+0

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

0
425
3

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

Guest Dec 5, 2014

#2
+20549
+5

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

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

heureka  Dec 5, 2014
#1
+5

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

Guest Dec 5, 2014
#2
+20549
+5

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

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

heureka  Dec 5, 2014
#3
+94087
0

This factorization could be handy to remember.  Thanks Heureka.

Melody  Dec 5, 2014