Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 23, 2011"
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− | { | + | 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>. |
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+ | 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. | ||
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+ | 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>. | ||
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+ | 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>. |
Latest revision as of 13:07, 28 June 2011
Problem
AoPSWiki:Problem of the Day/June 23, 2011
Solutions
Since the first term is , the two polynomials that it factors into must either be
and
, or
and
.
Due to the constant term being , the two factors must both have
or
as their constants.
Starting off with and
, with product
, we need a way to get rid of the
and
terms in the product.
Adding a into
, so it becomes
, gives a product of
.
To get rid of the and the
term, we change
to
. The new product is
, which is what we are trying to factor.
Therefore, .