Question

How do you factor `3t^3-7t^2-3t+7`?

Answer 1

`(3t-7)(t-1)(t+1)`

Explanation:

`"split the expression into 2 groups"`

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

`"factorise each pair"`

`color(red)(t^2)(3t-7)color(red)(-1)(3t-7)`

`"factor out " (3t-7)`

`(3t-7)(color(red)(t^2-1))`

`t^2-1" is a "color(blue)"difference of squares"`

`(3t-7)(t-1)(t+1)larr" factors of difference of squares"`

`rArr3t^3-7t^2-3t+7=(3t-7)(t-1)(t+1)`

Answer 2

`3t^3-7t^2-3t+7=(t-1)(t+1)(3t-7)`

Explanation:

let

`f(t)=3t^3-7t^2-3t+7`

All cubics have at least one real root, so using the factor theorem we will try to find one of teh roots.

`f(1)=3xx1-7xx1-3xx1=7=3-7-3+7=0`

so`" "t=1" is root "=>(t-1)" is a factor"`

`:.f(t)=3t^3-7t-3t+7=(t-1)(at^2+bt+c)`

from comparing coefficients

`"coeff. "t^2`

`LHS=3`

RHS=a#

`=>a=3`

so we now have:

`3t^3-7t^2-3t+7=(t-1)(3t^2+bt+c)`

compare the constant term

`LHS=+7`

`RHS=-1xxc=-c`

`:.c=-7`

`3t^3-7t^2-3t+7=(t-1)(3t^2+bt-7)`

compare coeff. `" "t^2`

`LHS=-7`

`RHS=-3+b`

`b-3=-7=>b=-4`

`3t^3-7t^2-3t+7=(t-1)(3t^2-4t-7)`

the question now is: can we factorise the quadratic?

`3t^2-4t-7`

`3xx-7=-21`

factors of`-21` that add to`-4" "`are `-7" " & " "+3`

can be factorised

replacing the middle term with these two factors we proceed as follows:

`3t^2-4t-7=3t^2-7t+3t-7`

`=(3t^2-73t-7)+(3t-7)`

`t(3t-7)+(3t-7)`

`(3t-7)(t+1)`

so the final factorisation

`3t^3-7t^2-3t+7=(t-1)(3t-7)(t+1)`

`3t^3-7t^2-3t+7=(t-1)(t+1)(3t-7)`