Question

Find the value of: $\frac{x^4-81}{ 2x2+13x+15 }\times\frac{2x+3}{x^3+27}\times\frac{[(x-3)^2+3x](x+5)}{(x+3)^2-6x}$

• A. (x2 - 9)
• B. (x – 3)
• C. (x + 5)
• D. (2x + 3)

Answer 1

Answer: (B)

(x – 3)

Answer 2

The correct option is (B): (x – 3).
$\frac{x^4-81}{ 2x2+13x+15 }\times\frac{2x+3}{x^3+27}\times\frac{[(x-3)^2+3x](x+5)}{(x+3)^2-6x}$
$\frac{(x^2-9)(x^2+9)}{(x+5)(2x+3)}\times\frac{2x+3}{(x+3)(x^2+9-3x)}\times\frac{[x^2+9-6x+3x](x+5)}{x^2+9+6x-6x}{}$
$\frac{(x-3)(x+3)(x^2+9)}{1}\times\frac{1}{(x+3)(x^2+9-3x)}\times\frac{(x^2+9-3x)}{(x^2+9)}$
=(x-3).