Question

question 4 of 10
divide.
\\(\frac{8y^4 + 3y^3}{2y^3}\\)

Explanation:

Step1: Split the first fraction

$\frac{36x^4 - 32x^3}{4x^2} = \frac{36x^4}{4x^2} - \frac{32x^3}{4x^2}$

Step2: Simplify first fraction terms

$\frac{36}{4}x^{4-2} - \frac{32}{4}x^{3-2} = 9x^2 - 8x$

Step3: Split the second fraction

$\frac{8y^4 + 3y^3}{2y^3} = \frac{8y^4}{2y^3} + \frac{3y^3}{2y^3}$

Step4: Simplify second fraction terms

$\frac{8}{2}y^{4-3} + \frac{3}{2}y^{3-3} = 4y + \frac{3}{2}$

Answer:

$\frac{36x^4 - 32x^3}{4x^2} = 9x^2 - 8x$
$\frac{8y^4 + 3y^3}{2y^3} = 4y + \frac{3}{2}$