Question

Find each quotient. All final answers must be in simplest form.56x4y5 - 49x3y6 - 35x2y3 / 7x?y2 =A8x^2y^3 - 7y^4 - 5yB8x^2 y^3 - 7xy^4 + 5yC 8x^2y^3 - 7xy^4 - 5yD 8x^2y^3 + 7xy^4 - 5y

Answer 1

Answer

Option C is correct

8x²y³ - 7xy⁴ - 5y

Explanation

The equation to be solved or reduced is

(56x⁴y⁵ - 49x³y⁶ - 35x²y³)/7x²y²

More properly written, the equation is

$\frac{56x^4y^5-49x^3y^6-35x^2y^3}{7x^2y^2}$

To solve this, we will break the division down and have each term carry the denominator and use the laws of indices to reduce the powers

$\begin{gathered} \frac{56x^4y^5-49x^3y^6-35x^2y^3}{7x^2y^2} \\ =\frac{56x^4y^5}{7x^2y^2}-\frac{49x^3y^6}{7x^2y^2}-\frac{35x^2y^3}{7x^2y^2} \\ =\frac{56}{7}x^{4-2}y^{5-2}-\frac{49}{7}x^{3-2}y^{6-2}-\frac{35}{7}x^{2-2}y^{3-2} \\ =8x^2y^3-7x^1y^4-5x^0y^1 \\ \text{Note that} \\ x^0=1 \\ x^1=x \\ y^1=y \\ 8x^2y^3-7x^1y^4-5x^0y^1 \\ =8x^2y^3-7xy^4-5^{}y^{} \end{gathered}$

Hope this Helps!!!