SOLUTION: Factor by grouping: a. 27t^3 + 45t^2 - 3t - 5 b. x^3 - 3x^2 - 4x + 12 Thanks!

Algebra ->  Rational-functions -> SOLUTION: Factor by grouping: a. 27t^3 + 45t^2 - 3t - 5 b. x^3 - 3x^2 - 4x + 12 Thanks!       Log On


   

Question 834414: Factor by grouping:
a. 27t^3 + 45t^2 - 3t - 5
b. x^3 - 3x^2 - 4x + 12
Thanks!
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Number (a) could be handled , 27t%5E3-3t%2B45t%5E2-5
3%289t%5E3-t%29%2B5%289t%5E2-1%29
3t%289t%5E2-1%29%2B5%289t%5E2-1%29
%283t%2B5%29%289t%5E2-1%29
%283t%2B5%29%283t%2B1%29%283t-1%29---- completely factored

Number (b) almost the same method, x%5E3+-+3x%5E2+-+4x+%2B+12
but look at the factorization of the coefficients.
x%5E3-3x%5E2-2%2A2x%2B2%2A2%2A3
x%5E3-3x%5E2-4x%2B4%2A3
x%5E3-3x%5E2%2B4%2A3-4x
x%5E3-3x%5E2%2B4%283-x%29
x%5E2%28x-3%29%2B4%283-x%29
I really would like a binomial in the form (x-k), where k is the constant.
x%5E2%28x-3%29%2B%28-1%294%28x-3%29
%28x-3%29x%5E2-4%28x-3%29
%28x-3%29%28x%5E2-4%29
%28x-3%29%28x-2%29%28x%2B2%29----- completely factored