Answer:
[tex]\frac{2y^{3}}{2y^{4}+16y^{3}}=\frac{1}{(y+8)}[/tex]
Step-by-step explanation:
The problem is [tex]\frac{2y^{3}}{2y^{4}+16y^{3}}=[/tex]
To solve this problem let us simplify the denominator by taking the greatest common factor of it
∵ The denominator is [tex]2y^{4}+16y^{3}[/tex]
∵ 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 [tex]y^{4}[/tex] and y³ is 3
∴ The greatest common factor of [tex]y^{4}[/tex] and y³ is y³
∴ The greatest common factor of [tex]2y^{4}[/tex] and 16y³ is 2y³
Divide each term of the denominator by the greatest common factor
∵ [tex]2y^{4}[/tex] ÷ 2y³ = y
∵ 16y³ ÷ 2y³ = 8
- The factorization of the denominator is 2y³(y + 8)
∴ [tex]2y^{4}+16y^{3}[/tex] = 2y³(y + 8)
Substitute it in the fraction
∴ [tex]\frac{2y^{3}}{2y^{4}+16y^{3}}=\frac{2y^{3}}{2y^{3}(y+8)}[/tex]
- Simplify the right hand side by dividing up and down by 2y³
∴ [tex]\frac{2y^{3}}{2y^{4}+16y^{3}}=\frac{1}{(y+8)}[/tex]
