Difference between revisions of "Sum and difference of powers"

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The '''sum and difference of powers''' are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.
 
The '''sum and difference of powers''' are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.
  
==Factored Forms of Sums and Differences of Powers==
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==Sums of Powers==
*<math>a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+ab^{n-2}+b^{n-1}</math>
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<math>a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-\ldots-ab^{2n-1}+b^{2n})</math>
*<math>a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-\ldots-ab^{2n-1}+b^{2n})</math>
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==Differences of Powers==
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If p is a positive integer and x and y are real numbers,
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<math>x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)</math>
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For example,
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<math>x^2-y^2=(x-y)(x+y)</math>
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<math>x^3-y^3=(x-y)(x^2+xy+y^2)</math>
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<math>x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)</math>
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Note that the number of terms in the ''long'' factor is equal to the exponent in the expression being factored.
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An amazing thing happens when x and y differ by 1, say, x = y+1. Then x-y = 1 and
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<math>x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}</math>
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<math>=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p</math>.
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For example,
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<math>(y+1)^2-y^2=(y+1)+y</math>
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<math>(y+1)^3-y^3=(y+1)^2+(y+1)y+y^2</math>
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<math>(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3</math>
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If we also know that <math>y\geq 0</math> then
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<math>2y\leq (y+1)^2-y^2\leq 2(y+1)</math>
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<math>3y^2\leq (y+1)^3-y^3\leq 3(y+1)^2</math>
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<math>4y^3\leq (y+1)^4-y^4\leq 4(y+1)^3</math>
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<math>(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p</math>
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==See Also==
 
==See Also==
 
* [[Factoring]]
 
* [[Factoring]]

Revision as of 09:04, 8 July 2008

The sum and difference of powers are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.

Sums of Powers

$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-\ldots-ab^{2n-1}+b^{2n})$

Differences of Powers

If p is a positive integer and x and y are real numbers,

$x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)$

For example,

$x^2-y^2=(x-y)(x+y)$

$x^3-y^3=(x-y)(x^2+xy+y^2)$

$x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)$

Note that the number of terms in the long factor is equal to the exponent in the expression being factored.

An amazing thing happens when x and y differ by 1, say, x = y+1. Then x-y = 1 and

$x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}$

$=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p$.

For example,

$(y+1)^2-y^2=(y+1)+y$

$(y+1)^3-y^3=(y+1)^2+(y+1)y+y^2$

$(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3$

If we also know that $y\geq 0$ then

$2y\leq (y+1)^2-y^2\leq 2(y+1)$

$3y^2\leq (y+1)^3-y^3\leq 3(y+1)^2$

$4y^3\leq (y+1)^4-y^4\leq 4(y+1)^3$

$(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p$

See Also

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