Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Divide. Divide \(4+11 q+6 q^{2}\) by \(2 q+1\)

Short Answer

Expert verified
The quotient after dividing \(4+11q+6q^2\) by \(2q+1\) is \(3q + 2\).

Step by step solution

01

Set up the long division

First, arrange the equation so that the divisor \(2q + 1\) is on the outside of the division symbol and the dividend \(4+11q+6q^2\) is under the division symbol. Make sure to write the terms of the dividend in decreasing order of the terms' degrees (degree of a term is the power of variable in the term), so instead of \(4+11q+6q^2\), write it as \(6q^2 + 11q + 4\).
02

Divide the highest degree terms

Divide \(6q^2\), the term with the highest degree in the dividend, by \(2q\), the term with the highest degree in the divisor. The result of this division \(3q\), is the first term of the quotient.
03

Multiply and Subtract

Multiply the divisor \(2q + 1\) by the first term of the quotient \(3q\). Subtract the result from the dividend and bring down the next term. This will provide you a new polynomial.
04

Repeat the Process

Apply the same division process to the new polynomial. Divide the highest degree term of the new polynomial by the highest degree term of the divisor. The result of this will be the next term of the quotient. Continue this process until every term in the dividend has been brought down.
05

Write down the final quotient

All the terms obtained from each division make up the final quotient.

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 Vaia!

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free