Find the partial fraction decomposition of the rational function.

[wat2000]

5x^2 + 8/x3 + x2

I got a 5x^2 + 8 = A/x^2 + B/x + 1
A(x+1) + B(x^2)
(Ax+A) + (Bx^2)
(Ax + Bx^2) + A
5x^2 + 8 = (A + B) x + (A)x^2
5x^2+x+8=(A+B) + (A)
This is about as far as I can get but I think I made a mistake somewhere but I don't know where? Can someone help me?

 

[Mark44]

5x^2 + 8/x3 + x2

First off, whenever your write rational expressions on a single line, and the numerator or denominator has multiple terms, use parentheses around the entire numerator and/or denominator.

The expression you wrote would normally be interpreted as 5x² + (8/x³) + x².

The right way to write this is (5x² + 8)/(x³ + x²).

Or better yet, here it is in LaTeX.
[tex]\frac{5x^2 + 8}{x^3 + x^2}[/tex]

I got a 5x^2 + 8 = A/x^2 + B/x + 1

That is not correct for two reasons: 1) you omitted the denominator on the left side; 2) the right side needs three terms, not two.

(5x² + 8)/(x³ + x²) = A/x + B/x² + C/(x + 1)

Now, multiply both sides of the equation by x²(x + 1) to get
5x² + 8 = Ax(x + 1) + B(x + 1) + Cx²
Solve for A, B, and C.

A(x+1) + B(x^2)
(Ax+A) + (Bx^2)
(Ax + Bx^2) + A
5x^2 + 8 = (A + B) x + (A)x^2
5x^2+x+8=(A+B) + (A)
This is about as far as I can get but I think I made a mistake somewhere but I don't know where? Can someone help me?

 

[wat2000]

Thanks!

 

[wat2000]

I got Is -7/x + 8/x^2 + 12/x+1. I don't think that's right though?

 

[wat2000]

I used 5x^2+8=(A+C)x^2 + (A+B)x + B
5x^2 +0x+8=(A+C) + (A+B) + B
then i put it on the matrix function on my calculator to get this answer.

 

[HallsofIvy]

Why in the world would you use "the matrix function on my calculator" to solve such easy equations? And why do you have [itex]x^2[/itex] and x on the left but not on the right of your equation?

A+ C= 5, A+ B= 0, B= 8.

So A+ 8= 0, A= -8.

Then -8+ C= 5, C= 13.

 

[wat2000]

Just doing what my professor said to do. he said we won't have enough time to do these equations by hand on the test so he taught my class how to use the matrix function on the calculator. Also Friday was the first time I had even heard of one of theses kind of equations so I am very new at this.