Question

Simplify: $\frac{49(x^4-2x^3-15x^2)}{14x(x-5)}$

• A. $\frac{7(x+3)}{2x}$
• B. $\frac{2x(x+3)}{7}$
• C. $\frac{7x(x+3)}{2}$
• D. $\frac{7(x+3)}{2}$

Answer 1

Answer: (C)

$\frac{7x(x+3)}{2}$

Given, $\frac{49(x^4-2x^3-15x^2)}{14x(x-5)}$

=$\frac{7x^2(x^2-2x-15)}{2x(x-5)}$

=$\frac{7x[x^2-5x+3x-15]}{2(x-5)}$

= $\frac{7x[x(x-5)+3(x-5)]}{2(x-5)}$

=$\frac{7x(x-5)(x+3)}{2(x-5)}$

= $\frac{7x(x+3)}{2}$