SOLUTION: please help me solve these problems! Thanks! x^3-3x2-x+3=0 y^2+2y-21=-2y

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Question 167537: please help me solve these problems! Thanks!
x^3-3x2-x+3=0
y^2+2y-21=-2y

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E3-3x2-x%2B3=0

Factor just the first two terms:
         x%5E3-3x%5E2 
         factor out x%5E2 
         x%5E2%28x-3%29

Factor just the last two terms
         (((-x+3}}}  
         factor out -1
         -1%28x-3%29

Substitute x%5E2%28x-3%29 for the
first two terms and -1%28x-3%29
for the last two terms in

x%5E3-3x2-x%2B3=0

x%5E2%28x-3%29-1%28x-3%29=0

Now factor the whole parentheses %28x-3%29
out of both:

%28x-3%29%28x%5E2-1%29=0

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

%28x-3%29%28x%5E2-1%5E2%29=0

So it factors as %28x-1%29%28x%2B1%29:

%28x-3%29%28x-1%29%28x%2B1%29=0

Use the zero-factor principle:


 
---------------------------

y%5E2%2B2y-21=-2y

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

y%5E2%2B2y-21=-2y

y%5E2%2B4y-21=0

Think of two whole numbers which when multiplied gives -21
and when added gives matrix%281%2C2%2C%22%2B%22%2C4%29.  They are matrix%281%2C2%2C%22%2B%22%2C7%29
and -3.  So the left side factors as

%28y%2B7%29%28y-3%29=0

Use the zero-factor principle:

matrix%282%2C3%2C+y%2B7=0%2C+%22%7C%22%2C+y-3=0%2C+y=-7%2C++%22%7C%22%2C++y=3%29  

Edwin