Question

What is the product?

Answer 1

Answer:

Option A is correct.

Step-by-step explanation:

We need to find the product of

$\frac{(x^2-16)}{(2x+8)} * \frac{(x^3-2x^2+x)}{(x^2+3x-4)}$

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 term

x(x^2-2x+1) = x(x-1)^2 = x(x-1)(x-1)

(x^2+3x-4) factorizing this term

x^2+4x-x-4 = x(x+4)-1(x+4)

= (x-1)(x+4)

Now, Putting these simplified terms in the given equation:

$\frac{(x-4)(x+4)}{2(x+4)}*\frac{x(x-1)(x-1)}{(x-1)(x+4)}$

Now cancelling the same terms that are in numerator and denominator

$=\frac{(x-4)}{2}*\frac{x(x-1)}{(x+4)}\\=\frac{(x-4)(x)(x-1)}{2(x+4)}\\=\frac{x(x-4)(x-1)}{2(x+4)}$

So, Option A is correct.

Answer 2

Answer:

=x(x-4)(x-1)/2(x+4)

Step-by-step explanation:

=x^2-4^2/2(x+4) * x^3-2x^2+x/x^2+3x-4

=(x+4)(x-4)/2(x+4) * x(x^2-2x+1)/x^2+3x-4

Factor x^2-2x+1 using the perfect square root

=(x+4)(x-4)/2(x+4) * x(x-1)^2/x^2+3x-4

Factor x^2+3x-4 using AC method.

=(x+4)(x-4)/2(x+4) * x(x-1)^2/(x-1)(x+4)

Cancel the common factor of x+4 and x-1

=(x-4)/2(x+4) * x(x-1)/1

=(x-4)x(x-1)/2(x+4)

Reorder the terms

=x(x-4)(x-1)/2(x+4)