Suppose that Q(x) = (x-a)(x-b), where a and b are not equal. Also let \frac{P(x)}{Q(x)} be a...

Question:

Suppose that {eq}Q(x) = (x-a)(x-b) {/eq}, where a and b are not equal. Also let {eq}\frac{P(x)}{Q(x)} {/eq} be a proper rational function so that {eq}\frac{P(x)}{Q(x)} = \frac{A}{x-a} + \frac{B}{x-b} {/eq}

(a) Show that {eq}A = \frac{P(a)}{Q'(a)} {/eq} and {eq}B = \frac{P(b)}{Q'(b)} {/eq}

(b) Use this result to find out the partial fraction decomposition for {eq}\frac{3x - 2}{x^2 - 4x -12} {/eq}

Proving Values of Arbitrary Constants:

In this particular problem, we have an expression for partial fraction decomposition of the rational function with the arbitrary constants.

We'll simply use the general rule of decomposition to simplify the fractions using the least common denominator rule and then equate each factor equal to zero to compute the value of each arbitrary constant.

To prove the values, we need to compute the derivative of the denominator at the obtained value after equating factors equal to zero respectively.

Answer and Explanation:

(a)

The given proper rational function is:

{eq}\displaystyle \frac{P(x)}{Q(x)}=\frac{P(x)}{(x-a)(x-b)} {/eq}

Here, the given factors of the denominator are:

{eq}\displaystyle Q(x) = (x-a)(x-b) {/eq}

The derivative of both sides of the above expression with respect to {eq}x {/eq} is:

{eq}\begin{align*} \displaystyle \frac{\mathrm{d}(Q(x)) }{\mathrm{d} x}& =\frac{\mathrm{d} (x-a)(x-b)}{\mathrm{d} x}\\ Q'(x)& =(x-a)\frac{\mathrm{d} (x-b)}{\mathrm{d} x}+(x-b)\frac{\mathrm{d} (x-a)}{\mathrm{d} x}\\ & =(x-a)(1-0)+(x-b)(1-0)\\ & =(x-a)+(x-b)\\ \end{align*} {/eq}

Plug {eq}x=a {/eq} in the above expression to get the value of {eq}Q'(a) {/eq}.

{eq}\begin{align*} \displaystyle Q'(a)& =(a-a)+(a-b)\\ & =0+(a-b)\\ &=a-b \end{align*} {/eq}

Plug {eq}x=b {/eq} in the above expression to get the value of {eq}Q'(b) {/eq}.

{eq}\begin{align*} \displaystyle Q'(b)&= (b-a)+(b-b)\\ &= (b-a)+(0)\\ &=b-a \end{align*} {/eq}

The given expression for the partial fraction decomposition of the rational function is:

{eq}\displaystyle \frac{P(x)}{Q(x)} = \frac{A}{x-a} + \frac{B}{x-b} {/eq}

Plug the value of {eq}Q(x) {/eq} in the above expression and simplify the fractions of the right-hand side of the above expression.

{eq}\begin{align*} \displaystyle \frac{P(x)}{(x-a)(x-b)}& = \frac{A(x-b)+B(x-a)}{(x-a)(x-b)}\\ P(x)& = A(x-b)+B(x-a)\\ \end{align*} {/eq}

Plug {eq}x-b=0 {/eq} in the above expression and simplify the value of the arbitrary constant {eq}B {/eq}.

{eq}\begin{align*} \displaystyle P(b)& = A(0)+B(b-a)\\ P(b)& = 0+B(b-a)\\ \frac{P(b)}{b-a}& = B\\ \end{align*} {/eq}

The denominator of the above expression is the same as the function {eq}Q'(b) {/eq}.

Therefore,

{eq}B =\displaystyle \frac{P(b)}{Q'(b)} {/eq}

Hence, it proved.

Plug {eq}x-a=0 {/eq} in the above expression and simplify the value of the arbitrary constant {eq}A {/eq}.

{eq}\begin{align*} \displaystyle P(a)& = A(a-b)+B(0)\\ P(a)& = A(a-b)+0\\ \frac{P(a)}{a-b}& = A\\ \end{align*} {/eq}

The denominator of the above expression is the same as the function {eq}Q'(a) {/eq}.

Therefore,

{eq}A = \displaystyle \frac{P(a)}{Q'(a)} {/eq}

Hence, it proved.

(b)

The given rational function is:

{eq}\displaystyle \frac{3x - 2}{x^2 - 4x -12} {/eq}

Simplifying the denominator of the above rational function, we get:

{eq}\begin{align*} \displaystyle x^2 - 4x -12&=x^2 - 6x+2x -12\\ &=x(x - 6)+2(x -6)\\ &=(x+2)(x -6)\\ \end{align*} {/eq}

Here, we have:

{eq}P(x)=3x - 2\\ Q(x)=(x+2)(x -6)\\ a=-2\\ b=6\\ {/eq}

According to the part (a), we have:

{eq}Q'(x)=(x+2)+(x -6)\\ Q'(a)=Q'(-2)= -8\\ Q'(b)=Q'(6)=8 {/eq}

The values of the function {eq}P(x) {/eq} at {eq}a=-2 {/eq} and {eq}b=6 {/eq} are:

{eq}P(a)=P(-2)= -8\\ P(b)=P(6)=16 {/eq}

The partial fraction decomposition of the given rational function is:

{eq}\begin{align*} \displaystyle \frac{3x - 2}{x^2 - 4x -12}&=\frac{3x - 2}{(x+2)(x -6)}\\ &=\frac{A}{(x+2)}+\frac{B}{(x -6)}\\ \end{align*} {/eq}

From part (a), we have:

{eq}A = \displaystyle \frac{P(a)}{Q'(a)}\\ B =\displaystyle \frac{P(b)}{Q'(b)}\\ {/eq}

Plug the respective values in the above expression to get the exact values of the arbitrary constants.

{eq}\begin{align*} \displaystyle A &= \displaystyle \frac{P(-2)}{Q'(-2)}\\ &= \displaystyle \frac{-8}{-8}\\ &=1 \end{align*} {/eq}

{eq}\begin{align*} \displaystyle B &=\displaystyle \frac{P(6)}{Q'(6)}\\ &=\displaystyle \frac{16}{8}\\ &=2 \end{align*} {/eq}

Therefore,

{eq}\begin{align*} \displaystyle \frac{3x - 2}{x^2 - 4x -12}=\frac{1}{(x+2)}+\frac{2}{(x -6)} \end{align*} {/eq}

Learn more about this topic:

Partial Fraction Decomposition: Rules & Examples

from High School Algebra I: Help and Review

Chapter 3 / Lesson 25
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