1
$\begingroup$

Can someone comment on the following as to whether this partial fractions can be true or not? I'm concerned I've come across a trick question.

1) $\frac{x(x^2+4)}{x^2-4} = \frac{A}{x+2} + \frac{B}{x-2}$

Reason would lead me to believe the answer is yes, but mathematically I'm unsure if I'm correct?

If we take the original function...

$\frac{x(x^2+4)}{x^2-4}$

... multiply out the brackets and use long division, I get...

$x + \frac{8x}{x^2-4}$

Factorising the denominator gives us (x+2)(x-2), so in partial fractions we get...

2) $\frac{x(x^2+4)}{x^2-4} = x + \frac{A}{x+2} + \frac{B}{x-2}$

However, reason tells me that since $x = \frac{x(x+2)}{x+2}$ then equation 1, the value of A can be $x(x+2)$ plus the value of A in equation 2. i.e. Thus converting equation 2 to be...

2.1) $\frac{x(x^2+4)}{x^2-4} = \frac{x(x+2) + A}{x+2} + \frac{B}{x-2}$

So basically what I'm asking is can the partial fractions be altered in these ways? Or is there a trick I'm missing somewhere?

$\endgroup$
3
$\begingroup$

No: we cannot have $$\dfrac{x(x^2 + 4)}{x^2-4} = \dfrac A{x-2} +\dfrac B{x+2}$$

Why not? $$\dfrac{x^3 + 4x}{x^2 - 2} \neq \dfrac{A(x+2) + B(x-2)}{x^2 - 4}$$

The original numerator is a polynomial of degree $3$. The "transformed" numerator is of degree at most $1$.

Recall that $A, B$ represent constants only, not polynomials (of degree greater than $0)$.

Before invoking partial fractions, we want to get the degree of the numerator less than the degree of the denominator. We can do this, as you do, by polynomial long division. $$\frac{x(x^2 + 4)}{x^2 - 4} = \frac{x^3 + 4x}{x^2 - 4}= x + \dfrac{8x}{x^2 - 4}$$

Now the degree in the numerator is less than the degree in the denominator of our remaining rational function. So we can now use partial fractions.

$$x + \dfrac{8x}{x^2 - 4} = x + \dfrac A{x-2} + \dfrac B{x+2}$$ where $$A(x+2) + B(x-2) = 8x$$

$\endgroup$
  • $\begingroup$ Ah thank you. I did not realise that A and B had to represent constants. Thanks! $\endgroup$ – Wolff Mar 2 '14 at 18:59
  • $\begingroup$ You are welcome, @Wolff! $\endgroup$ – Namaste Mar 2 '14 at 18:59
1
$\begingroup$

After Partial Fraction Decomposition, the degree of the denominator of each fraction(not the integral part) must be greater than that of the numerator

and for terms of the form $\displaystyle\frac a{(x+b)^n}$ can be represented as

$$\frac{A_1}{x+b}+\frac{A_2}{(x+b)^2}+\cdots+\frac{A_n}{(x+b)^n}$$

if the degrees of the numerator of the original expression, is greater than that of the denominator by $d(\ge0)$ degree

we shall have integral terms $c_0+c_1x+\cdots+c_dx^d$

$\endgroup$
  • 1
    $\begingroup$ @lab-bjattacharjee : The question is simply "Determine whether the statement is true or false" and asks whether the original equation can be put in the form $\frac{A}{x+2} + \frac{B}{x-2}$ Although it's on the topic of partial fractions it doesn't actually state to use partial fraction decomposition. If I substitute values of x after finding the values of A and B, then I get the same result for the original function and equation 2.1. $\endgroup$ – Wolff Mar 2 '14 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.