Question

Factor by grouping 8y3–4y2–14y+7

Answer 1

`\huge \boxed{ \boxed{ \red{ \sf (4 {y}^{ {2} } - 7)(2x - 1)}}}`

Explanation:

to understand this

you need to know about:

• factoring
• PEMDAS

given:

• 
  `\sf 8 {y}^(3) - 4 {y}^(2) - 14y + 7`

to do:

• factoring

tips and formulas:

how to factor

1. factor out common constants and terms
2. group

let's do:

`step - 2 : solve`

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

Answer 2

`(2y - 1)( {4y}^(2) - 7)`

Explanation:

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

`{4y}^(2) (2y - 1) - 7(2y - 1)`

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

`(2y - 1)( {4y}^(2) - 7)`

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