Question

25 (a square) - 4 (b square) + 28bc - 49 (c square)

Answer 1

Answer:

(5a - [2b - 7c]) and (5a + [2b + 7c])

Step-by-step explanation:

Factor 25a^2 - 4b^2 + 28bc - 49c^2.

Note that - 4b^2 + 28bc - 49c^2 involves the variables b and c, whereas 25a^2 has only one variable.  Thus, try to rewrite - 4b^2 + 28bc - 49c^2 as the square of a binomial:

- 4b^2 + 28bc - 49c^2 = -(4b^2 - 28bc + 49c^2), or

                                        -(2b - 7c)^2.

Thus, the original  25a^2 - 4b^2 + 28bc - 49c^2  looks like:

                               [5a]^2 - [2b - 7c]^2

Recall that a^2 - b^2 is a special product, the product of (a + b) and (a - b).  Applying this pattern to the problem at hand, we conclude:

Thus,   [5a]^2 - [2b - 7c]^2 has the factors (5a - [2b - 7c]) and (5a + [2b + 7c])