Respuesta :

AL2006
(a). 
The product of two binomials is sometimes called FOIL.
It stands for ...

       the product of the First terms                (3j  x  3j)
plus
       the product of the Outside terms          (3j  x  5)
plus
       the product of the Inside terms            (-5  x  3j)
plus
       the product of the Last terms                (-5  x  5)

FOIL works for multiplying ANY two binomials (quantities with 2 terms).

Here's another tool that you can use for this particular problem (a).
It'll also be helpful when you get to part-c .

Notice that the terms are the same in both quantities ... 3j and 5 .
The only difference is they're added in the first one, and subtracted
in the other one.

Whenever you have     

              (the sum of two things) x (the difference of the same things)

the product is going to be

                 (the first thing)²  minus  (the second thing)² .

So in (a), that'll be      (3j)² - (5)²  =  9j² - 25 .

You could find the product with FOIL, or with this easier tool.
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(b).
This is the square of a binomial ... multiplying it by itself.  So it's
another product of 2 binomials, that both happen to be the same:

                            (4h + 5) x (4h + 5)  .

You can do the product with FOIL, or use another little tool:

The square of a binomial        (4h + 5)²    is ...

         the square of the first term               (4h)²
plus
         the square of the last term                (5)²
plus
         double the product of the terms      2 · (4h · 5)
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(c).
Use the tool I gave you in part-a . . . twice .

The product of the first 2 binomials is           (g² - 4) .

The product of the last 2 binomials is also    (g² - 4) .

Now you can multiply these with FOIL,
or use the squaring tool I gave you in part-b .

a. (3j - 5)(3j + 5)
    3j(3j + 5) - 5(3j + 5)
    3j(3j) + 3j(5) - 5(3j) - 5(5)
    9j² + 15j - 15j - 25
    9j² - 25

b. (4h + 5)²
    (4h + 5)(4h + 5)
    4h(4h + 5) + 5(4h + 5)
    4h(4h) + 4h(5) + 5(4h) + 5(5)
    16h² + 20h + 20h + 25
    16h² + 40h + 25

c. (g - 2)²(g + 2)²
    (g - 2)(g - 2)(g + 2)(g + 2)
    (g(g - 2) - 2(g - 2))(g(g + 2) + 2(g + 2))
    (g(g) - g(2) - 2(g) + 2(2))(g(g) + g(2) + 2(g) + 2(2))
    (g² - 2g - 2g + 4)(g² + 2g + 2g + 4)
    (g² - 4g + 4)(g² + 4g + 4)
    g²(g² + 4g + 4) - 4g(g² + 4g + 4) + 4(g² + 4g + 4)
    g²(g²) + g²(4g) + g²(4) - 4g(g²) - 4g(4g) - 4g(4) + 4(g²) + 4(4g) + 4(4)
    g⁴ + 4g³ + 4g² - 4g³ - 16g² - 16g + 4g² + 16g + 16
    g⁴ + 4g³ - 4g³ + 4g² - 16g² + 4g² - 16g + 16g + 16
    g⁴ - 12g² + 4g² + 16
    g⁴ - 8g² + 16