Another Partial Fractions question

[jonnburton]

Hi everyone,

I have got stuck on another partial fractions question and wondered whether anyone could help.

The question is:

Express \(\displaystyle \frac{2x^2 +3x +12}{(2x-1)(x+3)}\) in the form \(\displaystyle A + \frac{B}{2x-1} + \frac{C}{x+3}\), where A, B and C are constants to be determined.

My first thoughts were this is an improper fraction and I needed to do long division in order to simplify it, but after seeing that the denominator is not a linear expression, it can't be long divided...

So then, noticing that the denominator multiplies out to \(\displaystyle 2x^2 - 7x -3\), I thought of cancelling the \(\displaystyle 2x^2\), which gives:

\(\displaystyle \frac{1}{-7x -3} + \frac {3x + 12}{(2x-1)(x+3)}\)


However, having looked at the answer in the book which is \(\displaystyle 1 + \frac{4}{2x-1}- \frac{3}{x+3}\), I can't see how what I have done brings me any closer to this solution.

 

[Deleted member 4993]

Hi everyone,

I have got stuck on another partial fractions question and wondered whether anyone could help.

The question is:

Express \(\displaystyle \frac{2x^2 +3x +12}{(2x-1)(x+3)}\) in the form \(\displaystyle A + \frac{B}{2x-1} + \frac{C}{x+3}\), where A, B and C are constants to be determined.

My first thoughts were this is an improper fraction and I needed to do long division in order to simplify it, but after seeing that the denominator is not a linear expression, it can't be long divided...

So then, noticing that the denominator multiplies out to \(\displaystyle 2x^2 - 7x -3\),... incorrect ... that should be 2x² + 5x -3

I thought of cancelling the \(\displaystyle 2x^2\), which gives:

Thatcannot be done - you need to review basic algebra skills before you start these problems

\(\displaystyle \frac{1}{-7x -3} + \frac {3x + 12}{(2x-1)(x+3)}\)


However, having looked at the answer in the book which is \(\displaystyle 1 + \frac{4}{2x-1}- \frac{3}{x+3}\), I can't see how what I have done brings me any closer to this solution.

First, you need to notice that the orders of the polynomials in the numerator and the denominators are same

so you need reduce the order of the numerator and write the fraction as follows:

\(\displaystyle 2x^2 +3x +12 \ = \ (2x^2 + 5x - 3) - (2x - 15)\)

so

\(\displaystyle \displaystyle \frac{2x^2 +3x +12}{(2x-1)(x+3)} \ \)

\(\displaystyle \displaystyle = \ \frac{(2x^2 +5x -3) \ - \ (2x - 15)}{(2x-1)(x+3)}\)

\(\displaystyle \displaystyle \ = \frac{(2x-1)(x+3) - (2x- 15)}{(2x-1)(x+3)} \ \)

\(\displaystyle \displaystyle = \ 1 + \frac{ - 2x + 15}{(2x-1)(x+3)}\) ...........................so now we have A = 1

Now reduce \(\displaystyle \displaystyle \frac{ - 2x + 15)}{(2x-1)(x+3)}\) into \(\displaystyle \displaystyle \left [ \frac{B}{(2x-1)} + \frac{C}{(x+3)}\right ]\)

 

[jonnburton]

First, you need to notice that the orders of the polynomials in the numerator and the denominators are same

so you need reduce the order of the numerator and write the fraction as follows:

\(\displaystyle 2x^2 +3x +12 \ = \ (2x^2 + 5x - 3) - (2x - 15)\)

so

\(\displaystyle \displaystyle \frac{2x^2 +3x +12}{(2x-1)(x+3)} \ \)

\(\displaystyle \displaystyle = \ \frac{(2x^2 +5x -3) \ - \ (2x - 15)}{(2x-1)(x+3)}\)

\(\displaystyle \displaystyle \ = \frac{(2x-1)(x+3) - (2x- 15)}{(2x-1)(x+3)} \ \)

\(\displaystyle \displaystyle = \ 1 + \frac{ - 2x + 15}{(2x-1)(x+3)}\) ...........................so now we have A = 1

Now reduce \(\displaystyle \displaystyle \frac{ - 2x + 15)}{(2x-1)(x+3)}\) into \(\displaystyle \displaystyle \left [ \frac{B}{(2x-1)} + \frac{C}{(x+3)}\right ]\)

Thank you very much for that explanation, Subhotosh. I understand how this is done now!

 

[HallsofIvy]

Hi everyone,

I have got stuck on another partial fractions question and wondered whether anyone could help.

The question is:

Express \(\displaystyle \frac{2x^2 +3x +12}{(2x-1)(x+3)}\) in the form \(\displaystyle A + \frac{B}{2x-1} + \frac{C}{x+3}\), where A, B and C are constants to be determined.

My first thoughts were this is an improper fraction and I needed to do long division in order to simplify it, but after seeing that the denominator is not a linear expression, it can't be long divided...

You understand now, do you not, that this is incorrect? Any "improper" algebraic fraction (numerator of higher degree than denominator) can be "long divided". You were probably thinking of "synthetic division" which is just a simplified way to divide in the special case that the divisor is linear.

So then, noticing that the denominator multiplies out to \(\displaystyle 2x^2 - 7x -3\), I thought of cancelling the \(\displaystyle 2x^2\), which gives:

\(\displaystyle \frac{1}{-7x -3} + \frac {3x + 12}{(2x-1)(x+3)}\)

NO! \(\displaystyle \frac{3+ 2}{3+ 1}\) is NOT equal to \(\displaystyle \frac{2}{1}\). You can only cancel factors.

However, having looked at the answer in the book which is \(\displaystyle 1 + \frac{4}{2x-1}- \frac{3}{x+3}\), I can't see how what I have done brings me any closer to this solution.

It didn't because what you did was just bad algebra.

 

[jonnburton]

You understand now, do you not, that this is incorrect? Any "improper" algebraic fraction (numerator of higher degree than denominator) can be "long divided". You were probably thinking of "synthetic division" which is just a simplified way to divide in the special case that the divisor is linear.


NO! \(\displaystyle \frac{3+ 2}{3+ 1}\) is NOT equal to \(\displaystyle \frac{2}{1}\). You can only cancel factors.

It didn't because what you did was just bad algebra.

I was under the impression that long division could only be performend when the divisor is linear - all of the examples in my textbook have been of this form. I will have to look into long division using non-linear divisors.

But I definitely see that what I did was bad algebra, and I should have known better.

 

[MarkFL]

...
Now reduce \(\displaystyle \displaystyle \frac{-2x+15}{(2x-1)(x+3)}\) into \(\displaystyle \displaystyle \left [ \frac{B}{(2x-1)} + \frac{C}{(x+3)}\right ]\)

Once you have it in this form graciously supplied by Subhotosh Khan, you may employ the Heaviside cover-up method to determine \(\displaystyle B\) and \(\displaystyle C\)

To find \(\displaystyle B\), observe that the root of \(\displaystyle 2x-1\) is \(\displaystyle \dfrac{1}{2}\). So "cover up" the factor \(\displaystyle 2x-1\) on the left, and evaluate what is left for \(\displaystyle x=\dfrac{1}{2}\) giving:

\(\displaystyle B=\dfrac{-1+15}{\dfrac{7}{2}}=4\)

To find \(\displaystyle B\), observe that the root of \(\displaystyle x+3\) is \(\displaystyle -3\). So "cover up" the factor \(\displaystyle x+3\) on the left, and evaluate what is left for \(\displaystyle x=-3\) giving:

\(\displaystyle C=\dfrac{6+15}{-6-1}=-3\)

Hence:

\(\displaystyle \displaystyle \frac{-2x+15}{(2x-1)(x+3)}=\frac{4}{2x-1}-\frac{3}{x+3}\)

 

[jonnburton]

Once you have it in this form graciously supplied by Subhotosh Khan, you may employ the Heaviside cover-up method to determine \(\displaystyle B\) and \(\displaystyle C\)

To find \(\displaystyle B\), observe that the root of \(\displaystyle 2x-1\) is \(\displaystyle \dfrac{1}{2}\). So "cover up" the factor \(\displaystyle 2x-1\) on the left, and evaluate what is left for \(\displaystyle x=\dfrac{1}{2}\) giving:

\(\displaystyle B=\dfrac{-1+15}{\dfrac{7}{2}}=4\)

To find \(\displaystyle B\), observe that the root of \(\displaystyle x+3\) is \(\displaystyle -3\). So "cover up" the factor \(\displaystyle x+3\) on the left, and evaluate what is left for \(\displaystyle x=-3\) giving:

\(\displaystyle C=\dfrac{6+15}{-6-1}=-3\)

Hence:

\(\displaystyle \displaystyle \frac{-2x+15}{(2x-1)(x+3)}=\frac{4}{2x-1}-\frac{3}{x+3}\)

Thanks a lot for going through that way of doing things MarkFL.

 

[vincent_nyoni]

x^4/x^4-1

(x^4)/(x^4-1)

 

[stapel]

I was under the impression that long division could only be performend when the divisor is linear - all of the examples in my textbook have been of this form.

It sounds as though your textbook has been misleading. No, just as you can long-divide a number by a more-than-one-digit value, so also you can long-divide a polynomial by a more-than-degree-one expression. You can see some examples online.