Question 1.41: Noting that x³ + 2x² — 11x — 52 = (x — 4)(x² + 6x + 13), exp......

The denominator has already been factorized. The linear factor, x — 4, produces a partial fraction of the form \frac{A}{x-4}.

The quadratic factor, x² + 6x + 13, will not factorize further into two linear factors.
Thus this factor generates a partial fraction of the form \frac{B x+C}{x^2+6 x+13}. Hence

Multiplying by {x — 4) and {x² + 6x + 13) produces

3 x^2+11 x+14=A\left(x^2+6 x+13\right)+(B x+C)(x-4)          (1.11)

The constants A, B and C can now be found.
Putting x = 4 into Equation (1.11) gives

106 = 4(53)
A = 2

Equating the coefficients of x² gives

3 = A + B
B = 1

Equating the constant term on both sides gives

14 = A(13) — 4C
C = 3

Hence