5. Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve.

3x-1/x(x+3)^3(2x^2+1)

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3x-1/x(x+3)^3(2x^2+1)

Answer is (A). For a term with an irreducible [tex]n[/tex]-degree polynomial in the denominator, its numerator must have at most a polynomial of degree [tex]n-1[/tex].

[tex]\dfrac{5x-2}{(x-1)^2}=\dfrac a{x-1}+\dfrac b{(x-1)^2}[/tex]

[tex]5x-2=a(x-1)+b[/tex]

Since [tex]5x-2=5x-5+3=5(x-1)+3[/tex], it follows that [tex]a=5[/tex] and [tex]b=3[/tex], so

[tex]\dfrac{5x-2}{(x-1)^2}=\dfrac5{x-1}+\dfrac3{(x-1)^2}[/tex]

[tex]\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac{x^3+x+x+2}{x^2(x^2+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac{cx+d}{x^2+1}[/tex]

[tex]x^3+x^2+x+2=ax(x^2+1)+b(x^2+1)+(cx+d)x^2[/tex]

[tex]x^3+x^2+x+2=(a+c)x^3+(b+d)x^2+ax+b[/tex]

When [tex]x=0[/tex], you find that [tex]b=2[/tex]. It's also clear that [tex]a=1[/tex]. So the remaining constants must be [tex]1+c=1\implies c=0[/tex] and [tex]2+d=1\implies d=-1[/tex], and so you get

[tex]\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac1x+\dfrac2{x^2}-\dfrac1{x^2+1}[/tex]

[tex]\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}[/tex]

[tex]\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)[/tex]

When [tex]x=3[/tex], you're left with

[tex]147=49a_1\implies a_1=\dfrac{147}{49}=3[/tex]

When [tex]x=\sqrt2[/tex] or [tex]x=-\sqrt2[/tex], you're left with

[tex]\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}[/tex]

Adding the two equations together gives [tex]-10=2a_5[/tex], or [tex]a_5=-5[/tex]. Subtracting them gives [tex]2\sqrt2=2\sqrt2a_4[/tex], [tex]a_4=1[/tex].

Now, you have

[tex]5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)[/tex]

[tex]5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)[/tex]

[tex]2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)[/tex]

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that [tex]a_2x^4=2x^4[/tex] and [tex]a_3(-3)(-2)=6a_3=-6[/tex]. These alone tell you that you must have [tex]a_2=2[/tex] and [tex]a_3=-1[/tex].

So the partial fraction decomposition is

[tex]\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}[/tex]

See below.

Step-by-step explanation:

First, distribute:

[tex]=\frac{1}{x(x+1)}[/tex]

Now, perform partial fraction decomposition. This is only two factors, so we only need linear functions:

[tex]\frac{1}{x(x+1)} =\frac{A}{x}+\frac{B}{x+1}[/tex]

Now, multiply everything by x(x+1):

[tex]1=A(x+1)+B(x)[/tex]

Now, solve for each variable. Let's let x=-1:

[tex]1=A(-1+1)+B(-1)[/tex]

[tex]1=0A-B=-B[/tex]

[tex]B=-1[/tex]

Now, let's let x=0:

[tex]1=A(0+1)+B(0)[/tex]

[tex]A=1[/tex]

So:

[tex]\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{(x+1)}[/tex]

Double Check:

[tex]\frac{1}{x}-\frac{1}{(x+1)}=\frac{(x+1)}{x(x+1)}-\frac{x}{x(x+1)}[/tex]

[tex]=\frac{x-x+1}{x(x+1)} =\frac{1}{x^2+x}[/tex]

9x^2-2x+6

=9(x^2+8)-9*8-2x+6

=9(x^2+8)-2x-66

=9(x^2+8)-2(x-5)-76

so ans is A

[tex](\frac{(3x + 4)}{(x + 2)})^{2} = \frac{(3x + 4)^{2}}{(x + 2)^{2}} = \frac{(3x + 4)(3x + 4)}{(x + 2)(x + 2)} = \frac{9x^{2} + 12x + 12x +16}{x^{2} + 2x + 2x + 4} = \frac{9x^{2} + 24x + 16}{x^{2} + 4x + 4} = 9 - \frac{12x - 20}{x^{2} + 4x + 4}[/tex]

[tex]\frac{5x-4}{x(x^2+7)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+7} + \frac{Dx+E}{(x^2+7)^2}[/tex]

Step-by-step explanation:

Given the expression [tex]\frac{5x-4}{x(x^2+7)^2}[/tex], we are to re-write the expression in form of a partial fraction.

Before we write in form of a partial fraction, we need to note the expression at the denominator. Since the expression in parenthesis is a quadratic equation, the equivalent numerator must be a linear expression.

Also the quadratic equation is a repeated form since it is squared. This means that we are to repeat the quadratic equation twice when writing as a partial fraction.

[tex]\frac{5x-4}{x(x^2+7)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+7} + \frac{Dx+E}{(x^2+7)^2}[/tex]

From the above partial fraction, it can be seen that x² + 7 in parenthesis was repeated twice and their equivalent expressions at the numerator are both linear i.e Bx+E and Dx+ E where A, B, C, D and E are the unknown constant.

[tex]\dfrac{A}{x}+\dfrac{B}{x-6}[/tex]

Step-by-step explanation:

Given the function [tex]\dfrac{37}{x(x-6)}[/tex], to write the form of its partial fraction on decomposition, we will separate the two functions separated by an addition sign. The numerator of each function will be constants A and b and the denominator will be the individual factors of each function at the denominator. The partial fraction of the rational function is as shown below.

[tex]= \dfrac{37}{x(x-6)}\\\\= \dfrac{A}{x}+\dfrac{B}{x-6}[/tex]

Since we are not to solve for the constants, hence the partial fraction is [tex]\dfrac{A}{x}+\dfrac{B}{x-6}[/tex]

= A/x +B/(x -2) +C/(x -4)

To find A, multiply the expression by x and evaluate at x=0.

.. A = 24/((-2)*(-4)) = 3

To find B, multiply the expression by x-2 and evaluate at x=2.

.. B = (18*2^2 -68*2 +24)/(2*(2-4)) = -40/-4 = 10

To find C, multiply the expression by x-4 and evaluate at x=4.

.. C = (18*4^2 -68*4 +24)/(4(4 -2)) = 40/8 = 5

The partial fraction decomposition is

.. = 3/x +10/(x -2) +5/(x -4)

[tex](\frac{(3x + 4)}{(x + 2)})^{2} = \frac{(3x + 4)^{2}}{(x + 2)^{2}} = \frac{(3x + 4)(3x + 4)}{(x + 2)(x + 2)} = \frac{9x^{2} + 12x + 12x +16}{x^{2} + 2x + 2x + 4} = \frac{9x^{2} + 24x + 16}{x^{2} + 4x + 4} = 9 - \frac{12x - 20}{x^{2} + 4x + 4}[/tex]