Question

Factor the polynomial completely. 8x^4y – 16x^2y^2

Answer 1

Answer:

8\,x^2\,y\,(x-\sqrt{2} )(x+\sqrt{2} )

Step-by-step explanation:

Let's start by extracting all common factors from the two terms of this binomial. These common factors are: 8, $x^2$, and $y$.

The extraction renders:

$8\,x^2\,y\,(x^2-2)$

In the real number system, the binomial in parenthesis can still be factored out considering that 2 is the perfect square of $\sqrt{2}$, that is:

$2=(\sqrt{2} )^2$

We can then forwards with the factoring of this binomial using the factorization of a difference of squares as:

$(x^2-2) = (x^2-(\sqrt{2} )^2)=(x-\sqrt{2} )(x+\sqrt{2} )$

Thus giving the complete factorization as:

$8\,x^2\,y\,(x-\sqrt{2} )(x+\sqrt{2} )$

Answer 2

8x^4y – 16x^2y^2=

=8x^4y-16x^4y

=-8x^4y

=-8*(x^4y)