Answer:
ANSWER: (t-1)*(t-2)*(t-3)
Explanation:
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "t2" was replaced by "t^2". 1 more similar replacement(s).
STEP1:
Equation at the end of step 1
(((t3) - (2•3t2)) + 11t) - 6
STEP2:
Checking for a perfect cube
t3-6t2+11t-6 is not a perfect cube
Trying to factor by pulling out :
Factoring: t3-6t2+11t-6
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 11t-6
Group 2: -6t2+t3
Pull out from each group separately :
Group 1: (11t-6) • (1)
Group 2: (t-6) • (t2)
t3-6t2+11t-6
can be divided by 3 different polynomials,including by t-3
dividend t3 - 6t2 + 11t - 6
- divisor * t2 t3 - 3t2
remainder - 3t2 + 11t - 6
- divisor * -3t1 - 3t2 + 9t
remainder 2t - 6
- divisor * 2t0 2t - 6
remainder 0
Quotient : t2-3t+2 Remainder: 0
Factoring t2-3t+2
wich gives you (t-1)*(t-2)*(t-3)