Question 535799
{{{11x(4x-3)-6(4x-3)}}} has a common factor, present in both terms. Getting that common factor out is applying the distributive property backwards, like this
{{{11x(4x-3)-6(4x-3)=(11x-6)(4x-3)}}}
{{{2x(x-5)-(x-5)}}} Same thing here
{{{2x(x-5)-(x-5)=2x(x-5)-1*(x-5)=(2x-1)(x-5)}}}
{{{3x^3-4x^2+6x-8}}} This polynomial is best factored by grouping. There are four terms. Look for a factor common to two of those terms, like {{{3x}}}
and a factor common to the other two terms. If the polynomial was an exercise in factoring, it should work. (just don't expect it to work for every polynomial.
{{{3x^3-4x^2+6x-8=(3x^3+6x)-(4x^2+8)=3x(x^2+2)-4(x^2+2)}}}
Do you see now another common factor? Keep factoring.
{{{3x(x^2+2)-4(x^2+2)=(3x-4)(x^2+2)}}} and that's your polynomial fully factored.
{{{xy+2x-y-2}}} is another 4-term factor-by-grouping problem. By the way, there are two ways to group. They both work (unless you make a mistake). You may see only one way, but that would be enough.
{{{xy+2x-y-2=(xy-y)+(2x-2)=y(x-1)+2(x-1)=(y+2)(x-1) }}}
{{{2x^2+2x-24}}}is a tougher factoring problem. If it can ve facored with nice integer numbers, it's going to end up being
{{{2x^2+2x-24=(2x+a)(x+b)}}} with {{{a*b=-24}}}
That gives you a bunch of choices for a and b because there are several pairs if factors whose products yield 24, and then you have a choice of which one to give the minus sign to. You know that the product term in x is
{{{2bx+ax=(2b+a)x=2x}}}, so {{{2b+a=2}}}
I would try pairs like 3 and 8 (3x8=24) or 4 and 6 (4x6=24) rather than 2 and 12  or 1 and 24, because you need to add to 2, which is a small number.
Trying 8 and -3: 2x8-3=13 does not work,
but 2x(-3)+8=2 does. Lucky first guess!
Let's verify
{{{(2x+8)(x-3)=2x^2-6x+8x-24=2x^2+2x-24}}} We got it!