Question

2y 4 +16y 3 2y 3 ​ =start fraction, 2, y, cubed, divided by, 2, y, start superscript, 4, end superscript, plus, 16, y, cubed, end fraction, equals

Answer 1

`(2y^(3))/(2y^(4)+16y^(3))=(1)/((y+8))`

Explanation:

The problem is
`(2y^(3))/(2y^(4)+16y^(3))=`

To solve this problem let us simplify the denominator by taking the greatest common factor of it

∵ The denominator is
`2y^(4)+16y^(3)`

∵ The factors of 2 are 1, 2

∵ The factors of 16 are 1, 2, 4, 8, 16

∴ The common factors of 2 and 16 are 1 and 2

∵ The greatest one is 2

∴ The greatest common factor of 2 and 16 is 2

The greatest common factor of a variable is the variable with the smallest exponent

∵ The smallest exponent of
`y^(4)` and y³ is 3

∴ The greatest common factor of
`y^(4)` and y³ is y³

∴ The greatest common factor of
`2y^(4)` and 16y³ is 2y³

Divide each term of the denominator by the greatest common factor

∵
`2y^(4)` ÷ 2y³ = y

∵ 16y³ ÷ 2y³ = 8

- The factorization of the denominator is 2y³(y + 8)

∴
`2y^(4)+16y^(3)` = 2y³(y + 8)

Substitute it in the fraction

∴
`(2y^(3))/(2y^(4)+16y^(3))=(2y^(3))/(2y^(3)(y+8))`

- Simplify the right hand side by dividing up and down by 2y³

∴
`(2y^(3))/(2y^(4)+16y^(3))=(1)/((y+8))`