p3 help!

CKS

Express as partial fractions:

(-2x^2 + 13) / (2x+1)(x^2 + 2x + 7)

Reply 1

El Stevo

-(2x^2)+13 / (2x+1)((x^2)+2x+7) = A/(2x+1) + B/((x^2)+2x+7)

-(2x^2)+13 = A((x^2)+2x+7) + B(2x+1)

solve to find A and B from there, first using x=(-1/2) to find A...

Reply 2

CKS

Thanks! How about:

( x^3+ x^2 - 2x + 4 ) / (x^2 - 4)

Reply 3

El Stevo

((x^3)+(x^2)-2x+4) / ((x^2)-4) is the same as ((x^3)+(x^2)-2x+4) / ((x+2)(x-2))

(x^3)+(x^2)-2x+4 / (x+2)(x-2) = A/(x+2) + B/(x-2)

(x^3)+(x^2)-2x+4 = A(x-2) + B(x+2)

then solve to find A using x=-2, solve to find B using x=2...

Reply 4

username9816

CKS

Express (x^3 + x^2 - 2x + 4) / (x^2 - 4) As Partial Fractions.

(x^3 + x^2 - 2x + 4) / (x^2 - 4)
= (x^3 + x^2 - 2x + 4) / [(x + 2)(x - 2)]
= A/(x + 2) + B/(x - 2)
= [A(x - 2) + B(x + 2)] / [(x + 2)(x - 2)]

Where A and B are constants to be determined.

Equating the numerators gives:
A(x - 2) + B(x + 2) = x^3 + x^2 - 2x + 4

Let x = -2:
---> -4A = -8 + 4 + 4 + 4 = 4
---> A = -1

Let x = 2:
---> 4B = 8 + 4 - 4 + 4 = 12
---> B = 3

Hence: (x^3 + x^2 - 2x + 4) / (x^2 - 4) = 3/(x - 2) - 1/(x + 2)