Skip to main content
Ch. 3 - Polynomial and Rational Functions
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 3 - Polynomial and Rational FunctionsProblem 40
Chapter 4, Problem 40

Use synthetic division and the Remainder Theorem to find the indicated function value. f(x)=6x4+10x3+5x2+x+1; f(−2/3)

Verified step by step guidance
1
Identify the polynomial function and the value at which you need to evaluate it: \(f(x) = 6x^{4} + 10x^{3} + 5x^{2} + x + 1\) and you want to find \(f\left(-\frac{2}{3}\right)\).
Set up synthetic division using the value \(-\frac{2}{3}\) as the divisor. Write down the coefficients of the polynomial in descending order of powers: 6, 10, 5, 1, and 1.
Perform synthetic division by bringing down the first coefficient (6), then multiply it by \(-\frac{2}{3}\), add the result to the next coefficient, and continue this process across all coefficients.
The final number you obtain after completing synthetic division is the remainder, which by the Remainder Theorem equals \(f\left(-\frac{2}{3}\right)\).
Interpret the remainder as the value of the function at \(x = -\frac{2}{3}\), completing the evaluation without directly substituting into the polynomial.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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 at specific values.
Recommended video:
05:10
Higher Powers of i

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, which provides a quick way to evaluate polynomials without direct substitution.
Recommended video:
05:10
Higher Powers of i

Polynomial Evaluation

Polynomial evaluation involves finding the value of a polynomial function at a given 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 compute f(c) quickly.
Recommended video:
Guided course
05:13
Introduction to Polynomials
Related Practice
Textbook Question

Use the Intermediate Value Theorem to show that each polynomial has a real zero between the given integers. f(x)=3x3−8x2+x+2; between 2 and 3

Textbook Question

For Exercises 40–46, (a) List all possible rational roots or rational zeros. (b) Use Descartes's Rule of Signs to determine the possible number of positive and negative real roots or real zeros. (c) Use synthetic division to test the possible rational roots or zeros and find an actual root or zero. (d) Use the quotient from part (c) to find all the remaining roots or zeros. f(x)=x3+3x2−4f(x) = x^3 + 3x^2 - 4

Textbook Question

Solve each polynomial inequality in Exercises 1–42 and graph the solution set on a real number line. Express each solution set in interval notation. x3≥9x2x^3≥9x^2

3
views
Textbook Question

Use Descartes' Rule of Signs to explain why 2x4+6x2+8=02x^4 + 6x^2 + 8 = 0 has no real roots.

Textbook Question

Find the horizontal asymptote, if there is one, of the graph of each rational function. h(x)=12x3/(3x2+1)

3
views
Textbook Question

An equation of a quadratic function is given. a) Determine, without graphing, whether the function has a minimum value or a maximum value. b) Find the minimum or maximum value and determine where it occurs. c) Identify the function's domain and its range. f(x)=3x2−12x−1