Respuesta :

Answer:

[tex](2y - 1)( {4y}^{2} - 7)[/tex]

Step-by-step explanation:

1) Factor out common terms in the first two terms, then in the last two terms.

[tex] {4y}^{2} (2y - 1) - 7(2y - 1)[/tex]

2) Factor out the common term 2y - 1.

[tex](2y - 1)( {4y}^{2} - 7)[/tex]

Therefor the answer is, ( 2y - 1 ) ( 4y² - 7 ).

Answer:

[tex] \huge \boxed{ \boxed{ \red{ \sf (4 {y}^{ {2} } - 7)(2x - 1)}}}[/tex]

Step-by-step explanation:

to understand this

you need to know about:

  • factoring
  • PEMDAS

given:

  • [tex] \sf 8 {y}^{3} - 4 {y}^{2} - 14y + 7[/tex]

to do:

  • factoring

tips and formulas:

how to factor

  1. factor out common constants and terms
  2. group

let's do:

[tex] step - 1 : define \\ \sf{8y}^{3} - {4y}^{2} - 14y + 7[/tex]

[tex] step - 2 : solve[/tex]

  1. [tex] \sf factor \: - 7 \: and \: 4 {y}^{2} \: from \: the \: expression : \\ \sf 4 {y}^{ {2} } (2y -1 ) - 7(2y - 1)[/tex]
  2. [tex] \sf \: group : \\ \sf (4 {y}^{ {2} } - 7)(2x - 1)[/tex]

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