Partial Fraction Decomposition Concepts

Partial Fraction Decomposition Concepts

Assessment

Interactive Video

Mathematics

9th - 12th Grade

Hard

Created by

Jackson Turner

FREE Resource

This video tutorial introduces partial fraction decomposition, a method for expressing a rational expression as a sum or difference of simpler fractions, which is useful in calculus. It explains the conditions required for decomposition, such as the degree of the numerator being less than the denominator, and the need for factoring the denominator. The tutorial demonstrates setting up the equation with variables for numerators and solving for these variables using substitution. Finally, it shows how to substitute the values back into the fractions to complete the decomposition.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the primary purpose of partial fraction decomposition in calculus?

To simplify complex fractions for easier integration

To solve differential equations

To perform matrix operations

To find the roots of polynomials

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What must be true about the degrees of the numerator and denominator to perform partial fraction decomposition?

The degree of the numerator must be greater than the denominator

The degree of the denominator must be zero

The degree of the numerator must be less than the denominator

The degrees must be equal

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in partial fraction decomposition after ensuring the degree condition is met?

Performing long division

Factoring the numerator

Factoring the denominator

Combining like terms

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

When the denominator has distinct linear factors, what type of numerators do the partial fractions have?

Quadratic expressions

Constant values

Exponential expressions

Linear expressions

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you solve for the constants in the numerators of the partial fractions?

By differentiating both sides

By integrating both sides

By performing long division

By selecting convenient values of x

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What value of x is chosen to eliminate the term with 'a' in the equation?

x = -3

x = 3

x = 1

x = 0

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the value of 'b' when x is set to 3?

b = 1

b = -2

b = 2

b = 0

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