Resolve the following into partial fractions (a)4(x-4)/x²-3x-3 , (b)x²+9x+8/x²+x-6 ,(c) 5x²-2x-19/(x-1)²(x+3)?

(a) To resolve the fraction 4(x-4)/(x²-3x-3) into partial fractions, we need to factor the denominator first.

The denominator x²-3x-3 can be factored as (x-3)(x+1).

Now, we can write the fraction as:

4(x-4)/((x-3)(x+1))

To resolve this into partial fractions, we need to find the constants A and B such that:

4(x-4)/((x-3)(x+1)) = A/(x-3) + B/(x+1)

To find A and B, we can multiply both sides of the equation by (x-3)(x+1):

4(x-4) = A(x+1) + B(x-3)

Expanding both sides:

4x - 16 = Ax + A + Bx - 3B

Now, we can equate the coefficients of x on both sides:

4 = A + B
And the constant terms on both sides:

-16 = A - 3B

Solving these two equations simultaneously, we find:

A = -4
B = 8

Therefore, the partial fraction decomposition of 4(x-4)/(x²-3x-3) is:

4(x-4)/((x-3)(x+1)) = -4/(x-3) + 8/(x+1)

(b) To resolve the fraction (x²+9x+8)/(x²+x-6) into partial fractions, we need to factor the denominator first.

The denominator x²+x-6 can be factored as (x-2)(x+3).

Now, we can write the fraction as:

(x²+9x+8)/((x-2)(x+3))

To resolve this into partial fractions, we need to find the constants A and B such that:

(x²+9x+8)/((x-2)(x+3)) = A/(x-2) + B/(x+3)

To find A and B, we can multiply both sides of the equation by (x-2)(x+3):

(x²+9x+8) = A(x+3) + B(x-2)

Expanding both sides:

x² + 9x + 8 = Ax + 3A + Bx - 2B

Now, we can equate the coefficients of x on both sides:

1 = A + B
And the constant terms on both sides:

8 = 3A - 2B

Solving these two equations simultaneously, we find:

A = 5
B = -4

Therefore, the partial fraction decomposition of (x²+9x+8)/(x²+x-6) is:

(x²+9x+8)/((x-2)(x+3)) = 5/(x-2) - 4/(x+3)

(c) To resolve the fraction (5x²-2x-19)/((x-1)²(x+3)) into partial fractions, we need to factor the denominator first.

The denominator (x-1)²(x+3) is already factored.

Now, we can write the fraction as:

(5x²-2x-19)/((x-1)²(x+3))

To resolve this into partial fractions, we need to find the constants A, B, and C such that:

(5x²-2x-19)/((x-1)²(x+3)) = A/(x-1) + B/(x-1)² + C/(x+3)

To find A, B, and C, we can multiply both sides of the equation by (x-1)²(x+3):

(5x²-2x-19) = A(x-1)(x+3) + B(x+3) + C(x-1)²

Expanding both sides:

5x² - 2x - 19 = A(x²+2x-3) + B(x+3) + C(x²-2x+1)

Now, we can equate the coefficients of x² on both sides:

5 = A + C
And the coefficients of x on both sides:

-2 = 2A - 2C + B
And the constant terms on both sides:

-19 = -3A + B + C

Solving these three equations simultaneously, we find:

A = 2
B = -1
C = 3

Therefore, the partial fraction decomposition of (5x²-2x-19)/((x-1)²(x+3)) is:

(5x²-2x-19)/((x-1)²(x+3)) = 2/(x-1) + (-1)/(x-1)² + 3/(x