Further mathematics:partial fractions?

I’ll go through one of these.

1) (3x+2) / [(x+1)(x^2 +x +2)]

First note that x^2 +x +2 cannot be factored. (If it could be factored, then that would be the first step).

(3x+2) / [(x+1)(x^2 +x +2)] = A/(x+1) + (Bx+C) / (x^2 +x +2)

Bx+C is used on the second term because x^2 +x +2 is of the second degree, so the term on top must be one degree lower.

Now multiply both sides by (x+1)(x^2 +x +2)

3x + 2 = A(x^2 +x +2) + (Bx+C)(x+1)

This is true for any value of x.

If you let x = -1, you can find A.

3(-1) + 2 = A((-1)^2 +(-1) +2) + (B(-1) +C)(1-1)

-3 + 2 = A(1-1+2) + 0

-1 = A(2)

A = -1/2

Replace A with -1/2

3x + 2 = (-1/2)(x^2 +x +2) + (Bx+C)(x+1)

Now let’s try x = 0

3(0) + 2 = (-1/2)(0^2 +0 +2) + (B(0)+C)(0+1)

2 = (-1/2)(2) + C(1)

2 = -1 + C

C = 3

Replace C with 3.

3x + 2 = (-1/2)(x^2 +x +2) + (Bx+3)(x+1)

Now pick a third value for x (one that you haven’t already used). Plug it in and solve for C.

After you’ve found C, go back to the equation

A/(x+1) + (Bx+C) / (x^2 +x +2)

and put in the values you just found for A, B, and C.

If numerator is 2x + a million and denominator is 3x – a million, then fraction is in least difficult sort. Neither might nicely be factored. the sole thank you to simplify fractions is with the aid of canceling aspects. of course x(2 + a million) = 3x, no longer 2x + a million; x(3 + a million) = 4x, no longer 3x – a million.