Factorise $6x^{3}-23x^{2}+29x-12$.
(A) Let $p(x) = 6x^{3}-23x^{2}+29x-12$.
By the Factor Theorem,we test values to find a root.
For $x = 1$,$p(1) = 6(1)^{3}-23(1)^{2}+29(1)-12 = 6-23+29-12 = 0$.
Since $p(1) = 0$,$(x-1)$ is a factor.
Dividing $6x^{3}-23x^{2}+29x-12$ by $(x-1)$,we get the quotient $6x^{2}-17x+12$.
Now,factorise the quadratic $6x^{2}-17x+12$:
$6x^{2}-9x-8x+12 = 3x(2x-3)-4(2x-3) = (3x-4)(2x-3)$.
Thus,the factors are $(x-1)(2x-3)(3x-4)$.
By the Factor Theorem,we test values to find a root.
For $x = 1$,$p(1) = 6(1)^{3}-23(1)^{2}+29(1)-12 = 6-23+29-12 = 0$.
Since $p(1) = 0$,$(x-1)$ is a factor.
Dividing $6x^{3}-23x^{2}+29x-12$ by $(x-1)$,we get the quotient $6x^{2}-17x+12$.
Now,factorise the quadratic $6x^{2}-17x+12$:
$6x^{2}-9x-8x+12 = 3x(2x-3)-4(2x-3) = (3x-4)(2x-3)$.
Thus,the factors are $(x-1)(2x-3)(3x-4)$.
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