linear divisors. Here’s how it is done:

• Set up the division by arranging the dividend and divisor polynomials in descending order of powers.
• Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
• Multiply the entire divisor by this new term and subtract the result from the dividend.
• Bring down the next term of the dividend and repeat the process until all terms have been dealt with.

This method provides both the quotient and the remainder simultaneously. If we used polynomial long division for \(x^2 + x + 1\) divided by \(x - 1\), we would obtain \(x + 2\) as the quotient, and 3 as the remainder, confirming the result obtained from synthetic division. While it may take more time compared to synthetic division for simple cases, it is indispensable for divisions involving higher degree or non-linear divisors.