Question 167537
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{{{x^3-3x2-x+3=0}}}

Factor just the first two terms:
         {{{x^3-3x^2}}} 
         factor out {{{x^2}}} 
         {{{x^2(x-3)}}}

Factor just the last two terms
         (((-x+3}}}  
         factor out {{{-1}}}
         {{{-1(x-3)}}}

Substitute {{{x^2(x-3)}}} for the
first two terms and {{{-1(x-3)}}}
for the last two terms in

{{{x^3-3x2-x+3=0}}}

{{{x^2(x-3)-1(x-3)=0}}}

Now factor the whole parentheses {{{(x-3)}}}
out of both:

{{{(x-3)(x^2-1)=0}}}

The binomial in the second parentheses is the 
difference or two squares, as you see:

{{{(x-3)(x^2-1^2)=0}}}

So it factors as {{{(x-1)(x+1)}}}:

{{{(x-3)(x-1)(x+1)=0}}}

Use the zero-factor principle:

{{{matrix(2,5, x-3=0, "|",  x-1=0, "|",    x+1=0,
          x=3, "|",  "x=1", "|",    x=-1)}}}
 
---------------------------

{{{y^2+2y-21=-2y}}}

Get {{{0}}} on the right by adding {{{2y}}} to both sides of:

{{{y^2+2y-21=-2y}}}

{{{y^2+4y-21=0}}}

Think of two whole numbers which when multiplied gives {{{-21}}}
and when added gives {{{matrix(1,2,"+",4)}}}.  They are {{{matrix(1,2,"+",7)}}}
and {{{-3}}}.  So the left side factors as

{{{(y+7)(y-3)=0}}}

Use the zero-factor principle:

{{{matrix(2,3, y+7=0, "|", y-3=0, y=-7,  "|",  y=3)}}}  

Edwin</pre>