<< Chapter < Page | Chapter >> Page > |
Find a partial fraction decomposition of the given expression.
We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus,
We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.
Notice we could easily solve for by choosing a value for that will make the term equal 0. Let and substitute it into the equation.
Now that we know the value of substitute it back into the equation. Then expand the right side and collect like terms.
Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.
Solve for using equation (1) and solve for using equation (3).
Thus, the partial fraction decomposition of the expression is
Could we have just set up a system of equations to solve [link] ?
Yes, we could have solved it by setting up a system of equations without solving for first. The expansion on the right would be:
So the system of equations would be:
Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor.
Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.
The partial fraction decomposition of when has a repeated irreducible quadratic factor and the degree of is less than the degree of is
Write the denominators in increasing powers.
Given a rational expression that has a repeated irreducible factor, decompose it.
Notification Switch
Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications?