Copyright

Find the zeros of 10 ( s + 1 ) ( s 2 + 4 s + 2 ) ( s 2 ( s + 4 ) ( s + 3 ) using partial...

Question:

Find the zeros of {eq}\frac{10(s+1)(s^2+4s+2)}{ (s^2(s+4)(s+3)}{/eq} using partial fractions?

Partial Fractions

Suppose we want to represent a complicated fraction as a sum of simpler fractions then we make use of Partial fraction expansion.

Partial fraction decomposition is the process where we add or subtract simpler rational expressions to get the original expression which is a much complicated rational expression.

Answer and Explanation:

{eq}\displaystyle \frac{10\left(s+1\right)\left(s^2+4s+2\right)}{\left(s^2\left(s+4\right)\left(s+3\right)\right)} \\ \displaystyle \mathrm{Expand}\:10\left(s+1\right)\left(s^2+4s+2\right) \\ \displaystyle \frac{10s^3+50s^2+60s+20}{s^2\left(s+4\right)\left(s+3\right)} \\ \displaystyle \mathrm{Create\:the\:partial\:fraction\:template\:using\:the\:denominator\:}s^2\left(s+4\right)\left(s+3\right) \\ \displaystyle \frac{10s^3+50s^2+60s+20}{s^2\left(s+4\right)\left(s+3\right)}=\frac{a_0}{s}+\frac{a_1}{s^2}+\frac{a_2}{s+4}+\frac{a_3}{s+3} \\ \displaystyle Multiply\:equation\:by\:the\:denominator \\ \displaystyle \frac{s^2\left(10s^3+50s^2+60s+20\right)\left(s+4\right)\left(s+3\right)}{s^2\left(s+4\right)\left(s+3\right)}=\frac{a_0s^2\left(s+4\right)\left(s+3\right)}{s}+\frac{a_1s^2\left(s+4\right)\left(s+3\right)}{s^2}+\frac{a_2s^2\left(s+4\right)\left(s+3\right)}{s+4}+\frac{a_3s^2\left(s+4\right)\left(s+3\right)}{s+3} \\ \displaystyle 10s^3+50s^2+60s+20=a_0s\left(s+4\right)\left(s+3\right)+a_1\left(s+4\right)\left(s+3\right)+a_2s^2\left(s+3\right)+a_3s^2\left(s+4\right) {/eq}

{eq}\displaystyle \mathrm{Solve\:the\:unknown\:parameters\:by\:plugging\:the\:real\:roots\:of\:the\:denominator:}\:0,\:-4,\:-3 {/eq}

{eq}\displaystyle \mathrm{Plug\:in}\:s=0\:\mathrm{into\:the\:equation} \\ \displaystyle 10\cdot \:0^3+50\cdot \:0^2+60\cdot \:0+20=a_0\cdot \:0\cdot \left(0+4\right)\left(0+3\right)+a_1\left(0+4\right)\left(0+3\right)+a_2\cdot \:0^2\left(0+3\right) \\ \displaystyle 20=12a_1 \\ \displaystyle \frac{12a_1}{12}=\frac{20}{12} \\ \displaystyle a_1=\frac{5}{3} {/eq}

{eq}\displaystyle \mathrm{Plug\:in}\:s=-4\:\mathrm{into\:the\:equation} \\ \displaystyle 10\left(-4\right)^3+50\left(-4\right)^2+60\left(-4\right)+20=a_0\left(-4\right)\left(\left(-4\right)+4\right)\left(\left(-4\right)+3\right)+a_1\left(\left(-4\right)+4\right)\left(\left(-4\right)+3\right)+10\left(-4\right)^3+50\left(-4\right)^2+60\left(-4\right)+20=a_0\left(-4\right)\left(\left(-4\right)+4\right)\left(\left(-4\right)+3\right)+a_1\left(\left(-4\right)+4\right)\left(\left(-4\right)+3\right)+a_2\left(-4\right) \\ \displaystyle \mathrm{Expand} \\ \displaystyle -60=-16a_2 \\ \displaystyle a_2=\frac{15}{4} {/eq}

{eq}\displaystyle \mathrm{Plug\:in}\:s=-3\:\mathrm{into\:the\:equation} \\ \displaystyle 10\left(-3\right)^3+50\left(-3\right)^2+60\left(-3\right)+20=a_0\left(-3\right)\left(\left(-3\right)+4\right)\left(\left(-3\right)+3\right)+a_1\left(\left(-3\right)+4\right)\left(\left(-3\right)+3\right)\left(\left(-3\right)+4\right)\left(\left(-3\right)+3\right)+a_2\left(-3\right)^2\left(\left(-3\right)+3\right)+a_3\left(-3\right)^2\left(\left(-3\right)+4\right) \\ \displaystyle Expand \\ \displaystyle 20=9a_3 \\ \displaystyle a_3=\frac{20}{9} {/eq}

Thus

{eq}\displaystyle a_1=\frac{5}{3},\:a_2=\frac{15}{4},\:a_3=\frac{20}{9} {/eq}

{eq}\displaystyle \mathrm{Plug\:in\:the\:solutions\:to\:the\:known\:parameters} \\ \displaystyle 10s^3+50s^2+60s+20=a_0s\left(s+4\right)\left(s+3\right)+\frac{5}{3}\left(s+4\right)\left(s+3\right)+\frac{15}{4}s^2\left(s+3\right)+\frac{20}{9}s^2\left(s+4\right) \\ \displaystyle 10s^3+50s^2+60s+20=a_0s^3+7a_0s^2+12a_0s+\frac{215s^3}{36}+\frac{785s^2}{36}+\frac{35s}{3}+20 \\ \displaystyle 10s^3+50s^2+60s+20=s^3\left(a_0+\frac{215}{36}\right)+s^2\left(7a_0+\frac{785}{36}\right)+s\left(12a_0+\frac{35}{3}\right)+20 \\ \displaystyle \mathrm{Solve}\:12a_0+\frac{35}{3}=60\:\mathrm{for}\:a_0 \\ \displaystyle a_0=\frac{145}{36} \\ \displaystyle \frac{\frac{145}{36}}{s}+\frac{\frac{5}{3}}{s^2}+\frac{\frac{15}{4}}{s+4}+\frac{\frac{20}{9}}{s+3} \\ Simplify \\ \displaystyle \frac{5}{3s^2}+\frac{20}{9\left(s+3\right)}+\frac{15}{4\left(s+4\right)}+\frac{145}{36s} {/eq}

Therefore

{eq}\displaystyle \frac{10\left(s+1\right)\left(s^2+4s+2\right)}{\left(s^2\left(s+4\right)\left(s+3\right)\right)}=\frac{5}{3s^2}+\frac{20}{9\left(s+3\right)}+\frac{15}{4\left(s+4\right)}+\frac{145}{36s} {/eq}

Now to find zeros we have to equate above expression to zero.

{eq}\displaystyle \frac{5}{3s^2}+\frac{20}{9\left(s+3\right)}+\frac{15}{4\left(s+4\right)}+\frac{145}{36s}=0 \\ \displaystyle \mathrm{Find\:Least\:Common\:Multiplier\:of\:}3s^2,\:9\left(s+3\right),\:4\left(s+4\right),\:36s \\ =\displaystyle 36s^2\left(s+3\right)\left(s+4\right) \\ \displaystyle \mathrm{Multiply\:by\:LCM=}36s^2\left(s+3\right)\left(s+4\right) \\ \displaystyle \frac{5}{3s^2}\cdot \:36s^2\left(s+3\right)\left(s+4\right)+\frac{20}{9\left(s+3\right)}\cdot \:36s^2\left(s+3\right)\left(s+4\right)+\frac{15}{4\left(s+4\right)}\cdot \:36s^2\left(s+3\right)\left(s+4\right)+\frac{145}{36s}\cdot \:36s^2\left(s+3\right)\left(s+4\right)=0\cdot \:36s^2\left(s+3\right)\left(s+4\right) \\ \displaystyle \mathrm{Simplify} \\ \displaystyle 60\left(s+3\right)\left(s+4\right)+80s^2\left(s+4\right)+135s^2\left(s+3\right)+145s\left(s+3\right)\left(s+4\right)=0 \\ \displaystyle Solve \\ \displaystyle 360\left(s+1\right)\left(s^2+4s+2\right)=0 \\ \displaystyle \mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right) \\ \displaystyle s+1=0:\quad s=-1 \\ \displaystyle s^2+4s+2=0 \\ \displaystyle \mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:} x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \displaystyle s=\frac{-4+\sqrt{4^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}= -2+\sqrt{2} \\ \displaystyle s=\frac{-4-\sqrt{4^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}= -2-\sqrt{2} \\ \displaystyle The\:solutions\:are \\ \displaystyle s=-1,\:s=-2+\sqrt{2},\:s=-2-\sqrt{2} \\ \displaystyle \mathrm{Verify\:Solutions} \\ \displaystyle Find\:undefined\:\left(singularity\right)\:points \\ \displaystyle \mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{5}{3s^2}+\frac{20}{9\left(s+3\right)}+\frac{15}{4\left(s+4\right)}+\frac{145}{36s}\mathrm{\:and\:compare\:to\:zero} \\ \displaystyle \mathrm{The\:following\:points\:are\:undefined} \\ \displaystyle s=0,\:s=-3,\:s=-4 \\ \displaystyle \mathrm{Combine\:undefined\:points\:with\:solutions:} \\ \displaystyle Thus \\ \displaystyle s=-1,\:s=-2+\sqrt{2},\:s=-2-\sqrt{2} {/eq}


Learn more about this topic:

Loading...
Partial Fractions: Rules, Formula & Examples

from High School Algebra I: Help and Review

Chapter 3 / Lesson 26
25K

Related to this Question

Explore our homework questions and answers library