If g(y)=20y3+38y2–23, use synthetic division to find g(
–
2).

To solve \( g(-2) \) using synthetic division for the function \( g(y) = 20y^3 + 38y^2 - 23 \), let's break it down step-by-step.

### Step 1: Set up the coefficients
The polynomial is \( 20y^3 + 38y^2 + 0y - 23 \). The coefficients of the terms are:
- \( 20 \) for \( y^3 \),
- \( 38 \) for \( y^2 \),
- \( 0 \) for \( y \),
- \( -23 \) for the constant.

So, the coefficients are \( [20, 38, 0, -23] \).

### Step 2: Perform synthetic division
We will divide these coefficients by \( y = -2 \) using synthetic division.

\[
\begin{array}{r|rrrr}
-2 & 20 & 38 & 0 & -23 \\
   &    & -40 & 4 & -8 \\
\hline
   & 20 & -2 & 4 & -31 \\
\end{array}
\]

### Step 3: Interpret the result
After performing synthetic division, the last value in the bottom row is the remainder, which gives \( g(-2) \).

So, \( g(-2) = -31 \).

Would you like further clarification or additional details on this process?

Here are 5 related questions to think about:
1. How is synthetic division different from long division for polynomials?
2. What does the remainder in synthetic division represent?
3. Can synthetic division be used to factor polynomials?
4. How would you use synthetic division for polynomials with higher degrees?
5. What are some applications of synthetic division in calculus?

**Tip:** Synthetic division only works when dividing by a linear term of the form \( (y - c) \), where \( c \) is a constant.

If g(y) = 20y^3 + 38y^2 – 23, use synthetic division to find g(-2).

This problem demonstrates how to use synthetic division to solve g(y) = 20y^3 + 38y^2 - 23 and find g(-2). Follow a step-by-step breakdown of the polynomial division process and understand key concepts such as the remainder theorem. Ideal for students in Grades 9-12 studying algebra.

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