To factorize:

`t^3-13t^2+46t-48`

First arrange the expression so that it can simplify:

`= t^3 -13t^2+46t-27-21` (Note that -27-21=-48)

Now rearrange the order so that it can be factorized more readily:

`t^3 -27 -13t^2 +46t-21`

`` Now we can factorize:

`(t^3-27) - (13t^2 -46t +21)` (Note the negative (-) symbol)

`(t-3)(t^2+3t+9)-(13t-7)(t-3)`

Note we have a difference of two cubes (`t^3 and 3^3)` and then a trinomial the factors of which (`1 times 13)` (first term) and (`7 times 3 =21` )(third term) render a middle term of `-46`

Factorize by taking out the common factor:

`(t-3)[(t^2+3t +9) - (13t-7)(1)]`

Remove the brackets in side :

`(t-3)(t^2+3t+9-13t+7)`

Simplify by adding like terms:

`(t-3)(t^2-10t+16)`

Factorize using the factors of the first term `(1 times 3)` and the factors of the third term which render the middle term of -10 (`2 times 8)` :

`(t-3)(t-2)(t-8)`

**Ans:**

**x= (t-3)(t-2)(t-8)**

Define function

x=f(t)=`t^3-13t^2+46t-48`

Now ,find zeros of f(t) ,which possible may be `+-1,+-2,+-3,+-4,+-6,+-8,...`

`f(2)=2^3-13xx2^2+46xx2-48`

`=8-52+92-48=0`

Thus zero of f(t) is 2

By factor theorem , t-2 will be one factor of f(t)

`x=f(t)=t^3-13t^2+46t-48`

`=t^3-2t^2-11t^2+22t+24t-48`

`=t^2(t-2)-11t(t-2)+24(t-2)`

`=(t-2)(t^2-11t+24)`

`=(t-2)(t^2-8t-3t+24)`

`=(t-2)(t(t-8)-3(t-8))`

`=(t-2)(t-3)(t-8)`

Thus

x=(t-2)(t-3)(t-8)