/*! 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 15 Use long division to divide. \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 15

Use long division to divide. \(\left(x^{4}+5 x^{3}+6 x^{2}-x-2\right) \div(x+2)\)

Short Answer

Expert verified
The result of the long division is \( x^{3} + 3x^{2} + 1 \).

Step by step solution

01

Set up the Long Division

Arrange the integral in the form of long division: \[ \begin{array}{c|cccc}x+2 & x^{4} & +5x^{3} & +6x^{2} & -x & -2\end{array} \] Note that you can change the order of the terms according to the exponent, in decreasing order.
02

Start the Division

Divide the first term of the dividend by the divisor, x^4 divided by x is \(x^3\). So, place \(x^3\) at the top as the first term of our result. Multiply \((x+2)\) by \(x^3\) and subtract it from the dividend. Bring down the next term if the subtraction does not yield zero.\[ \begin{array}{c|ccccc}x+2 & x^{4} & +5x^{3} & +6x^{2} & -x & -2 \& -\(x^{4}\) & -\(2x^{3}\) \ \cline{2-3}& & \(3x^{3}\) & \(+6x^{2}\)\end{array} \]
03

Continue the Division

Proceed with the division by repeating Step 2: Divide the first term ( \(3x^3\) ) by \(x\) gives \(3x^2\). Multiply \((x+2)\) by \(3x^2\) and subtract.\[ \begin{array}{c|ccccc}x+2 & x^{4} & +5x^{3} & +6x^{2} & -x & -2 \& -\(x^{4}\) & -\(2x^{3}\) \ \cline{2-3}& & \(3x^{3}\) & \(+6x^{2}\) \& & -\(3x^{3}\) & -\(6x^{2}\) \ \cline{3-4}& & & \(0x^{2}\) & \(+x\)\end{array} \]
04

Complete the Division

Repeat Step 2. However, since the first term is zero, we simply bring down the next non-zero term, which is \(+x\). So, divide \(x\) by \(x\) gives \(1\). Multiply \((x+2)\) by \(1\) and subtract. The remainder is zero, so the division finishes.\[ \begin{array}{c|ccccc}x+2 & x^{4} & +5x^{3} & +6x^{2} & -x & -2 \& -\(x^{4}\) & -\(2x^{3}\) \ \cline{2-3}& & \(3x^{3}\) & \(+6x^{2}\) \& & -\(3x^{3}\) & -\(6x^{2}\) \ \cline{3-4}& & & \(0x^{2}\) & \(+x\) \& & & -\(x\) & -\(2\) \ \cline{4-5}& & & & \(0\)\end{array} \]

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

Solve the inequality and graph the solution on the real number line. \((x+2)^{2} \leq 25\)

Use a graphing utility to graph the equation. Use the graph to approximate the values of \(x\) that satisfy each inequality. \(y=\frac{1}{2} x^{2}-2 x+1 \quad\) (a) \(y \leq 0 \quad\) (b) \(y \geq 7\)

Between two consecutive zeros, a polynomial must be entirely ______________ or entirely __________________.

Determine whether each value of \(x\) is a solution of the inequality. Inequality. \(x^{2}-x-12 \geq 0 \quad\) (a) \(x=5 \quad\) (b) \(x=0\) (c) \(x=-4\) (d) \(x=-3\)

A rectangular parking lot with a perimeter of 440 feet is to have an area of at least 8000 square feet. Within what bounds must the length of the rectangle lie?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Calculus

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.