DW123's question at Yahoo Answers regarding partial fraction decomposition

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SUMMARY

The discussion focuses on the process of partial fraction decomposition of the function f(x) = (-10x-7) / (2x²-17x+21). The numerator and denominator are factored to yield f(x) = - (10x + 7) / ((2x - 3)(x - 7)). Using the Heaviside cover-up method, the values A and B are determined to be 4 and -7, respectively. The final decomposition is expressed as f(x) = 4 / (2x - 3) - 7 / (x - 7).

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MarkFL
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Hello DW123,

We are give to decompose:

$\displaystyle f(x)=\frac{-10x-7}{2x^2-17x+21}$

Our first step is to factor the numerator and denominator:

$\displaystyle f(x)=-\frac{10x+7}{(2x-3)(x-7)}$

Now, we will assume the decomposition will take the form:

$\displaystyle -\frac{10x+7}{(2x-3)(x-7)}=\frac{A}{2x-3}+\frac{B}{x-7}$

Using the Heaviside cover-up method, we may find the value of $A$ by covering up the factor $(2x-3)$ on the left side, and evaluate what's left where $x$ takes on the value of the root of the covered up factor, i.e., $\displaystyle x=\frac{3}{2}$. Hence:

$\displaystyle A=-\frac{10\cdot\frac{3}{2}+7}{\frac{3}{2}-7}=4$

Likewise, we find the value of $B$ by covering up the factor $(x-7)$ on the left side, and evaluate what's left for $x=7$:

$\displaystyle B=-\frac{10\cdot7+7}{2\cdot7-3}=-7$

And so we conclude that:

$\displaystyle f(x)=\frac{-10x-7}{2x^2-17x+21}=\frac{4}{2x-3}-\frac{7}{x-7}$

To DW123 or any other guests viewing this topic, partial fraction decomposition is sometimes taught in pre-calculus, but usually in calculus as a means of integrating functions, so if you have other related questions, please feel free to post them in:

http://www.mathhelpboards.com/f21/

http://www.mathhelpboards.com/f10/
 
Last edited:

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