Question

Factorizar P(m,n)=m^5 +(m^4)n +n^5

Answer 1

Explicación paso a paso:

${m}^{5} + ( {m}^{4} )n + {n}^{5} \\ {m}^{5} + {m}^{4} n + {n}^{5} \\ =$

Answer 2

Podemos extraer un término

$P(m,n)=m^5 +(m^4)n +n^5\\\\P(m,n)=n^5[\dfrac{m^{5}}{n^{5}} +\dfrac{m^{4}}{n^{4}}+1]$

asumamos que:

$\dfrac{m}{n}=x$

$P(m,n)=n^5[\dfrac{m^{5}}{n^{5}} +\dfrac{m^{4}}{n^{4}}+1]\\\\P(m,n)=n^5[x^{5}+x^{4}+1]$

sumemos y restamos:

$x^{2}$

$P(m,n)=n^5[x^{5}+x^{4}+1+x^{2}-x^{2}]\\\\P(m,n)=n^5[x^{5}-x^{2}+x^{4}+x^{2}+1]\\\\P(m,n)=n^5[x^{2}(x^{3}-1)+x^{4}+x^{2}+1]\\\\P(m,n)=n^5[x^{2}(x-1)(x^{2}+x+1)+ x^{4}+x^{2}+1]$

recordemos la propiedad de argand

$P(m,n)=n^5[x^{2}(x-1)(x^{2}+x+1)+ x^{4}+x^{2}+1]\\\\P(m,n)=n^5[x^{2}(x-1)(x^{2}+x+1)+(x^{2}+x+1)(x^{2}-x+1)]\\\\P(m,n)=n^5(x^{2}+x+1)(x^{2}(x-1)+(x^{2}-x+1))\\\\P(m,n)=n^5(x^{2}+x+1)(x^{3}+x+1)$

pero:

$\dfrac{m}{n}=x$

$P(m,n)=n^5(x^{2}+x+1)(x^{3}+x+1)\\\\P(m,n)=n^5[(\dfrac{m}{n}^{2})+\dfrac{m}{n}+1][(\dfrac{m}{n}^{3})+\dfrac{m}{n}+1]\\\\P(m,n)=n^5[\dfrac{m^{2}+mn+n^{2}}{n^{2}}][\dfrac{m^{3}+mn^{2} +n^{3}}{n^{3}}]$

$P(m,n)=(m^{2}+mn+n^{2})(m^{3}+mn^{2}+n^{3})$

AUTOR SmithValdez

Viva ronaldooo