document.write( "Question 716251: How do you factor the equation y^3+y^2-4y-4? \n" ); document.write( "

Algebra.Com's Answer #439786 by jsmallt9(3759) $\"\"$ $\"About$

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$\"y%5E3%2By%5E2-4y-4\"$
\n" ); document.write( "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.

\n" ); document.write( "First we will rewrite the expression as additions so we can use the Commutative and Associative properties to change the order and grouping freely:
\n" ); document.write( " $\"y%5E3%2By%5E2%2B%28-4y%29%2B%28-4%29\"$

\n" ); document.write( "Next we split the expression into two equal-sized groups:
\n" ); document.write( " $\"%28y%5E3%2By%5E2%29%2B%28%28-4y%29%2B%28-4%29%29\"$
\n" ); document.write( "Next we factor out the greatest common factor (GCF) from each group. Notes:

This is one of the rare times when we do factor out a GCF of 1!
If the first terms of each group have different signs, then factor out the negative GCF from the group whose first term is negative

The GCF of the first group is $\"y%5E2\"$ 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:
\n" ); document.write( " $\"y%5E2%28y%2B1%29%2B%28-4%29%28y%2B1%29\"$

\n" ); document.write( "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:
\n" ); document.write( " $\"%28y%2B1%29%28y%5E2%2B%28-4%29%29\"$
\n" ); document.write( "or
\n" ); document.write( " $\"%28y%2B1%29%28y%5E2-4%29\"$

\n" ); document.write( "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 $\"a%5E2-b%5E2+=+%28a%2Bb%29%28a-b%29\"$ pattern:
\n" ); document.write( " $\"%28y%2B1%29%28y%2B2%29%28y-2%29\"$
\n" ); document.write( "The expression is now fully factored. \n" ); document.write( "

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