# Partial fraction decomposition of the rational expression

## Homework Statement

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

(x2 – 7x + 16)/[(x + 2)(x2 – 4x + 5)]

## The Attempt at a Solution

[A/(x+2)] + [(Bx+C)/(x2-4x+5)]

x2-7x+16= A(X2-4x+5)+(Bx+C)(x+2)
X2-7x+16=[A+B]x2+[-4A+B+C]x+[5A+C]

A+B=1 => B=1-A
-4A+B+C=-7 => Will need to plugin later
5A+C=16 => C=16-5A

-4A+B+C=-7 => 4A+1-A+16-5A= -10A=17 A=(-17/10)

Now I plug A back into the others

B=1-(-17/10) => B=(27/10)

C=16-5(-17/10) => C= (49/2)

Result: [(-17/10)/(x+2)] + [((27x/10)+(49/2))/(x2-4x+5)]

However, I am being told the answer is wrong. What is it that I am doing wrong?

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Mark44
Mentor

## Homework Statement

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

(x2 – 7x + 16)/[(x + 2)(x2 – 4x + 5)]

## The Attempt at a Solution

[A/(x+2)] + [(Bx+C)/(x2-4x+5)]

x2-7x+16= A(X2-4x+5)+(Bx+C)(x+2)
X2-7x+16=[A+B]x2+[-4A+B+C]x+[5A+C]
I see a couple of mistakes on the line above, in the coefficients of x and the constant term.
A+B=1 => B=1-A
-4A+B+C=-7 => Will need to plugin later
5A+C=16 => C=16-5A

-4A+B+C=-7 => 4A+1-A+16-5A= -10A=17 A=(-17/10)

Now I plug A back into the others

B=1-(-17/10) => B=(27/10)

C=16-5(-17/10) => C= (49/2)

Result: [(-17/10)/(x+2)] + [((27x/10)+(49/2))/(x2-4x+5)]

However, I am being told the answer is wrong. What is it that I am doing wrong?

I realize there is a mistake, but you could please elaborate upon my mistake without giving me the answer? Your response did not make it obvious or clear for me. Maybe a more indepth response would help me better understand my own mistake. Thank you in advance for taking the time to help me with this problem!

Is it suppose to be X2-7x+16=[A+B]x2+[-4A+2B+C]x+[5A+2C]

Is that correct?

Mark44
Mentor
That's what I get

That's what I get
But now I am stuck because I can only get one value.

A+B=1 => A=1+B

-4A+2B+C=-7 =>

5A+2C=16 =>

Mark44
Mentor
But now I am stuck because I can only get one value.

A+B=1 => A=1+B
Mistake above. A = 1 - B
-4A+2B+C=-7 =>

5A+2C=16 =>
You have three equations in three unknowns, so you should be able to solve for them.

Mistake above. A = 1 - B

You have three equations in three unknowns, so you should be able to solve for them.
That was my question. How am I suppose to solve for 3 unknowns when I cannot cancel anything out. I need atleast 2 known variables to solve any of the equations. Do I subtract equations or some sort of method?

Use one of the three equations to get a substitution formula. For example, you have A=1-B so far. By substituting into the other two equations, you should be able to reduce your system down to 2 equations and 2 unknowns. Go from there...

## Homework Statement

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

(x2 – 7x + 16)/[(x + 2)(x2 – 4x + 5)]

## The Attempt at a Solution

[A/(x+2)] + [(Bx+C)/(x2-4x+5)]

You need to change the Bx+C. That polynomial can be reduced. So change it to a B and a C.

Mark44
Mentor

## Homework Statement

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

(x2 – 7x + 16)/[(x + 2)(x2 – 4x + 5)]

## The Attempt at a Solution

[A/(x+2)] + [(Bx+C)/(x2-4x+5)]

You need to change the Bx+C. That polynomial can be reduced. So change it to a B and a C.
NO! mistalopez has the correct decomposition.

NO! mistalopez has the correct decomposition.
Are you sure? We learned to break down polynomials to factors first, before making it a Bx+c. We learned that the only time you don't use that is when it does not come out evenly in a factor.

Mark44
Mentor
Yes, I'm sure. x2 -4x + 5 can't be factored into linear factors with real coefficients.

Oh my god, how stupid of me. I read that wrong.
I'm so sorry. >.<

Mark44
Mentor
Oh my god, how stupid of me. I read that wrong.
I'm so sorry. >.<
We all screw up from time to time...