Long Factoring

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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:

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

The coefficients reminded me of an arithmetical phenomenon:

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

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

Already knowing the answer, it was easy to demonstrate the factoring:

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

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

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

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

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

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