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