Problem 21 - Precalculus with Calculus Previews

Question

Proceed as in Example 3 and use the Remainder Theorem to find \(f(c)\) for the given value of \(c\). $$ f(x)=15 x^{3}+17 x^{2}-30 ; c=\frac{1}{5} $$

Step-by-Step Solution

Answer

The value of \(f\left(\frac{1}{5}\right)\) is \(-\frac{146}{5}\).

Step 1: Recall the Remainder Theorem

The Remainder Theorem states that if a polynomial \(f(x)\) is divided by \(x - c\), the remainder of this division is \(f(c)\). This means, to find \(f(c)\), we simply substitute \(c\) into the polynomial \(f(x)\).

Step 2: Substitute \(c\) into \(f(x)\)

Substitute \(c = \frac{1}{5}\) into the given polynomial \(f(x) = 15x^3 + 17x^2 - 30\). This yields: \(f\left(\frac{1}{5}\right) = 15\left(\frac{1}{5}\right)^3 + 17\left(\frac{1}{5}\right)^2 - 30\).

Step 3: Calculate \(\left(\frac{1}{5}\right)^3\)

Calculate \(\left(\frac{1}{5}\right)^3 = \frac{1}{125}\).

Step 4: Calculate \(\left(\frac{1}{5}\right)^2\)

Calculate \(\left(\frac{1}{5}\right)^2 = \frac{1}{25}\).

Step 5: Compute Each Term of \(f\left(\frac{1}{5}\right)\)

Now, compute each term with the substituted values: \(15 \times \frac{1}{125} = \frac{15}{125} = \frac{3}{25}\); \(17 \times \frac{1}{25} = \frac{17}{25}\); and the constant term is \(-30\).

Step 6: Combine the Terms

Sum all terms: \(f\left(\frac{1}{5}\right) = \frac{3}{25} + \frac{17}{25} - 30\). Simplify the expression: Combine \(\frac{3}{25}\) and \(\frac{17}{25}\) to get \(\frac{20}{25} = \frac{4}{5}\). Thus, \(f\left(\frac{1}{5}\right) = \frac{4}{5} - 30\).

Step 7: Simplify the Result

The final step is to simplify \(\frac{4}{5} - 30\). Convert \(30\) to a fraction with denominator 5: \(30 = \frac{150}{5}\). Therefore, \(f\left(\frac{1}{5}\right) = \frac{4}{5} - \frac{150}{5} = -\frac{146}{5}\).

Key Concepts

Polynomial Function

A polynomial function is a mathematical expression involving a sum of powers of one or more variables multiplied by coefficients. For example, the given polynomial \( f(x) = 15x^3 + 17x^2 - 30 \) is a third-degree polynomial in terms of the variable \( x \). Here, each term includes a coefficient (like 15 or 17) and a power of \( x \) (like \( x^3 \) or \( x^2 \)). Polynomials are fundamental in algebra because they are used to model various real-world situations.

• The highest power of the variable in a polynomial indicates its degree. In our example, \( x^3 \) is the term with the highest degree, making this a cubic polynomial.
• The standard form of polynomials arranges terms in descending powers.
• The coefficients play an essential role in shaping the polynomial's graph.

Understanding polynomial functions is crucial as they form the basis for further studies in algebra, calculus, and numerous applications in science and engineering.

Substitution Method

The substitution method is a straightforward technique where you "substitute" or replace a variable with a given value. This is especially useful in evaluating polynomials at specific points. In our problem, we use the Remainder Theorem, which tells us that by substituting \( c \) into \( f(x) \), we effectively find the remainder when \( f(x) \) is divided by \( x - c \).
To apply this method:

• First, identify the polynomial and the value to substitute. In our case, the polynomial is \( f(x) = 15x^3 + 17x^2 - 30 \), and \( c = \frac{1}{5} \).
• Substitute \( \frac{1}{5} \) for \( x \) into each term of the polynomial separately.
• After substitution, compute each term individually and then combine them for the final result.

This approach simplifies solving polynomial equations by focusing on the point of interest, making it a favorite among students and educators for its simplicity and effectiveness.

Polynomial Division

Polynomial division is similar to long division but is applied to polynomials. It's a way to divide one polynomial by another, like dividing numbers. However, when using the Remainder Theorem, we don't physically perform the division. Instead, polynomial division helps understand the relationship between the dividend, divisor, quotient, and remainder.
In the Remainder Theorem, instead of completing the division, we learn that the expression \( f(c) \) itself is the remainder of the polynomial \( f(x) \) divided by \( x - c \).

• This connection means that for the equation \( f(x) = (x - c)Q(x) + R \), \( f(c) = R \) when \( Q(x) \) is the quotient and \( R \) is the remainder.
• By substituting \( c \) in place of \( x \) in \( f(x) \), we bypass the need for detailed polynomial division and directly acquire the remainder.
• This is particularly efficient when checking roots or evaluating polynomials at specific points.

Thus, while full polynomial division has its uses in mathematics, the substitution via the Remainder Theorem offers a quick path to the remainder when the divisor is of the form \( x-c \), saving time and effort.