Question

resolve into impartial fractions:
2x^2 + 7x + 23/(x-1)(x+3)^2

Answer 1

\[\frac{2x^{2}+7x+23}{(x-1)(x+3)^{2}}\]
not much you can do with that ??

Answer 2

evidently there is a lot that you can do with it because it does have an answer.

Answer 3

what are you asked to do? simplify

Answer 4

resolve into partial fractions?

Answer 5

\[\frac{2x^{2}+7x+23}{(x-1)(x+3)^{2}}=\frac A{x-1}+\frac B{x+3}+\frac C{(x+3)^2}+\frac D{(x+3)^3}\]

Answer 6

\[\frac{2x^{2}+7x+23}{(x-1)(x+3)^{2}}=\frac A{x-1}+\frac B{x+3}+\frac C{(x+3)^2}+\frac D{(x+3)^3}\]which implies that\[A(x+3)^3+b(x-1)(x+3)^2+C(x-1)(x+3)+D(x-1)=2x^2+7x+23\]now you gotta expand the left side and compare coefficients, which is going to suck

Answer 7

to clarify:\[A(x+3)^3+B(x-1)(x+3)^2+C(x-1)(x+3)+D(x-1)\]\[=2x^2+7x+23\]now expand the left and collect like terms

Answer 8

why do you need the (x+3)^3 term?
\[\rightarrow \frac{A}{x-1}+\frac{B}{(x+3)^{2}} = \frac{2}{x-1}-\frac{5}{(x+3)^{2}}\]

Answer 9

yeah, why do you need the (x+3)^3 term?

Answer 10

yeah, what they said.?

Answer 11

You don't need it,
\[\begin{array}{l} \frac{23+7 x+2 x^2}{(-1+x) (3+x)^2}=\frac{A}{x-1}+\frac{B}{x+3}+\frac{C}{(x+3)^2} \\ \text{Multiply both sides by }(x-1) (x+3)^2\text{ and simplify:} \\ 2 x^2+7 x+23=A (x+3)^2+(x-1) \left(C+B (x+3)\right) \\ \text{expand and collect in terms of powers of }x \\ 2 x^2+7 x+23=9 A-3 B-C+\left(A+B\right) x^2+\left(6 A+2 B+C\right) x \\ \text{equate coefficients on both sides, giving }3\text{ equations in }3\text{ unknowns} \\ 23=9 A-3 B-C \\ 7=6A+2 B+C \\ 2=A+B \\ \text{Solve using matrices or substitution}\\\end{array}\]

Answer 12

oh I misread, sorry

Answer 13

that's okay Turning Test.
Thanks for your help everyone!