/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Explain how to perform long divi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 57

Explain how to perform long division of polynomials. Use \(2 x^{3}-3 x^{2}-11 x+7\) divided by \(x-3\) in your explanation.

Short Answer

Expert verified
The result of the division \(2x^{3} - 3x^{2} -11x + 7\) by \(x - 3\) is \(2x^{2} + x - 8 + \frac{1}{x - 3}\).

Step by step solution

01

Setup the Division

Arrange the division in long division format: Set \(x - 3\) outside and \(2x^{3} - 3x^{2} - 11x + 7\) inside the division symbol.
02

Division of the First Term

Start by dividing the first term of the dividend, \(2x^{3}\), by the leading term of the divisor, \(x\). This gives us \(2x^{2}\). Write down the result above the division line.
03

Multiply and Subtract

Multiply the result from Step 2 by the divisor \(x - 3\). This gives \(2x^{3} - 6x^{2}\). Subtract this from the dividend and write down the result, \((-3x^{2} - 2x^{3}) - 11x + 7 = x^{2} - 11x + 7\).
04

Repeat Division and Subtraction

Repeat Steps 2 and 3 for \(x^{2}\). The result is \(x - 3\) above the division line. After subtraction, the result is \(-8x + 7\).
05

Continue Until Remainder Degree is Less Than Divisor

Repeat Steps 2 and 3 for \(-8x\). The result is \(-8\) above the division line. After subtraction, the remainder is \(1\), which degree is less than the divisor's degree.
06

Write the Final Answer

Combine the results from above the division line to form the final quotient, \(2x^{2} + x - 8\), with a remainder of \(+1\). The final answer is typically written with the remainder over the divisor, yielding: \(2x^{2} + x - 8 + \frac{1}{x - 3}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equation for \(f\) is given by the simplified expression that results after performing the indicated operation. Write the equation for \(f\) and then graph the function. $$\frac{x-\frac{1}{x}}{x+\frac{1}{x}}$$

a. If \(y=k x^{2},\) find the value of \(k\) using \(x=2\) and \(y=64\). b. Substitute the value for \(k\) into \(y=k x^{2}\) and write the resulting equation. c. Use the equation from part (b) to find \(y\) when \(x=5\).

The heat generated by a stove element varies directly as the square of the voltage and inversely as the resistance. If the voltage remains constant, what needs to be done to triple the amount of heat generated?

Use a graphing utility to graph $$f(x)=\frac{x^{2}-4 x+3}{x-2} \text { and } g(x)=\frac{x^{2}-5 x+6}{x-2}$$ What differences do you observe between the graph of \(f\) and the graph of \(g\) ? How do you account for these differences?

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The inequality \(\frac{x-2}{x+3}<2\) can be solved by multiplying both sides by \(x+3,\) resulting in the equivalent inequality \(x-2<2(x+3)\).
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.