Reduce to a partial fraction?

Question

15 +5x + 5x^2 - 4x^3
x^2 (x^2 +5)

Yes the denominator is

a) x+2(3X^2)=3x^2+6X^2
b) x+2(-4x) =4x^2 -8x
c) x+2 (3) = 3x +6
d) x+2( -2) = - 2
(x-2)

3x^3 +6x^2 -4x^2-8x+3x+6-2= 3x^3 + 2x^2 - 5x+4

Answer 1

I assume that you mean to write (15 + 5x + 5x² - 4x³)/(x²(x² + 5)). In this case, the partial fraction deomposition is as follows:

(15 + 5x + 5x² - 4x³)/(x²(x² + 5)) = A/x² + B/x + (Cx+D)/(x²+5)

-4x³ + 5x² + 5x + 15 = A(x²+5) + Bx(x²+5) + (Cx+D)x²

-4x³ + 5x² + 5x + 15 = (B+C)x³ + (A+D)x² + 5Bx + 5A

5A = 15
5B = 5
A+D = 5
B+C = -4

A = 3
B = 1
C = -5
D = 2

(15 + 5x + 5x² - 4x³)/(x²(x² + 5)) = 3/x² + 1/x + (-5x+2)/(x² + 5)

Answer 2

15 +5x + 5x² - 4x^3 / x² (x² +5)

= A/x + B/x² + (Cx + D)/ (x²)(x² + 5)
= [A(x)(x² + 5) + (x² + 5) + (Cx + D) (x²)] / [(x²)(x² + 5)]
= [A(x^3 + 5x) + Bx² + 5B + Cx^3 + Dx²] / [(x²)(x² + 5)]
≡Ax^3 + 5Ax + Bx² + 5B + Cx^3 + Dx²

15 ≡ 5B → B = 3.

5x ≡ x(5A) → A = 1.

5x² ≡ x²(B + D)
→ 5 = B + D
→ 5 = 3 + D
D = 2.

-4x^3 = x^3(A + C)
→ -4 = A + C
→ -4 = 1 + C
C = - 5.

15 +5x + 5x² - 4x^3 / x² (x² +5) = 1/x + 3/x² + (-5x + 2)/ x²(x² + 5)

Answer 3

Partial fractions do not apply to polynomials, only to rational functions. Do you have a denominator around there somewhere?

Answer 4

Do you mean:
(a) to be factorized?
(b) 1/x^2/(x^2+5) = (ax+b)/x^2 +(cx+d)/(x^2+5), where a, b, c, d to be found ???
not clear! revise your question.