Question

Factor by grouping 8y3–4y2–14y+7

Answer 1

Answer:

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

Step-by-step 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:

$step - 1 : define \\ \sf{8y}^{3} - {4y}^{2} - 14y + 7$

$step - 2 : solve$

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

Answer:

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

Step-by-step 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 ).