*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **ˆ’ x + 1 + (Bx + C )x or
3 = (A + B )x2 + (âˆ’A + C )x + A. We get A+B = 0
âˆ’A + C = 0 A=3
From A = 3 and A + B = 0, we get B = âˆ’3. From âˆ’A + C = 0, we get C = A = 3. We get x3
4. Since 4x3
x2 âˆ’ 2 3
3
3 âˆ’ 3x
= +2
2+x
âˆ’x
x x âˆ’x+1 isnâ€™t proper, we use long division and we get a quotient of 4x with a remainder of
3 8x
8x
8x. That is, x4x 2 = 4x + x2 âˆ’2 so we now work on resolving x2 âˆ’2 into partial fractions. The
2âˆ’
quadratic x2 âˆ’ 2, though it doesnâ€™t factor nicely, is, nevertheless, reducible. Solving x2 âˆ’ 2 = 0 8.6 Partial Fraction Decomposition 527 âˆš
be
gives us x = Â± 2, and each of these zeros mustâˆš of multiplicity one since Theorem 3.14
âˆš
enables us to now factor x2 âˆ’ 2 = x âˆ’ 2 x + 2 . Hence,
8x
8x
A
B
âˆš
âˆš=
âˆš+
âˆš
=
x2 âˆ’ 2
xâˆ’ 2 x+ 2
xâˆ’ 2 x+ 2
âˆš
âˆš
âˆš
Clearing fractions, we get 8x = A x + 2 + B x âˆ’ 2 or 8x = (A + B )x + (A âˆ’ B ) 2.
We get the system
A+B = 8
âˆš
(A âˆ’ B ) 2 = 0
âˆš
From (A âˆ’ B ) 2 = 0, we get A = B , which, when subst...

View
Full
Document