Question 716251
{{{y^3+y^2-4y-4}}}
Since the GCF is 1 and since there are too many terms for factoring patterns or for trinomial factoring, probably factoring by grouping will be easiest.<br>
First we will rewrite the expression as additions so we can use the Commutative and Associative properties to change the order and grouping freely: 
{{{y^3+y^2+(-4y)+(-4)}}}<br>
Next we split the expression into two equal-sized groups:
{{{(y^3+y^2)+((-4y)+(-4))}}}
Next we factor out the greatest common factor (GCF) from each group. Notes:<ul><li>This is one of the rare times when we do factor out a GCF of 1!</li><li>If the first terms of each group have different signs, then factor out the negative GCF from the group whose first term is negative</li></ul>The GCF of the first group is {{{y^2}}} and the GCF of the second group is 4. But since the first terms of the two groups do have different signs we will be factoring out -4 from the second group:
{{{y^2(y+1)+(-4)(y+1)}}}<br>
If we're lucky the "non-GCF" factors match at this point. (If they do not match, restart and change the order and grouping. If the "non-GCF" factors never match no matter what the order or grouping, then factoring by grouping will not work.) Fortunately the "non-GCF" factors match so we can it out of the two groups:
{{{(y+1)(y^2+(-4))}}}
or
{{{(y+1)(y^2-4)}}}<br>
Just like you keep reducing fractions until you can reduce them any more, you keep factoring until you can't factor any more. The first factor, y+1, will not factor any further. But the second factor is a difference of squares which we can factor using the {{{a^2-b^2 = (a+b)(a-b)}}} pattern:
{{{(y+1)(y+2)(y-2)}}}
The expression is now fully factored.