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Difference between revisions of "Sum and difference of powers"

(→‎Factorizations of Sums of Powers)
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<math>x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)</math>
 
<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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Note that the number of terms in the second factor is equal to the exponent in the expression being factored.
  
 
An amazing thing happens when <math>x</math> and <math>y</math> differ by <math>1</math>, say, <math>x = y+1</math>. Then <math>x-y = 1</math> and  
 
An amazing thing happens when <math>x</math> and <math>y</math> differ by <math>1</math>, say, <math>x = y+1</math>. Then <math>x-y = 1</math> and  
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<math>1^3+2^3+3^3+4^3=10^2</math>
 
<math>1^3+2^3+3^3+4^3=10^2</math>
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<math>1^3+2^3+3^3+4^3+\ldots+n^3=\left(\frac{n(n+1)}{2} \right) ^2=(1+2+3+4+\ldots+n)^2</math>
  
 
==Factorizations of Sums of Powers==
 
==Factorizations of Sums of Powers==
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If we have a difference of powers of degree <math>n</math>, then
 
If we have a difference of powers of degree <math>n</math>, then
  
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<cmath>x^n-y^n=(x-y)(x^n+x^ay+x^by^2...+y^n)</cmath>
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<cmath>x^n-y^n=(x-y)(x^{n - 1}+x^{n-2}y+x^{n-3}y^2...+y^{n - 1})</cmath>
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where <math>a, b, c,...</math> are equivalent to <math>n-1, n-2, n-3,...</math> respectively.
 
  
 
Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial <math>(x+y)^n</math>, except for the fact that the coefficient on each of the terms is <math>1</math>. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.
 
Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial <math>(x+y)^n</math>, except for the fact that the coefficient on each of the terms is <math>1</math>. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.
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* [[Difference of squares]], an extremely common specific case of this.
 
* [[Difference of squares]], an extremely common specific case of this.
 
* [[Binomial Theorem]]
 
* [[Binomial Theorem]]
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{{stub}}
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[[Category:Elementary algebra]]
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[[Category:Algebra]]

Latest revision as of 05:25, 6 May 2024

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

Sums of Odd 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 second 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:

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

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

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

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

Sum of Cubes

$1^3=1^2$

$1^3+2^3 =3^2$

$1^3 +2^3+3^3=6^2$

$1^3+2^3+3^3+4^3=10^2$

$1^3+2^3+3^3+4^3+\ldots+n^3=\left(\frac{n(n+1)}{2} \right) ^2=(1+2+3+4+\ldots+n)^2$

Factorizations of Sums of Powers

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

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

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

Note that all these sums of powers can be factorized as follows:

If we have a difference of powers of degree $n$, then

\[x^n-y^n=(x-y)(x^{n - 1}+x^{n-2}y+x^{n-3}y^2...+y^{n - 1})\]

Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial $(x+y)^n$, except for the fact that the coefficient on each of the terms is $1$. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.

- icecreamrolls8

See Also

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