How do you express (6x^2+8x+30)/(x^3-27)6x2+8x+30x3−27 in partial fractions?
1 Answer
Explanation:
Note that:
x^3-27 = (x-3)(x^2+3x+9)x3−27=(x−3)(x2+3x+9)
Sticking with Real coefficients, we are looking for a partial fraction decomposition of the form:
(6x^2+8x+30)/(x^3-27) = A/(x-3) + (Bx+C)/(x^2+3x+9)6x2+8x+30x3−27=Ax−3+Bx+Cx2+3x+9
color(white)((6x^2+8x+30)/(x^3-27)) = (A(x^2+3x+9)+(Bx+C)(x-3))/(x^3-27)6x2+8x+30x3−27=A(x2+3x+9)+(Bx+C)(x−3)x3−27
color(white)((6x^2+8x+30)/(x^3-27)) = ((A+B)x^2+(3A-3B+C)x-3C)/(x^3-27)6x2+8x+30x3−27=(A+B)x2+(3A−3B+C)x−3Cx3−27
So equating coefficients, we get the following system of linear equations:
{ (A+B=6), (3A-3B+C=8), (-3C=30) :}
From the third equation we find:
C = -10
Substituting this value of
3A-3B-10=8
Hence:
3A-3B=18
So:
A-B=6
Combining this with the first equation, we find:
A=6
B=0
So:
(6x^2+8x+30)/(x^3-27) = 6/(x-3) - 10/(x^2+3x+9)
