Welcome back to Professor Baker's Math Class! Today, we explored different methods for evaluating polynomials. Evaluating a polynomial means finding the value of the polynomial expression for a given value of the variable. We focused on two key techniques: Direct Substitution and Synthetic Substitution.
Direct Substitution
Direct substitution involves plugging the given value directly into the polynomial and simplifying using the order of operations (PEMDAS/BODMAS).
Let's consider an example: Evaluate $f(x) = 2x^3 + 5x^2 + 4x + 8$ when $x = -2$.
Using direct substitution:
$f(-2) = 2(-2)^3 + 5(-2)^2 + 4(-2) + 8$
$f(-2) = 2(-8) + 5(4) + 4(-2) + 8$
$f(-2) = -16 + 20 - 8 + 8$
$f(-2) = 4$
Therefore, the value of the polynomial at $x = -2$ is 4.
Classifying Polynomials
Before we dive further into evaluation, it's helpful to remember how to classify polynomials. Polynomials are classified by their degree (the highest power of the variable) and the number of terms.
- Linear: Degree 1 (e.g., $f(x) = 12 - 5x$)
- Quadratic: Degree 2 (e.g., $f(x) = \pi x^2$)
- Cubic: Degree 3 (e.g., $f(x) = x^3 - 3x^2 - 2x^3$)
- Monomial: One term
- Binomial: Two terms
- Trinomial: Three terms
So, for example, $f(x) = 12 - 5x$ is a Linear Binomial, and $f(x) = x^3 - 3x^2 - 2x^3$ is a Cubic Trinomial.
Homework
Time to practice! For homework, please complete problems #28-46 even on pages 333-334. You can choose either Direct Substitution to evaluate the polynomials. Remember to show your work clearly and double-check your calculations!
Keep up the great work, and see you in the next class!