# Long Factoring

• Dec 4th 2010, 04:31 AM
rcs
Long Factoring
x^4 + 2x^3 + 3x^2 + 2x + 1
x ^4 + x^2 + 2x^3 + 2x^2 + 2x + 1
(x^4 + 2x^3 + x^2) +( 2x^2 + 2x +1)
x^2 ( x^2 + 2x + 1) + 2 (x^2 + x) + 1
x^2(x+1)^2 + 2x (x+1) + 1
[x(x+1)+ 1]^2 + 2[x(x+1)] + 1
[x(x+1) + 1]^2
(x^2 + x + 1)^2 answer.
• Dec 4th 2010, 05:39 AM
Soroban
Hello, rcs!

Your work is absolutely correct! . . . Good work!

Quote:

$\displaystyle \text{Factor: }\:x^4 + 2x^3 + 3x^2 + 2x + 1$

The coefficients reminded me of an arithmetical phenomenon:

. . . $\displaystyle \begin{array}{ccc} 1^2 &=& 1 \\ 11^2 &=& 121 \\ 111^2 &=& 12321 \\ 1111^2 &=& 1234321 \\ \vdots && \vdots \end{array}$

So I immediately concluded that we have: .$\displaystyle (x^2 + x + 1)^2$

$\displaystyle x^4 + 2x^3 + 3x^2 + 2x + 1$

. . $\displaystyle =\;(x^4 + x^3 + x^2) + (x^3 + x^2 + x) + (x^2 + x + 1)$

. . $\displaystyle =\;x^2(x^2+x+1) + x(x^2+x+1) + (x^2+x+1)$

. . $\displaystyle =\;(x^2+x+1)(x^2+x+1)$

. . $\displaystyle =\;(x^2+x+1)^2$

• Dec 4th 2010, 01:18 PM
rcs
thank you
thank you so much Soroban. :) i was trying to solve ... thank God you checked my work. Love Live MHF!