Problem 34 - A Graphical Approach to Precalculus with Limits

Question

Use synthetic substitution to find \(P(k).\) $$k=-2 ; \quad P(x)=x^{2}+5 x+6$$

Step-by-Step Solution

Answer

Using synthetic substitution, \(P(-2) = 0\).

Step 1: Setup the coefficients

Identify the coefficients of the polynomial \(P(x) = x^2 + 5x + 6\). The coefficients are \(1\) (from \(x^2\)), \(5\) (from \(x\)), and \(6\) (constant term).

Step 2: Initialize the synthetic substitution

Write the coefficients \(1, 5, 6\) in a row. Place the value of \(k = -2\) to the left of them, outside a synthetic division-like box. Prepare to perform substitution by synthetic division.

Step 3: Perform synthetic substitution

1. Bring down the leading coefficient \(1\) directly below.2. Multiply \(-2\) by \(1\); write the result, \(-2\), under the next coefficient, \(5\). Add to get \(3\).3. Multiply \(-2\) by \(3\); write the result, \(-6\), under the next coefficient, \(6\). Add to get \(0\).

Step 4: Write the result

The result of the synthetic substitution is \(P(-2)\). The last number that you have obtained is the value of \(P(-2)\), which is \(0\).

Key Concepts

Polynomial Evaluation

Evaluating a polynomial means finding the value of a polynomial function at a specific point. In this case, we want to find the value of the polynomial \( P(x) = x^2 + 5x + 6 \) when \( x = k = -2 \). In general, polynomial evaluation involves substituting a value into a polynomial equation and simplifying to find the result.
When performing polynomial evaluation manually, it typically involves substituting the given value of \( x \) into the polynomial equation and calculating the final result. However, for larger polynomials, this direct substitution can become tedious, especially when using high values of \( x \) or when the polynomial degree is large.
This exercise introduces synthetic substitution as a more efficient alternative, leveraging techniques from synthetic division, which simplifies the calculations without expanding everything fully. This method can make evaluating polynomials quicker and less error-prone for both small and large polynomials.

Synthetic Division

Synthetic division is a shortcut method for dividing a polynomial by a linear divisor of the form \( x - c \). When evaluating polynomials, we adapt this method in what we call synthetic substitution to find polynomial values at certain points.
In synthetic division, rather than writing out all terms of the polynomial, we focus only on coefficients. These coefficients are manipulated through addition and multiplication using the factor being divided, which is \( c \) in \( x - c \). The inside numbers perform an efficient division process without the long division's complexity.
It is particularly beneficial when evaluating polynomials because it efficiently computes the remainder, which directly corresponds to the value of the polynomial at \( x = c \). In synthetic substitution, \( c \) is replaced by the point \( k \) we are interested in evaluating. The operations are executed step by step, ultimately yielding the polynomial's value at the desired point.

Coefficients Identification

Identifying coefficients is a crucial preliminary step in polynomial evaluation using synthetic substitution. The coefficients are the numerical factors paired with each term's power of \( x \). For the polynomial \( P(x) = x^2 + 5x + 6 \), we identify them as follows:

• The coefficient of \( x^2 \) is 1.
• The coefficient of \( x \) is 5.
• The constant term is 6, which is technically the coefficient of \( x^0 \).

The process of identifying these coefficients is necessary because they become the working numbers in synthetic substitution. These coefficients are moved into a sequence, which simplifies operations like addition and multiplication when evaluating the polynomial.
This brings clarity by using a single focus on each term's numerical value, rather than dealing with \( x \) terms repeatedly, thus reducing potential errors associated with algebraic manipulation.