Factorise $8x^{3} - 26x^{2} + 13x + 5$.

(N/A) Let $p(x) = 8x^{3} - 26x^{2} + 13x + 5$.
By the factor theorem,we test values to find a root. Let $x = 1$:
$p(1) = 8(1)^{3} - 26(1)^{2} + 13(1) + 5 = 8 - 26 + 13 + 5 = 0$.
Since $p(1) = 0$,$(x - 1)$ is a factor.
Now,divide $8x^{3} - 26x^{2} + 13x + 5$ by $(x - 1)$ using long division or synthetic division:
$8x^{3} - 26x^{2} + 13x + 5 = (x - 1)(8x^{2} - 18x - 5)$.
Next,factorize the quadratic $8x^{2} - 18x - 5$ by splitting the middle term:
$8x^{2} - 20x + 2x - 5 = 4x(2x - 5) + 1(2x - 5) = (4x + 1)(2x - 5)$.
Thus,the factors are $(x - 1)(4x + 1)(2x - 5)$.