Chapter 11.5, Problem 10AYU

To determine

To express: The expression $\frac{3x^4+x^2-2}{x^3+8}$ as the sum of a polynomial and a proper rational expression if it is an improper fraction.

Answer to Problem 10AYU

The expression $\frac{3x^4+x^2-2}{x^3+8}$ is an improper rational expression. It is expressed as $\frac{3x^4+x^2-2}{x^3+8}=3x+\frac{25}{6\left(x+2\right)}+\frac{-19x-56}{6\left(x^2-2x+4\right)}$ .

Explanation of Solution

Given information:

The expression $\frac{3x^4+x^2-2}{x^3+8}$ .

Formula used:

Improper rational expression of the form $\frac{P}{Q}$ where Q has non repeated linear factors of the form $\left(x-a_1\right)\left(x-a_2\right)\cdots \left(x-a_n\right)$ then $\frac{P}{Q}$ is expressed as $\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_n}{x-a_n}$ .

Calculation:

Consider the expression $\frac{3x^4+x^2-2}{x^3+8}$ .

Observe that the degree of the numerator of the rational expression is greater than the denominator so the provided expression is improper.

When a polynomial is divided by its factor then dividend is the product of divisor and quotient increased by remainder.

Apply the method of long division.

  $\begin{array}{l}{x}^3+8\begin{array}{c}\hfill 3x\\ \hfill \overline{)3x^4+x^2-2}\end{array}\\ \begin{array}{ccc}& & 3x^4+24x\end{array}\\ \begin{array}{ccc}& & \begin{array}{cc}& x^2-24x-2\end{array}\end{array}\end{array}$

Now, since $\text{Dividend}=\text{Quotient}\times \text{Divisor}+\text{Remainder}$ . Here, dividend is $5x^3+2x-1$ quotient is $5x$ , divisor is $x^2-4$ and remainder is $22x-1$ .

Therefore,

  $\frac{3x^4+x^2-2}{x^3+8}=3x+\frac{x^2-24x-2}{x^3+8}$

Factors of the denominator are of the form,

  $\begin{array}{c}{x}^3+8=x^3+2^3\\ =\left(x+2\right)\left(x^2-2x+4\right)\end{array}$ .

Recall that improper rational expression of the form $\frac{P}{Q}$ where Q has non repeated linear factors of the form $\left(x-a_1\right)\left(x-a_2\right)\cdots \left(x-a_n\right)$ then $\frac{P}{Q}$ is expressed as $\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_n}{x-a_n}$ .

Rewrite the expression $\frac{x^2-24x-2}{x^3+8}$ as,

  $\begin{array}{c}\frac{x^2-24x-2}{x^3+8}=\frac{A}{\left(x+2\right)}+\frac{Bx+C}{\left(x^2-2x+4\right)}\\ \frac{x^2-24x-2}{x^3+8}=\frac{A\left(x^2-2x+4\right)+\left(Bx+C\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ \frac{x^2-24x-2}{x^3+8}=\frac{A\left(x^2-2x+4\right)+\left(Bx+C\right)\left(x+2\right)}{x^3+8}\end{array}$

Therefore, $x^2-24x-2=A\left(x^2-2x+4\right)+\left(Bx+C\right)\left(x+2\right)$

Now, solve the above expression for values of A and B .

  $\begin{array}{c}{x}^2-24x-2=A\left(x^2-2x+4\right)+\left(Bx+C\right)\left(x+2\right)\\ =A{x}^2-2Ax+4A+B{x}^2+2Bx+Cx+2C\\ =\left(A+B\right)x^2+\left(-2A+2B+C\right)x+4A+2C\end{array}$

Now, equate the coefficients on both the side of equation to obtain $A+B=1$ , $-2A+2B+C=-24$ and $4A+2C=-2$ .

Now, solve the three equations involving three variables.

From equation $-2A+2B+C=-24$ isolate the value of B , $B=\frac{-24+2A-C}{2}$ .

Now, substitute $B=\frac{-24+2A-C}{2}$ and in the equation $A+B=1$ .

  $\begin{array}{c}A+\frac{-24+2A-C}{2}=1\\ 2A-24+2A-C=2\\ C=4A-26\end{array}$

Now substitute $C=4A-26$ in the equation $4A+2C=-2$ .

  $\begin{array}{c}4A+2\left(4A-26\right)=-2\\ 12A-52=-2\\ 12A=50\\ A=\frac{25}{6}\end{array}$

Now, substitute $A=\frac{25}{6}$ , in equation $A+B=1$ .

  $\begin{array}{c}\frac{25}{6}+B=1\\ B=-\frac{19}{6}\end{array}$

Now, substitute $A=\frac{25}{6}$ , in equation $C=4A-26$ .

  $\begin{array}{c}C=4\left(\frac{25}{6}\right)-26\\ C=-\frac{28}{3}\end{array}$

Now, substitute the values of A, B and C in the partial fraction decomposition.

  $\begin{array}{c}\frac{x^2-24x-2}{x^3+8}=\frac{\frac{25}{6}}{\left(x+2\right)}+\frac{\left(-\frac{19}{6}\right)x+\left(-\frac{28}{3}\right)}{\left(x^2-2x+4\right)}\\ \frac{x^2-24x-2}{x^3+8}=\frac{25}{6\left(x+2\right)}+\frac{-19x-56}{6\left(x^2-2x+4\right)}\end{array}$

Therefore, $\frac{3x^4+x^2-2}{x^3+8}=3x+\frac{x^2-24x-2}{x^3+8}$ is rewritten as,

  $\frac{3x^4+x^2-2}{x^3+8}=3x+\frac{25}{6\left(x+2\right)}+\frac{-19x-56}{6\left(x^2-2x+4\right)}$

Thus, the expression $\frac{3x^4+x^2-2}{x^3+8}$ is an improper rational expression. It is expressed as $\frac{3x^4+x^2-2}{x^3+8}=3x+\frac{25}{6\left(x+2\right)}+\frac{-19x-56}{6\left(x^2-2x+4\right)}$ .