Try the fastest way to create flashcards
Question

Use long division to divide.

(x3+4x23x12)÷(x3)(x^3+4x^2-3x-12)\div(x-3)

Solution

Verified
Answered 2 years ago
Answered 2 years ago
Step 1
1 of 5

We use long division to divide the expressions.

x+2;             x2+7x+18x3)   x3+4x23x12x3;x3+3x2x3;4x        7x23xx3;4 3+7x2+21xx3+;4x3          +18x12x3;4x3          +18x+54x3+;4x3                     +42\begin{array}{l} \phantom{x+2;}{~~~~~~~~~~~~~x^2+7x+18}\\ x-3\overline{\smash{)}~~~x^3+4x^2-3x-12}\\ \phantom{x-3;}\underline{-x^3+3x^2}\\ \phantom{{x-3;}4x}~~~~~~~~7x^2-3x\\ \phantom{{x-3;}4~^3+}\underline{-7x^2+21x}\\ \phantom{{x-3+;}4x^3~~~~~~~~~~+}{18x-12}\\ \phantom{{x-3;}4x^3~~~~~~~~~~+}\underline{-18x+54}\\ \phantom{{x-3+;}4x^3~~~~~~~~~~~~~~~~~~~~~+}{42}\\ \end{array}

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Thomas' Calculus 14th Edition by Christopher E Heil, Joel R. Hass, Maurice D. Weir

Thomas' Calculus

14th EditionISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir
10,142 solutions
Calculus I with Precalculus 3rd Edition by Bruce E. Edwards, Ron Larson

Calculus I with Precalculus

3rd EditionISBN: 9780840068330Bruce E. Edwards, Ron Larson
12,640 solutions
Calculus: Early Transcendentals 8th Edition by James Stewart

Calculus: Early Transcendentals

8th EditionISBN: 9781285741550 (6 more)James Stewart
11,081 solutions
Calculus: Early Transcendentals 9th Edition by Daniel K. Clegg, James Stewart, Saleem Watson

Calculus: Early Transcendentals

9th EditionISBN: 9781337613927 (3 more)Daniel K. Clegg, James Stewart, Saleem Watson
11,050 solutions

More related questions

1/4

1/7