Use synthetic division and the Remainder Theorem to find the indicated function value. f(x)=2x⁴−5x³−x²+3x+2; f(−1/2)

Verified step by step guidance

1. Identify the polynomial function and the value at which you need to evaluate it: \(f(x) = 2x^{4} - 5x^{3} - x^{2} + 3x + 2\) and you want to find \(f\left(-\frac{1}{2}\right)\).
2. Set up synthetic division using the divisor \(x - r\), where \(r = -\frac{1}{2}\). Write the coefficients of the polynomial in descending order of powers: \$2, -5, -1, 3, 2$.
3. Perform synthetic division by bringing down the first coefficient, then multiply it by \(r = -\frac{1}{2}\), add this result to the next coefficient, and repeat this process for all coefficients.
4. The final number you get after completing synthetic division is the remainder, which by the Remainder Theorem equals \(f\left(-\frac{1}{2}\right)\).
5. Interpret this remainder as the value of the function at \(x = -\frac{1}{2}\), completing the evaluation without directly substituting into the polynomial.

Key Concepts

Synthetic Division

Synthetic division is a shortcut method for dividing a polynomial by a linear binomial of the form (x - c). It simplifies the long division process by using only the coefficients of the polynomial, making calculations faster and less error-prone. This method is especially useful for evaluating polynomials and finding remainders.

Remainder Theorem

The Remainder Theorem states that when a polynomial f(x) is divided by (x - c), the remainder is equal to f(c). This means you can find the value of the polynomial at x = c by performing synthetic division and looking at the remainder, providing a quick way to evaluate polynomials without direct substitution.

Polynomial Evaluation

Polynomial evaluation involves finding the value of a polynomial function at a specific input. Instead of substituting the value directly into the polynomial expression, synthetic division combined with the Remainder Theorem offers an efficient alternative, especially for higher-degree polynomials, to determine f(c) quickly.