Question 834414
Number (a) could be handled ,  {{{27t^3-3t+45t^2-5}}}
{{{3(9t^3-t)+5(9t^2-1)}}}
{{{3t(9t^2-1)+5(9t^2-1)}}}
{{{(3t+5)(9t^2-1)}}}
{{{(3t+5)(3t+1)(3t-1)}}}---- completely factored


Number (b) almost the same method, {{{x^3 - 3x^2 - 4x + 12}}}
but look at the factorization of the coefficients.
{{{x^3-3x^2-2*2x+2*2*3}}}
{{{x^3-3x^2-4x+4*3}}}
{{{x^3-3x^2+4*3-4x}}}
{{{x^3-3x^2+4(3-x)}}}
{{{x^2(x-3)+4(3-x)}}}
I really would like a binomial in the form (x-k), where k is the constant.
{{{x^2(x-3)+(-1)4(x-3)}}}
{{{(x-3)x^2-4(x-3)}}}
{{{(x-3)(x^2-4)}}}
{{{(x-3)(x-2)(x+2)}}}----- completely factored