Question 161897
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FACTOR BY GROUPING

{{{3AX+3AY+4X+4Y}}}

In the first two terms only, {{{3AX+3AY}}}, we can factor out
{{{3A}}} and get {{{3A*(X+Y)}}}.


In the last two terms only, {{{matrix(1,2, "+", 4X+4Y )}}}, we can factor out
{{{matrix(1,1,"+4")}}} and get {{{matrix(1,1,"+4*(X+Y)")}}}.  So we have this:

{{{3A*(X+Y)+4(X+Y)}}}

Now we can take out the parentheses as the common factor
{{{(X+Y)}}} leaving the {{{3A}}} and the {{{matrix(1,1,"+4")}}},
so the final factored form is:

{{{(X+Y)(3A+4)}}} 

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FACTOR BY FINDING THE GREATEST COMMON FACTOR: 
{{{25y^5-15y^4+ 6y^3+5y^2}}}

We can factor {{{y^2}}} out of all the terms:

{{{y^2(25y^3-15y^2+ 6y+5)}}}

That's as far as we can go, because there is no way
to factor what is inside the parentheses by grouping:

Edwin</pre></b></font>