Question

\(p(x)=2 x^3+4 x^2+6 x+7 ; x=\frac{1}{2}\)

Short answer

The value of the given polynomial function \(p(x)\) when \(x = \frac{1}{2}\) is \(11 \frac{1}{4}\).

Step-by-step solution

Substitute \(x\) in the polynomial function

Substitute \(x = \frac{1}{2}\) in the polynomial function \(p(x)\).\nSo, \(p(\frac{1}{2}) = 2*(\frac{1}{2})^3 + 4*(\frac{1}{2})^2 + 6*\frac{1}{2} + 7\)

Simplify the powers

Simplify the terms with powers:\n\(p(\frac{1}{2}) = 2*(\frac{1}{8}) + 4*(\frac{1}{4}) + 3 + 7\)

Evaluate the multiplication

Perform the multiplication in each term. Doing so results in:\n \(p(\frac{1}{2}) = \frac{1}{4} + 1 + 3 + 7\)

Final evaluation

Finally, perform the addition:\n\(p(\frac{1}{2}) = \frac{1}{4} + 1 + 3 + 7 = 11 \frac{1}{4}\)

Key concepts

Substituting Values in Polynomials

Understanding how to substitute values into a polynomial is a critical skill in algebra that allows students to evaluate functions for different inputs. For example, given a polynomial function like \(p(x)=2x^3+4x^2+6x+7\), evaluating it for \(x=\frac{1}{2}\) requires replacing each instance of \(x\) with \(\frac{1}{2}\).

The key to successful substitution lies in carefully replacing the variable and ensuring that each term of the polynomial is treated separately. For example, in the given function \(p(x)\), when we substitute \(x\) with \(\frac{1}{2}\), it is critical to apply the substitution consistently across each term to avoid any errors, like so:
\(p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) + 7\).

This process turns the abstract polynomial into a concrete numerical expression which we can then simplify and calculate, moving us one step closer to finding the polynomial's value for the specified input.

Simplifying Polynomial Expressions

Once the substitution is done, the next step in evaluating a polynomial function is simplification. Simplifying polynomial expressions requires knowledge of basic arithmetic and algebraic properties, including the operations with powers and the order of operations (PEMDAS—Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction).

Looking at our current function after substitution, \(p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) + 7\), the first thing is to simplify the powers of \(\frac{1}{2}\), which are \(\left(\frac{1}{2}\right)^3\) and \(\left(\frac{1}{2}\right)^2\).

Calculating these terms, we end up with \(\frac{1}{8}\) and \(\frac{1}{4}\) respectively. Multiplying these by their coefficients simplifies our expression to a series of simple fractions and whole numbers that can be more easily managed: \(p\left(\frac{1}{2}\right) = \frac{1}{4} + 1 + 3 + 7\), setting us up for the final arithmetic stage of adding these values together to get the evaluated polynomial's value.

Polynomial Arithmetic

The last step in evaluating polynomials is performing arithmetic operations on the simplified expressions. This step is where you add, subtract, multiply, or divide the numbers you have obtained after substituting and simplifying values in the polynomial.

In our example, after simplifying, we have the expression \(p\left(\frac{1}{2}\right) = \frac{1}{4} + 1 + 3 + 7\). To find the final value, we perform addition: combining the fractional part with the whole numbers to obtain \(p\left(\frac{1}{2}\right) = 11\frac{1}{4}\).

Dealing with Fractions

When adding a fraction to whole numbers, it's vital not to overlook this part. Here, \(\frac{1}{4}\) is kept as is because there are no other fractions to combine it with. Finally, by adding up all the whole numbers and incorporating the fraction, you arrive at the function's total evaluated value for the substituted input. It's important to take these steps methodically to avoid small mistakes that could lead to incorrect outcomes.