Problem 56 - Precalculus Mathematics for Calculus

Question

Use the Factor Theorem to show that \(x-c\) is a factor of \(P(x)\) for the given value(s) of \(c\). $$P(x)=x^{4}+3 x^{3}-16 x^{2}-27 x+63, \quad c=3,-3$$

Step-by-Step Solution

Answer

Both \( x-3 \) and \( x+3 \) are factors of \( P(x) \).

Step 1: Understand the Factor Theorem

The Factor Theorem states that \( x-c \) is a factor of a polynomial \( P(x) \) if and only if \( P(c) = 0 \). This means we'll need to substitute the value of \( c \) into the polynomial and check the result.

Step 2: Substitute \( c = 3 \) into \( P(x) \)

Compute \( P(3) = (3)^4 + 3(3)^3 - 16(3)^2 - 27(3) + 63 \). Calculate each term separately: - \( 3^4 = 81 \) - \( 3(3)^3 = 81 \) - \( 16(3)^2 = 144 \) - \( 27(3) = 81 \) - Constant term = 63. Now sum them up: \( 81 + 81 - 144 - 81 + 63 = 0 \). Since \( P(3) = 0 \), \( x-3 \) is a factor.

Step 3: Substitute \( c = -3 \) into \( P(x) \)

Compute \( P(-3) = (-3)^4 + 3(-3)^3 - 16(-3)^2 - 27(-3) + 63 \). Calculate each term separately: - \((-3)^4 = 81 \) - \(3(-3)^3 = -81 \) - \(16(-3)^2 = 144 \) - \(-27(-3) = 81 \) - Constant term = 63. Now sum them up: \( 81 - 81 - 144 + 81 + 63 = 0 \). Since \( P(-3) = 0 \), \( x+3 \) is a factor.

Key Concepts

Polynomial Division

Polynomial division is a process similar to long division with numbers, but it deals with polynomials. When you perform polynomial division, you're essentially trying to find out how many times a smaller polynomial called the divisor can fit into the larger polynomial, which is known as the dividend.

Consider dividing one polynomial by another: you start by comparing the leading terms (the term with the highest power) of the dividend and the divisor.

• Multiply the entire divisor by the term obtained and subtract this product from the dividend.
• This will reduce the degree of the polynomial you are dividing.
• You repeat these steps until the degree of the remaining polynomial (remainder) is less than that of the divisor.

The purpose of polynomial division is often to verify factors or to simplify polynomials. In the exercise provided, polynomial division isn't explicitly used, but understanding it underscores why factors and roots correlate strongly: roots are values that cause the polynomial to be zero, indicating complete division by a factor with no remainder.

Roots of Polynomials

Roots of polynomials are the values of the variable that make the polynomial equal to zero. These are vital as they indicate where the graph of the polynomial will cross the x-axis.

Using the Factor Theorem from the exercise, if substituting a value for the variable into the polynomial results in zero, then that value is a root of the polynomial. An example from the solution involves substituting the values 3 and -3 into the polynomial.

• For P(3), the result was zero; hence, 3 is a root, and \(x - 3\) is a factor.
• Similarly, P(-3) equals zero; therefore, -3 is a root, and \(x + 3\) is a factor.

This approach gives an efficient means of identifying roots without graphing or continued guessing. Understanding roots is essential for solving polynomial equations, facilitating factored forms, and predicting polynomial behavior.

Synthetic Substitution

Synthetic substitution is a streamlined method of evaluating polynomials quickly by utilizing only number operations. It's a variant of synthetic division and is particularly handy when applying the Factor Theorem.

Compared to traditional substitution, where you plug a number into each term of the polynomial, synthetic substitution offers a compact step-by-step approach:

• Write down the coefficients of the polynomial in a row.
• Bring down the leading coefficient to the row below.
• Multiply it by the value you're substituting for \((c)\).
• Add this result to the next coefficient, writing the sum below.
• Repeat until all coefficients have been used.
• If the final result is zero, the value substituted is a root.

In the provided exercise with \(c = 3\) and \(c = -3\), synthetic substitution wouldn’t alter the answer but could have provided a quicker path to the conclusion that x-c or x+c is a valid factor. Students often prefer it for its efficiency, especially when confined to root checks.