Question

$(14x^{2}yz-28x^2{y}^2{z}^3+32y^2{z}^2)÷(-4xy)$

• A. $\frac{7}{2}yz+7xy{z}^2+8xyz$
• B. $-\frac{7}{2}xz+7xy{z}^3-\frac{8y{z}^2}{x}$
• C. $-\frac{7}{2}xz-7xy{z}^3+\frac{8y{z}^2}{x}$
• D. $\frac{7}{2}xz-7xy{z}^2-\frac{8y{z}^2}{x}$

Answer 1

Answer: (B)

$-\frac{7}{2}xz+7xy{z}^3-\frac{8y{z}^2}{x}$

$(14x^{2}yz-28x^2{y}^2{z}^3+32y^2{z}^2)$ $÷$ $(-4xy)$ $=\frac{14x^{2}yz}{(-4xy)}-\frac{28x^2{y}^2{z}^3}{(-4xy)}+\frac{32y^2{z}^2}{(-4xy)}$
$=\frac{(-7)xz}{2}+7xy{z}^3+\frac{(-8)y{z}^2}{x}$