Respuesta :

2(y³ -5) (x²y+7) This is the correct answer 

Given expression: [tex]2x^2y^4 - 10x^2y + 14y^3 - 70.[/tex]

We need to factor it.

[tex]\mathrm{Rewrite\:}14\mathrm{\:as\:}7\cdot \:2[/tex]

[tex]\mathrm{Rewrite\:}70\mathrm{\:as\:}35\cdot \:2[/tex]

[tex]\mathrm{Rewrite\:}10\mathrm{\:as\:}5\cdot \:2[/tex]

[tex]\mathrm{Factor\:out\:common\:term\:}2[/tex]

[tex]=2\left(x^2y^4+7y^3-5x^2y-35\right)[/tex]

Makeing it into two groups.

[tex]=\left(x^2y^4-5x^2y\right)+\left(7y^3-35\right)[/tex]

[tex]\mathrm{Factor\:out\:}7\mathrm{\:from\:}7y^3-35\mathrm{:\quad }7\left(y^3-5\right)[/tex]

[tex]\mathrm{Factor\:out\:}x^2y\mathrm{\:from\:}x^2y^4-5x^2y\mathrm{:\quad }x^2y\left(y^3-5\right)[/tex]

[tex]=7\left(y^3-5\right)+x^2y\left(y^3-5\right)[/tex]

[tex]\mathrm{Factor\:out\:common\:term\:}y^3-5[/tex]

[tex]=\left(y^3-5\right)\left(x^2y+7\right)[/tex]

[tex]2\left(x^2y^4+7y^3-5x^2y-35\right) =\left(y^3-5\right)\left(x^2y+7\right)[/tex].

Therefore, final factored form is [tex]2\left(y^3-5\right)\left(x^2y+7\right).[/tex]