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x^3+x^2+2x+2 factor by grouping so it ends up with two binomials.

 Dec 5, 2014

Best Answer 

 #2
avatar+26364 
+5

x^3+x^2+2x+2 factor by grouping so it ends up with two binomials.

$$\small{
\text{Formula:}
$
\boxed{a+ax+x^2+x^3 = (1+x)(a+x^2)}
\\\\
$
\text{Example: }
$\\$
a=2 \qquad 2+2x+x^2+x^3 = (1+x)(2+x^2) \\
a=1 \qquad 1+1x+x^2+x^3 = (1+x)(1+x^2)
$
}$$

 Dec 5, 2014
 #1
avatar
+5

You have to play around with it a bit (or a lot) using trial-and-error trying different possibilities until it works.

 

Check your answer here: http://m.wolframalpha.com/input/?i=factorize+x%5E3%2Bx%5E2%2B2x%2B2&x=0&y=0

 

 Dec 5, 2014
 #2
avatar+26364 
+5
Best Answer

x^3+x^2+2x+2 factor by grouping so it ends up with two binomials.

$$\small{
\text{Formula:}
$
\boxed{a+ax+x^2+x^3 = (1+x)(a+x^2)}
\\\\
$
\text{Example: }
$\\$
a=2 \qquad 2+2x+x^2+x^3 = (1+x)(2+x^2) \\
a=1 \qquad 1+1x+x^2+x^3 = (1+x)(1+x^2)
$
}$$

heureka Dec 5, 2014
 #3
avatar+118587 
0

This factorization could be handy to remember.  Thanks Heureka.

 Dec 5, 2014

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