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kodGreya [7K]
3 years ago
13

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

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
8 0

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])

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both terms have a factor of 2 (the 2 out in front of them). They're the ones that get canceled when dividing by 2:

(2 (6<em>x</em> - 1) + 2 (2<em>x</em> + 5)) / 2 = 2/2 (6<em>x</em> - 1) + 2/2 (2<em>x</em> - 5)

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Looking ahead, it turns out that the equation is solved by <em>x</em> = 7. This makes 6<em>x</em> - 1 = 41 and 2<em>x</em> + 5 = 19. So the equation is saying that, if you make these replacements,

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