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

##### 2 Answers

#### Answer:

#### 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:

#### Explanation:

let

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

so

from comparing coefficients

RHS=a#

so we now have:

**compare the constant term**

**compare coeff. #" "t^2#**

**the question now is: can we factorise the quadratic?**

factors of

can be factorised

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

so the final factorisation