Simplifying the Right Side of a Fraction Equation

[molly16]

Calculus Homework Help!

Homework Statement

Determine the values of A, B, and C if the following is true:

11x^2-14x+9/(3x-1)(x^2+1) = A/3x-1 + Bx+C/x^2+1

(Hint: Simplify the right side by combining fractions and comparing numerators)

Homework Equations

None

The Attempt at a Solution

I'm completely lost right now. Can someone please explain this?
I tried simplifying the right side. Then I cross multiplied the denominators and ended up with a huge equation that makes no sense.

 

[scurty]

Then I cross multiplied the denominators and ended up with a huge equation that makes no sense.

Can you show us the equation you got? From the sound of it you are on the right track. All you need to do is equate coefficients.

 

[lewdawgdude]

Yup, you got to get a common denominator on the right hand side. Then realize if the two fractions on the right and left hand side are equal, so the two numerators are equal. From there you can tell what the coefficients are on the right hand side from the left hand side.

 

[Ray Vickson]

Homework Statement

Determine the values of A, B, and C if the following is true:

11x^2-14x+9/(3x-1)(x^2+1) = A/3x-1 + Bx+C/x^2+1

(Hint: Simplify the right side by combining fractions and comparing numerators)

Homework Equations

None

The Attempt at a Solution

I'm completely lost right now. Can someone please explain this?
I tried simplifying the right side. Then I cross multiplied the denominators and ended up with a huge equation that makes no sense.

I cannot figure out what your expression is. It reads as $$11x^2-14x+\frac{9}{3x-1}(x^2+1),$$ but maybe you meant
$$11x^2-14x+\frac{9}{(3x-1)(x^2+1)}, \text{ or } \frac{11x^2-14x+9}{ (3x-1)(x^2+1)}.$$ If you want people to know what you mean you need to use brackets; for example, the last expression above can be entered in plain text as (11x^2-14x+9)/[(3x-1)(x^2+1)]. The same holds on the right: what you wrote means $$\frac{A}{3}x-1 + Bx+ \frac{C}{x^2}+1,$$ but I suspect that is not what you really want.

RGV

 

[Rayquesto]

Do you remember how to decompose a function into partial fractions?

 

[Rayquesto]

However, you could use long division to divide out the (x^2+1), but I bet you will run into one problem... the remainder. Come back when that happens. We will have your questions answered.

 

[molly16]

The original equation was:

(11x^2-14x+9)/[(3x-1)(x^2+1)] = (A)/(3x-1) + (Bx+C/x^2+1)

and I ended up with:
11x^4+19x^3-33x^2+27x = 3Ax^3-Ax^2+3Ax-A+3Bx^4-Bx^3+3Bx^2-Bx+3Cx^3-Cx^2+3Cx-C

I wasn't sure what to do after that.

 

[HallsofIvy]

The original equation was:

(11x^2-14x+9)/[(3x-1)(x^2+1)] = (A)/(3x-1) + (Bx+C/x^2+1)

I'll be it wasn't! In fact, I'll bet it was
(11x^2-14x+9)/[(3x-1)(x^2+1)] = (A)/(3x-1) + (Bx+C)/(x^2+1)
Do you see the difference?

and I ended up with:
11x^4+19x^3-33x^2+27x = 3Ax^3-Ax^2+3Ax-A+3Bx^4-Bx^3+3Bx^2-Bx+3Cx^3-Cx^2+3Cx-C

I wasn't sure what to do after that.

Now, as others have told you combine "like" coefficients and compare the two sides:
11x^4+ 19x^3- 33x^2+ 27x= 3Bx^4+ (3A- B+ 3C)x^3+ (-A+ 3B- C)x^2+ (3A- B+ 3C)x+ (-A- C)
so we must have 3B= 11, 3A- B+ 3C= 19, -A+ 3B- C= -33, 3A- B+ 3C= 27, and -A-C= 0, 5 equations for the three coefficients, A, B, C so it may well be "over-determined". I did not check to see if your calculations were correct.

Alternatively, you might just let x be any three numbers, say 0, 1, and -1, in order to get three equations for A, B, and C.

 

[molly16]

I'll be it wasn't! In fact, I'll bet it was
(11x^2-14x+9)/[(3x-1)(x^2+1)] = (A)/(3x-1) + (Bx+C)/(x^2+1)
Do you see the difference?

Oh oops. Sorry about that

Now, as others have told you combine "like" coefficients and compare the two sides:
11x^4+ 19x^3- 33x^2+ 27x= 3Bx^4+ (3A- B+ 3C)x^3+ (-A+ 3B- C)x^2+ (3A- B+ 3C)x+ (-A- C)
so we must have 3B= 11, 3A- B+ 3C= 19, -A+ 3B- C= -33, 3A- B+ 3C= 27, and -A-C= 0, 5 equations for the three coefficients, A, B, C so it may well be "over-determined". I did not check to see if your calculations were correct.

Alternatively, you might just let x be any three numbers, say 0, 1, and -1, in order to get three equations for A, B, and C.

I let x be 0, 1, and -1 and ended up with the equations -A-C=0, 4A+4B+4C=24 and -8A+8B-8C=-68. Then I used equation 1 and 2 to solve for B . I got B=6 but in our textbook (nelson calculus and vectors 12) it says that B=2. I'm also not sure how you would solve for A and C. I tried and got A=C but in the textbook they are not the same (A=5 and C=-4)

 

[Rayquesto]

Just use decomposition to partial fractions actually.

have this:

A/3x-1 + Bx+C/x^2+1 as your premise...

Now if 11x^2-14x+9/(3x-1)(x^2+1) = A/3x-1 + Bx+C/x^2+1, then

11x^2-14x+9=A(x^2+1) + (Bx+C)(3x-1)

11x^2-14x+9=Ax^2 + 1 + 3Bx^2 + Bx + 3Cx + C...

A + 3B=11
A + C=9
B + 3C=-14 so on and so fourth.

 

[scurty]

Just use decomposition to partial fractions actually.

have this:

A/3x-1 + Bx+C/x^2+1 as your premise...

Now if 11x^2-14x+9/(3x-1)(x^2+1) = A/3x-1 + Bx+C/x^2+1, then...

Ray, try to use parentheses correctly when helping other people so as to not confuse them! What Ray meant was:

A/(3x-1) + (Bx+C)/(x^2+1) as your premise...

Now if (11x^2-14x+9)/[(3x-1)(x^2+1)] = A/(3x-1) + (Bx+C)/(x^2+1), then...


Or to not confuse anybody at all..

$\displaystyle\frac{A}{3x-1} + \frac{Bx+C}{x^2+1}$ as your premise...

Now if $\displaystyle\frac{11x^2-14x+9}{(3x-1)(x^2+1)} = \frac{A}{3x-1} + \frac{Bx+C}{x^2+1}$, then...