please explain how to do this problem!?

Question

Use grouping to factor the polynomial
pg2 - pb2 - vg2 + vb2.

A. (g - b)(g - b)(p - v)
B. (g - b)(g - b)(p + v)
C. (g + b)(g + b)(p - v)
D. (g - b)(g + b)(p - v)

Answer 1

D is the correct answer
given expression
=p(g^2-b^2)-v(g^2-b^2)
=(g^2-b^2)(p-v)
={(g)^2-(b)^2}(p-v)
=(g-b)(g+b)(p-v)

Answer 2

When you have 4 terms like this you want to look at them and see what terms have something in common. For example the pg2-pb2 both have a p in them, also pg2 and vg2 both have a g2 in them so there are two different ways to do this problem depending on how you group the terms.

To keep is simple I'll just group them as they are written.
(pg2-pb2)+(-vg2+vb2) It is importent to keep the neg sign with the first v otherwise it may make your solution not match one of the choices. From here you just have to factor each part and then check to see if they can be factored again at all.

(pg2-pb2)+(-vg2+vb2)
p(g2-b2)-v(g2-b2)
Both terms have the same thing inside the parenthasis so you can factor that out.
(g2-b2)(p-v) now you have a difference of squares and a binomial
(g-b)(g+b)(p-v)

Answer 3

pg2 - pb2 - vg2 + vb2
=p(g^2-b^2)-v(g^2-b^2)
=(p-v)(g^2-b^2)
=(p-v)(g-b)(g+b)

D. (g - b)(g + b)(p - v)

Answer 4

pg2 - pb2 - vg2 + vb2 =
(pg² - vg²) + (vb² - pb²) =
g²(p - v) + b²(v - p) =
g²(p - v) - b²(p - v) =
(g² - b²)(p - v) =
(g + b)(g - b)(p - v) = D.

Answer 5

pg² - pb² - vg² + vb² =
p(g² - b²) - v(g² - b²) =
(p - v)(g² - b²) =
(p - v)(g + b)(g - b)

the 1st 2 terms have p in common; factor it out. the 2nd 2 terms have -v in common, factor it out.

then the 2 terms have (g² - b²) in common, factor that out.

finally factor (g² - b²) into product of sum and difference.

Answer 6

pg2-pb2-vg2+vb2
=p(g2-b2)-v(g2-b2)
=(g2-b2)(p-v)
=(g-b)(g+b)(p-v)
D is the answer.

Answer 7

=p(g^2-b^2)-v(g^2-b^2)
=(g^2-b^2)(p-v)
={(g)^2-(b)^2}(p-v)
=(g-b)(g+b)(p-v)