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

2 Answers
Jul 19, 2017

Answer:

#(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)#

Jul 19, 2017

Answer:

#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)#