Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 23, 2011"

Latest revision as of 13:07, 28 June 2011

Since the first term is $x^5$ , the two polynomials that it factors into must either be $x^2$ and $x^3$ , or $x$ and $x^4$ .

Due to the constant term being $1$ , the two factors must both have $1$ or $-1$ as their constants.

Starting off with $x^2+1$ and $x^3+1$ , with product $x^5+x^3+x^2+1$ , we need a way to get rid of the $x^3$ and $x^2$ terms in the product.

Adding a $-x^2$ into $x^3+1$ , so it becomes $x^3-x^2+1$ , gives a product of $(x^3-x^2+1)(x^2+1)=x^5-x^4+x^3+1$ .

To get rid of the $-x^4$ and the $x^3$ term, we change $x^2+1$ to $x^2+x+1$ . The new product is $x^5+x+1$ , which is what we are trying to factor.

Therefore, $\boxed{x^5+x+1=(x^3-x^2+1)(x^2+x+1)}$ .

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 {{:AoPSWiki:Problem of the Day/June 23, 2011}}
 ==Solutions==
-{{potd_solution}}
+Since the first term is <math>x^5</math>, the two polynomials that it factors into must either be <math>x^2</math> and <math>x^3</math>, or <math>x</math> and <math>x^4</math>.
+Due to the constant term being <math>1</math>, the two factors must both have <math>1</math> or <math>-1</math> as their constants.
+Starting off with <math>x^2+1</math> and <math>x^3+1</math>, with product <math>x^5+x^3+x^2+1</math>, we need a way to get rid of the <math>x^3</math> and <math>x^2</math> terms in the product.
+Adding a <math>-x^2</math> into <math>x^3+1</math>, so it becomes <math>x^3-x^2+1</math>, gives a product of <math>(x^3-x^2+1)(x^2+1)=x^5-x^4+x^3+1</math>.
+To get rid of the <math>-x^4</math> and the <math>x^3</math> term, we change <math>x^2+1</math> to <math>x^2+x+1</math>. The new product is <math>x^5+x+1</math>, which is what we are trying to factor.
+Therefore, <math>\boxed{x^5+x+1=(x^3-x^2+1)(x^2+x+1)}</math>.