{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PartialFractions - 3 The denominator contains irreducible...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
A quick summary of partial fractions Math 408C: Di/erential and Integral Calculus Unique Numbers 55420, 55425, and 55430 We apply the technique of partial fractions to a rational function F ( x ) = P ( x ) Q ( x ) ; where both P and Q are polynomials. We assume that long division has already been done, leaving us with deg( P ) < deg( Q ) . The next step is to factor the denominator Q ( x ) . There are then four possible cases. 1. The denominator is the product of linear factors, none of which are repeated, say Q ( x ) = ( ax + p )( bx + q )( cx + r )( dx + s ) : In this case, the partial fraction decomposition takes the form F ( x ) = A ax + p + B bx + q + C cx + r + D dx + s ; where A; B; C; : : : are constants that we can solve for. 2. The denominator is the product of linear factors, some of which are repeated, say Q ( x ) = ( ax + p ) 3 ( bx + q )( cx + r )( dx + s ) : In this case, the partial fraction decomposition takes the form F ( x ) = A 1 ax + p + A 2 ( ax + p ) 2 + A 3 ( ax + p ) 3 + B bx + q + C cx + r + D dx + s ; where A 1 ; A 2 ; A 3 and B; C; D : : : are again constants we solve for.
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3. The denominator contains irreducible quadratic factors, none of which are repeated. (We say a quadratic is irreducible if it cannot be factored using real numbers; for example, x 2 +1 is irreducible.) If, say, Q ( x ) = ( ax 2 + p 2 )( bx + q )( cx + r )( dx + s ) ; with ax 2 + p 2 irreducible, then the partial fraction decomposition will be F ( x ) = A 1 x + A 2 ax 2 + p 2 + B bx + q + C cx + r + D dx + s : 4. The denominator contains a repeated irreducible quadratic factor, say Q ( x ) = ( ax 2 + p 2 ) 3 ( bx + q )( cx + r )( dx + s ) : In this case, the partial fraction decomposition takes the form F ( x ) = A 1 x + A 2 ax 2 + p 2 + A 3 x + A 4 ( ax 2 + p 2 ) 2 + A 5 x + A 6 ( ax 2 + p 2 ) 3 + B bx + q + C cx + r + D dx + s :...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern