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.
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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.
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: .$\displaystyle (x^2 + x + 1)^2$
Already knowing the answer, it was easy to demonstrate the factoring:
$\displaystyle x^4 + 2x^3 + 3x^2 + 2x + 1$
. . $\displaystyle =\;(x^4 + x^3 + x^2) + (x^3 + x^2 + x) + (x^2 + x + 1)$
. . $\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) $
. . $\displaystyle =\;(x^2+x+1)^2$
thank you so much Soroban. :) i was trying to solve ... thank God you checked my work. Love Live MHF!
