# PAR TIAL FRAC TION + DECOMPOSITION. Let’s add the two fractions below. We need a common denominator: In this section we are going to learn how to take.

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PAR TIAL FRAC TION + DECOMPOSITION

Let’s add the two fractions below. We need a common denominator: In this section we are going to learn how to take this answer and “decompose” it meaning break down into the fractions that were added together to get this answer

We start by factoring the denominator. There could have been a fraction for each factor of the denominator but we don’t know the numerators so we’ll call them A and B. Now we’ll clear this equation of fractions by multiplying every term by the common denominator. (x+3)(x-2)

This equation needs to be true for any value of x. We pick an x that will “conveniently” get rid of one of the variables and solve for the other. “The Convenient x Method for Solving” Let x = -3 B(0) = 0

Now we’ll “conveniently” choose x to be 2 to get rid of A and find B. Let x = 2 A(0) = 0 43

Summary of Partial Fraction Decomposition When Denominator Factors Into Linear Factors (Factors of first degree) Factor the denominator Set fraction equal to sum of fractions with each factor as a denominator using A, B,B, etc. for numerators Clear equation of fractions Use “convenient” x method to find A, B, etc. Next we’ll look at repeated factors and quadratic factors

Partial Fraction Decomposition With Repeated Linear Factors When the denominator has a repeated linear factor, you need a fraction with a denominator for each power of the factor. (x-1)(x+2) 2 Let x = 1

To find B we put A and C in and choose x to be any other number. Let x = 0 1313 -2 2 3 Let x = -2

Partial Fraction Decomposition With Quadratic Factors When the denominator has a quadratic factor (that won’t factor), you need a fraction with a linear numerator. (x+1)(x 2 +4) The convenient x method doesn’t work as nicely on these kind so we’ll use the “equating coefficients” method. First multiply everything out.

Look at the kinds of terms on each side and equate coefficients (meaning put the coefficients = to each other) Look at x 2 terms: 0 = A + B No x terms on left Look at x terms: 0 = B + C Look at terms with no x’s: 1 = 4A + C Solve these. Substitution would probably be easiest. A = - B C = - B 1 = 4(-B) + (-B) No x 2 terms on left

Partial Fraction Decomposition With Repeated Quadratic Factors When the denominator has a repeated quadratic factor (that won’t factor), you need a fraction with a linear numerator for each power. (x 2 +4) 2 multiply out equate coefficients of various kinds of terms (next screen)

Look at x 3 terms: 1 = A Look at x 2 terms: 1 = B Look at x terms: 0 = 4A+C Look at terms with no x: 0 = 4B+D 0 = 4(1)+C-4 =C 0 = 4(1)+D-4 = D 1 1 -4 1 1

Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.www.mathxtc.com Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au

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