- #1

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[tex]\frac{7}{3s^{2}(3s+1)}[/tex]

Can this be decomposed, and how?

Can this be decomposed, and how?

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- Thread starter ko_kidd
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- #1

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[tex]\frac{7}{3s^{2}(3s+1)}[/tex]

Can this be decomposed, and how?

Can this be decomposed, and how?

- #2

symbolipoint

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3s^2

\]

[/tex] and [tex] \[

3s + 1

\]

[/tex]

- #3

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[tex]\frac{A}{3s^{2}}[/tex] + [tex]\frac{B}{3s+1}[/tex]

which would equal out to

[tex]\frac{A}{3s^{2}}[/tex] + [tex]\frac{B}{3s+1}[/tex] = [tex]\frac{7}{3s^{2}(3s+1)}[/tex]

then

7 = [tex]A(3s+1)[/tex] + [tex]3Bs^{2}[/tex]

--which is where I'm stuck, because eliminating "s" leaves me with A = 7, which doesn't make sense if I want to separate the fraction.

- #4

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- #5

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[tex]\frac{7}{3s^{2}(3s+1)}=\frac{As+B}{3s^{2}}+\frac{C}{3s+1}[/tex]

A=-21

B=7

C=21

A=-21

B=7

C=21

Last edited:

- #6

symbolipoint

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[tex] \[

\begin{array}{l}

\frac{a}{{3s^2 }} + \frac{b}{{3s + 1}} = performSteps = \frac{{(3s + 1)a + 3s^2 b}}{{3s^2 (3s + 1)}} \\

fromOriginalDeno\min ator,\;0s + 7 = 3s^2 b + 3sa + a \\

Answer:\;\;a = 7\quad b = \frac{{ - 7}}{s} \\

\end{array}

\]

[/tex]

- #7

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Hmm, I'll have to try to this in a second.

--wow the method gao xiong did works but I don't recall seeing this in my initial searches through the textbook or online.

Nice method, thanks.

Big-T, I don't understand where the a/s would come from if the highest polynomial is 9s^3 (or the highest term is cubic) when you combine the denominator.

--wow the method gao xiong did works but I don't recall seeing this in my initial searches through the textbook or online.

Nice method, thanks.

Big-T, I don't understand where the a/s would come from if the highest polynomial is 9s^3 (or the highest term is cubic) when you combine the denominator.

Last edited:

- #8

symbolipoint

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3s^2

\] [/tex]

is an irreducible quadratic expression and gave a suitable linear binomial expression with it.

- #9

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I have one more problem.

Would this:[tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]

simplify to something like

[tex]\frac{Ax+B}{x^{2}+13x+38} + \frac{C}{x}[/tex] = [tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]

- #10

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I have one more problem.

Would this:[tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]

simplify to something like

[tex]\frac{Ax+B}{x^{2}+13x+38} + \frac{C}{x}[/tex] = [tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]

well first look if [tex]\{x^{2}+13x+38}[/tex] has any roots over reals so the problem could get more simplified, since you could express that to some form of (x+-a)(x+-b) where a,b are the real roots.

then it would be of the form

[tex]\frac{A}{x+-a} + \frac{B}{x+-b}+\frac{C}{x}[/tex]

Last edited:

- #11

HallsofIvy

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[tex]\frac{7}{3s^{2}(3s+1)}[/tex]

Can this be decomposed, and how?

3s^2

\]

[/tex] and [tex] \[

3s + 1

\]

[/tex]

With that "s

[tex]\frac{7}{3s^2(3s+1)}= \frac{A}{s}+ \frac{B}{s^2}+ \frac{C}{3s+1}[/tex]

Multiplying through by the common denominator, [itex]7= As(3s+1)+ B(3s+1)+ Cs^2[/itex]. Taking s= 0, 7= B. Taking s= -1/3, 7= C/9 so C= 63. Finally, taking s= 1, 7= 4A+ 4B+ C= 4A+ 28+ 63. 4A= 7- 91= -84, A= -21.

- #12

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I don't doubt what you're saying, but how do [you] rationalize

a 1/s and 1/s^2 from one 1/s^2. I've never seen that before.

a 1/s and 1/s^2 from one 1/s^2. I've never seen that before.

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- #13

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With that "s^{2}", you are going to need both 1/s and 1/s^{2}.

[tex]\frac{7}{3s^2(3s+1)}= \frac{A}{s}+ \frac{B}{s^2}+ \frac{C}{3s+1}[/tex]

Multiplying through by the common denominator, [itex]7= As(3s+1)+ B(3s+1)+ Cs^2[/itex]. Taking s= 0, 7= B. Taking s= -1/3, 7= C/9 so C= 63. Finally, taking s= 1, 7= 4A+ 4B+ C= 4A+ 28+ 63. 4A= 7- 91= -84, A= -21.

You lost the 7 in the original numerator. The correct values are A = -3, B = 1, C = 9.

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