What is the product?(x^2-16)/(2x+8) x (x^3-2x^2+x)/(x^2+3x-4)a. x(x-4)(x-1)/2(x+4)b. x(x-1)/2c. (x+4)(x-4)/2x(x-1)d. (x-4)(x-1)/2x(x+4)

Accepted Solution

Answer:Option A is correct.Step-by-step explanation:We need to find the product of[tex]\frac{(x^2-16)}{(2x+8)} * \frac{(x^3-2x^2+x)}{(x^2+3x-4)}[/tex]We know (a^2-b^2) = (a+b)(a-b)so, (x^2-16) = (x)^2-(4)^2 = (x-4)(x+4)2x+8 Taking 2 common from this term:2x+8 = 2(x+4)(x^3-2x^2+x) Taking x common from this termx(x^2-2x+1) = x(x-1)^2 = x(x-1)(x-1)(x^2+3x-4) factorizing this termx^2+4x-x-4 = x(x+4)-1(x+4)= (x-1)(x+4)Now, Putting these simplified terms in the given equation:[tex]\frac{(x-4)(x+4)}{2(x+4)}*\frac{x(x-1)(x-1)}{(x-1)(x+4)}[/tex]Now cancelling the same terms that are in numerator and denominator[tex]=\frac{(x-4)}{2}*\frac{x(x-1)}{(x+4)}\\=\frac{(x-4)(x)(x-1)}{2(x+4)}\\=\frac{x(x-4)(x-1)}{2(x+4)}[/tex]So, Option A is correct.