Problem 39 - Algebra 2

Question

Find \(p(4)\) and \(p(-2)\) for each function. \(p(x)=\frac{1}{8} x^{3}-\frac{1}{4} x^{2}-\frac{1}{2} x+5\)

Step-by-Step Solution

Answer

\(p(4) = 7\) and \(p(-2) = 4\).

Step 1: Substitute 4 into the polynomial

To find \(p(4)\), substitute \(x = 4\) into the function \(p(x) = \frac{1}{8} x^3 - \frac{1}{4} x^2 - \frac{1}{2} x + 5\). Calculate: \(p(4) = \frac{1}{8} (4)^3 - \frac{1}{4} (4)^2 - \frac{1}{2} (4) + 5\).

Step 2: Evaluate powers and fractions for p(4)

First calculate the powers: \((4)^3 = 64\) and \((4)^2 = 16\). Then compute the fractions: \(\frac{1}{8} \times 64 = 8\), \(\frac{1}{4} \times 16 = 4\), and \(\frac{1}{2} \times 4 = 2\).

Step 3: Calculate expression for p(4)

Substitute the simplified fractions into the expression: \(p(4) = 8 - 4 - 2 + 5\). Add and subtract these values to get: \(p(4) = 7\).

Step 4: Substitute -2 into the polynomial

To find \(p(-2)\), substitute \(x = -2\) into the function \(p(x) = \frac{1}{8} x^3 - \frac{1}{4} x^2 - \frac{1}{2} x + 5\). Calculate: \(p(-2) = \frac{1}{8} (-2)^3 - \frac{1}{4} (-2)^2 - \frac{1}{2} (-2) + 5\).

Step 5: Evaluate powers and fractions for p(-2)

First calculate the powers: \((-2)^3 = -8\) and \((-2)^2 = 4\). Then compute the fractions: \(\frac{1}{8} \times (-8) = -1\), \(\frac{1}{4} \times 4 = 1\), and \(\frac{1}{2} \times (-2) = -1\).

Step 6: Calculate expression for p(-2)

Substitute the simplified fractions into the expression: \(p(-2) = -1 - 1 + 1 + 5\). Add and subtract these values to get: \(p(-2) = 4\).

Key Concepts

Polynomial Evaluation

Evaluating a polynomial simply means finding the result of the polynomial for a given value of its variable. Consider the polynomial function given by:

• \( p(x) = \frac{1}{8}x^3 - \frac{1}{4}x^2 - \frac{1}{2}x + 5 \)

To evaluate this polynomial for a specific value, substitute the value into the function in place of \( x \). This substitution will involve performing calculations with exponents and fractions.
For instance, to find \( p(4) \), replace every "\( x \)" in the function with 4 and solve the resulting expression. Similarly, substituting \(-2\) for \(x\) will help you find \( p(-2) \). Evaluating polynomials helps understand how the function behaves at specific points, and it is a fundamental skill in algebra and calculus.

Substitution Method

The substitution method is a straightforward way to evaluate polynomial functions. This involves replacing the variable with a specific number and solving the equation step by step. It may sound simple at first, but attention to detail is crucial.
When you substitute a number into a polynomial, you replace each instance of the variable with that number. Let's consider the function \( p(x) = \frac{1}{8}x^3 - \frac{1}{4}x^2 - \frac{1}{2}x + 5 \). To find \( p(4) \), substitute \( x = 4 \) so it becomes:

• \( \frac{1}{8}(4^3) - \frac{1}{4}(4^2) - \frac{1}{2}(4) + 5 \)

Breaking this down, calculate each power first, then apply the coefficients, and finally sum or subtract the terms. This method is helpful not only for polynomials but in solving any algebraic expressions where you need to find specific values related to its variables.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations. Understanding how to manipulate these expressions is key when working with polynomial functions. A polynomial like \( p(x) = \frac{1}{8}x^3 - \frac{1}{4}x^2 - \frac{1}{2}x + 5 \) is a fantastic example of an algebraic expression that uses exponents, coefficients, and constants.
Such expressions can be simplified or evaluated using basic arithmetic operations. When you work with algebraic expressions, you'll often employ methods such as substitution or factorization to solve problems or simplify expressions. Recognizing terms like coefficients (\( \frac{1}{8}, -\frac{1}{4}, -\frac{1}{2} \)) and variables is essential. The rules of algebra help ensure that you're following correct order of operations – an important aspect in achieving accurate results.

Cubic Polynomials

Cubic polynomials are polynomial expressions where the highest degree of the variable is three. These are expressed in the general form \( ax^3 + bx^2 + cx + d \), where \( a, b, c, \) and \( d \) are constants. In the exercise, we have a cubic polynomial represented by:

• \( p(x) = \frac{1}{8}x^3 - \frac{1}{4}x^2 - \frac{1}{2}x + 5 \)

This function has distinct components:

• The cubic term \( ax^3 \), in this case \( \frac{1}{8}x^3 \)
• The quadratic term \( bx^2 \), which is \( -\frac{1}{4}x^2 \)
• The linear term \( cx \), which is \( -\frac{1}{2}x \)
• The constant term \( d \), which is 5

Cubic polynomials are interesting in algebra as they can intersect the x-axis up to three times and have a distinct curve because of the cubic term. Analyzing these functions is useful in understanding more complex mathematical concepts and their applications in real-world problems.