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College Algebra : Polynomial Functions

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Example Question #1 : Partial Fractions

What is the partial fraction decomposition of the following:

$\frac{1}{x^{2}-3x+2}$

Possible Answers:

$=\frac{1}{x-1}-\frac{2}{x-2}$

$=\frac{3}{2x-1}+\frac{2}{x+9}$

$=-\frac{2}{x+3}+\frac{2}{x-10}$

$=\frac{1}{x+4}+\frac{2}{x-5}$

$=\frac{1}{x-2}-\frac{1}{x-1}$

Correct answer:

$=\frac{1}{x-2}-\frac{1}{x-1}$

Explanation:

Factor the denominator: $x^{2}-3x+2=(x-2)(x-1)$

$\frac{1}{x^{2}-3x+2}=\frac{1}{(x-2)(x-1)}$

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

Multiply both sides of the equation by (x-2)(x-1)

1=A(x-1)+B(x-2)

Let x=1 :

1=B(1-2)

B=-1

Let x=2 :

1=A(2-1)

A=1

$\frac{1}{(x-2)(x-1)}=\frac{1}{x-2}-\frac{1}{x-1}$

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Example Question #101 : Polynomial Functions

Find A and B in the expression

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

Possible Answers:

$A=4/3 \; ; \; B=2/3$

$A=1 \; ; \; B=4$

$A = 5/3 \; ; \; B = -2/3$

$A=x \; ; \; B=4$

Correct answer:

$A = 5/3 \; ; \; B = -2/3$

Explanation:

Take the expression $\frac{x+4}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}$ and multiply both sides by . This transforms the equation into the following:

x+4=A(x+2)+B(x-1)

To solve for A we want to get rid of the B. So we set x equal to 1:

$1+4 = A(1+3)+B(1-1)\rightarrow 5 = 3A \rightarrow A=5/3$

Similarly, we can get rid of A by setting x equal to -2:

$-2+4=A(-2+2)+B(-2-1)\rightarrow2=-3B\rightarrow B=-2/3$

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College Algebra : Polynomial Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Partial Fractions

Example Question #101 : Polynomial Functions

All College Algebra Resources

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