Question

Resolve into partial fractions
6x2-x+1÷x3+x2+x+1

Answer 1

Answer:How do you resolve x⁴/x²-3x+2 into partial fractions?

Solution: x⁴/x²-3x+2

= x^2 + 3x + 7 + [(27x-14)/(x^2–3x+2)]

= x^2 + 3x + 7 + [(15x-14)/(x-2)(x-1)]

= x^2 + 3x + 7 + A/(x-2) + B/(x-1)

Let us consider the factors in bold type:

[(15x-14)/(x-2)(x-1)] = A/(x-2) + B/(x-1) or

A(x-1) + B(x-2) = 15x - 14, or

x(A+B) - A - 2B = 15x - 14, or

A+B = 15 …(1)

A+2B = 14 …(2)

Subtract (1) from (2): B = -1.

From (1): A = 15+1 = 16.

Therefore, x⁴/x²-3x+2 = x^2 + 3x + 7 + 16/(x-2) - 1/(x-1). Answer.

Check: x^2 + 3x + 7 + 16/(x-2) - 1/(x-1)

= x^2 + 3x + 7 + [16(x-1) - 1(x-2)] / [x²-3x+2]

= x^2 + 3x + 7 + [15x-14]/[x²-3x+2]

= [[x^2 + 3x + 7]*[x²-3x+2] + [15x-14]]/[x²-3x+2]

= [x^4–3x^3+2x^2+3x^3–9x^2+6x+7x^2–21x+14+15x+18]/[x²-3x+2]

= [x^4+x^3(-3+3)+x^2(2–9+7)+x(6–21+15)+(14–14)]/[x²-3x+2]

= ]x^4 + 0 + 0 + 0 + 0]/[x²-3x+2]

= x^4/[x²-3x+2]. Correct.

Step-by-step explanation: