Question 854423
{{{15x^2 -2x -1=(3x-1)(5x+1)}}}
 
{{{a^3 +6a^2b +12ab^2 +8b^3=(a+2b)^3}}}
(I know you wrote {{{a^3 +6a^2b +12ab^3 +8b^3}}} , but I think there was a typo somewhere).
 
How did I get it?
 
The factorization of
{{{15x^2 -2x -1=(3x-1)(5x+1)}}}
is easier to calculate this way:
If {{{y=1/x}}}<-->{{{x=1/y}}} ,
{{{(15x^2 -2x -1)=(-x^2)*(-15+2/x+1/x^2)=(-1/y^2)*(y^2+2y-15)=(-1/y^2)*(y+5)*(y-3)=((y+5)/y)*((y-3)/(-y))=(1+5/y)(-1+3/y)=(1+3x)(-1+5x)}}}
 
If that is Greek to you, there is another, more traditional way.
To factorize something like {{{15x^2 -2x -1}}} , you multiply the coefficients at the ends
(in this case, 15 and -1, that multiply to give -15),
and look for pairs of factors of that result that will add to the middle coefficient (-2 in this case).
When you figure out that 3 and -5 are those factors,
you re-write the polynomial with two middle terms with those coefficients,
that is {{{3x}}} and {{{-5x}}} in place of the middle term {{{-2x}}} ,
and then you factorize by grouping:
{{{15x^2 -2x -1=15x^2+3x-5x -1=(15x^2-5x)+(3x-1)=5x*(3x-1)+1*(3x-1)=(5x+1)*(3x-1)}}}
 
With {{{a^3 +6a^2b +12ab^2 +8b^3}}} it is a question of realizing that
{{{a^3}}} is the cube of {{{a}}} and
{{{8b^3}}} is the cube of {{{2b}}} .
That should make you suspect that it could be the cube of a binomial:
{{{(a+2b)^3=a^3+3*a^2*(2b)+3*a*(2b)^2+b^3}}}
If you do the indicated operations,
{{{3*a^2*(2b)=6a^2b}}} and {{{3*a*(2b)^2=3*a*(4b^2)=12ab^2}}} ,
and you find that indeed {{{(a+2b)^3=a^3 +6a^2b +12ab^2 +8b^3}}}