Friday, March 10, 2017

`(3x + 1)/(2x^3 + 3x^2)` Write the partial fraction decomposition of the rational expression.

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


To decompose this to partial fractions, factor the denominator.


`2x^3 + 3x^2 = x^2(2x + 3)`


Then, write a fraction for each factor. For the repeated factor x, form a partial fraction for each exponent x from 1 to 2. And assign a variable for each numerators. 


`A/x`   ,   `B/x^2`   and    `C/(2x + 3)`


Add these three fractions and set it equal to the given fraction.


`(3x+1)/(x^2(2x+3)) = A/x +...

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


To decompose this to partial fractions, factor the denominator.


`2x^3 + 3x^2 = x^2(2x + 3)`


Then, write a fraction for each factor. For the repeated factor x, form a partial fraction for each exponent x from 1 to 2. And assign a variable for each numerators. 


`A/x`   ,   `B/x^2`   and    `C/(2x + 3)`


Add these three fractions and set it equal to the given fraction.


`(3x+1)/(x^2(2x+3)) = A/x + B/x^2+C/(2x+3)`


To solve for the values of A, B and C, eliminate the fractions in the equation. So multiply both sides by the LCD.


`x^2(2x+3) * (3x+1)/(x^2(2x+3)) = (A/x+B/x^2+C/(2x+3)) *x^2(2x+3)`


`3x+1=Ax(2x+3) + B(2x +3) + Cx^2`


Then, plug-in the roots of the factors.


For the factor x^2, its root is x=0


`3*0+1=A*0(2*0+3) + B(2*0+3)+C*0^2`


`1=3B`


`1/3=B`


For the factor (2x + 3), its root is x=-3/2.


`3(-3/2)+1=A(-3/2)(2(-3/2)+3)+B(2(-3/2)+3)+C(-3/2)^2`


`-7/2=A(-3/2)(0)+ B(0)+9/4C`


`-7/2=9/4C`


`-14/9=C`


To get the value of A, assign any value to x, and plug-in the values of B and C to:


`3x+1=Ax(2x+3) + B(2x +3) + Cx^2`


Let x = 1.


`3*1+1=A*1(2*1+3) + 1/3(2*1+3)+(-14/9)*1^2`


`4=5A + 5/3-14/9`


`4=5A+1/9`


`4-1/9=5A`


`35/9=5A`


`7/9=A`


So the partial fraction decomposition of the given rational expression is:


`(7/9)/x + (1/3)/x^2 + (-14/9)/(2x+3)`


This simplifies to:


`7/(9x) + 1/(3x^2)-14/(9(2x+3))`



To check, express them with same denominators.


`7/(9x) + 1/(3x^2)-14/(9(2x+3))=7/(9x) * (x(2x+3))/(x(2x+3)) + 1/(3x^2)*(3(2x+3))/(3(2x+3)) -14/(9(2x+3))*x^2/x^2`


`= (14x^2+21x)/(9x^2(2x+3)) + (6x+9)/(9x^2(2x+3)) - (14x^2)/(9x^2(2x+3))`


Now that they have same denominators, let's proceed to add/subtract them.


`= (14x^2+21x+6x+9-14x^2)/(9x^2(2x+3)) = (27x+9)/(9x^2(2x+3)) = (9(3x+1))/(9x^2(2x+3))= (3x+1)/(x^2(2x+3))`


`= (3x+1)/(2x^3+3x^2)`



Therefore,  `(3x+1)/(2x^3+3x^2)=7/(9x) + 1/(3x^2)-14/(9(2x+3))` .


No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Exposition A "decidedly amused" Bobby Gillian leaves the offices of Tolman & Sharp where he is given an envelope containing $1...