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Decomposing P ( x ) Q ( x ) When Q(x) Contains a nonrepeated irreducible quadratic factor

Find a partial fraction decomposition of the given expression.

8 x 2 + 12 x −20 ( x + 3 ) ( x 2 + x + 2 )

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,

8 x 2 + 12 x −20 ( x + 3 ) ( x 2 + x + 2 ) = A ( x + 3 ) + B x + C ( x 2 + x + 2 )

We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.

( x + 3 ) ( x 2 + x + 2 ) [ 8 x 2 + 12 x 20 ( x + 3 ) ( x 2 + x + 2 ) ] = [ A ( x + 3 ) + B x + C ( x 2 + x + 2 ) ] ( x + 3 ) ( x 2 + x + 2 )                                         8 x 2 + 12 x 20 = A ( x 2 + x + 2 ) + ( B x + C ) ( x + 3 )

Notice we could easily solve for A by choosing a value for x that will make the B x + C term equal 0. Let x = −3 and substitute it into the equation.

               8 x 2 + 12 x 20 = A ( x 2 + x + 2 ) + ( B x + C ) ( x + 3 )     8 ( 3 ) 2 + 12 ( 3 ) 20 = A ( ( 3 ) 2 + ( 3 ) + 2 ) + ( B ( 3 ) + C ) ( ( 3 ) + 3 )                                   16 = 8 A                                    A = 2

Now that we know the value of A , substitute it back into the equation. Then expand the right side and collect like terms.

8 x 2 + 12 x −20 = 2 ( x 2 + x + 2 ) + ( B x + C ) ( x + 3 ) 8 x 2 + 12 x −20 = 2 x 2 + 2 x + 4 + B x 2 + 3 B + C x + 3 C 8 x 2 + 12 x −20 = ( 2 + B ) x 2 + ( 2 + 3 B + C ) x + ( 4 + 3 C )

Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.

          2 + B = 8 (1) 2 + 3 B + C = 12 (2)         4 + 3 C = −20 (3)

Solve for B using equation (1) and solve for C using equation (3).

    2 + B = 8 (1)           B = 6 4 + 3 C = −20 (3)         3 C = −24           C = −8

Thus, the partial fraction decomposition of the expression is

8 x 2 + 12 x −20 ( x + 3 ) ( x 2 + x + 2 ) = 2 ( x + 3 ) + 6 x −8 ( x 2 + x + 2 )
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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 A first. The expansion on the right would be:

8 x 2 + 12 x −20 = A x 2 + A x + 2 A + B x 2 + 3 B + C x + 3 C 8 x 2 + 12 x −20 = ( A + B ) x 2 + ( A + 3 B + C ) x + ( 2 A + 3 C )

So the system of equations would be:

          A + B = 8 A + 3 B + C = 12       2 A + 3 C = −20

Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor.

5 x 2 −6 x + 7 ( x −1 ) ( x 2 + 1 )

3 x −1 + 2 x −4 x 2 + 1

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Decomposing P ( x ) Q ( x ) When Q(x) Has a repeated 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.

Decomposition of P ( x ) Q ( x ) When Q(x) Has a repeated irreducible quadratic factor

The partial fraction decomposition of P ( x ) Q ( x ) , when Q ( x ) has a repeated irreducible quadratic factor and the degree of P ( x ) is less than the degree of Q ( x ) , is

P ( x ) ( a x 2 + b x + c ) n = A 1 x + B 1 ( a x 2 + b x + c ) + A 2 x + B 2 ( a x 2 + b x + c ) 2 + A 3 x + B 3 ( a x 2 + b x + c ) 3 + + A n x + B n ( a x 2 + b x + c ) n

Write the denominators in increasing powers.

Given a rational expression that has a repeated irreducible factor, decompose it.

  1. Use variables like A , B , or C for the constant numerators over linear factors, and linear expressions such as A 1 x + B 1 , A 2 x + B 2 , etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as
    P ( x ) Q ( x ) = A a x + b + A 1 x + B 1 ( a x 2 + b x + c ) + A 2 x + B 2 ( a x 2 + b x + c ) 2 + +   A n + B n ( a x 2 + b x + c ) n
  2. Multiply both sides of the equation by the common denominator to eliminate fractions.
  3. Expand the right side of the equation and collect like terms.
  4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.

Questions & Answers

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rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
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the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
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This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
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Confunction Identity
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Practice Key Terms 2

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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