Question

Given 2x^3-4x^2-15x+5 divided by x^2-2x-8 as the function of f(x)

Find the partial fractional resolution of f(x)

Answer 1

First, factorize the denominator x^2-2x-8:

x^2 - 2x - 8 = (x-4)(x+2)

Now, express f(x) using partial fractions:

f(x) = (2x^3 - 4x^2 - 15x + 5) / (x^2 - 2x - 8)
f(x) = (2x^3 - 4x^2 - 15x + 5) / ((x-4)(x+2))

Now, we express f(x) in the form of partial fractions:

f(x) = A / (x-4) + B / (x+2)

Now, we solve for A and B by multiplying both sides by the denominator of the original expression:

2x^3 - 4x^2 - 15x + 5 = A(x+2) + B(x-4)

Expand and simplify:

2x^3 - 4x^2 - 15x + 5 = Ax + 2A + Bx - 4B

Group like terms:

2x^3 - 4x^2 - 15x + 5 = (A + B)x + 2A - 4B

Now, we have a system of equations:

A + B = 2
2A - 4B = -15

Solving this system of equations, you will find:

A = 3
B = -1

Therefore, the partial fraction resolution of f(x) is:

f(x) = 3 / (x-4) - 1 / (x+2)