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Ipatiy [6.2K]
3 years ago
9

Help? Please. I really need help

Mathematics
1 answer:
tia_tia [17]3 years ago
3 0

Answer:

D

Step-by-step explanation:

Any other product in Quadrant IV will be negative also, so Quadrant IV is part of my answer. The points (x, y), with xy < 0, lie in Quadrants II and IV.

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c^22c^2 + 13c +21 \\4c^2 + 13(2)c + 42\\(2c + 6)(2c+7)\\(c+3)(2c+7)


This is a way to factoring trinomials (there exist different equivalent methods).

Multiply the trinomial but the term accompanying  c^2. This is the second line. Then, you could take the square of the 4c^2, ant try to create a factor () () that will correspond to the expression in the second line. That is, we want 4c^2 + 13(2) c + 42 = (2c + ?) (2c + ?)

In ? we put the corresponding numbers that, if we multiply them we will obtain 42, and if we add them we will obtain 13. This numbers are 6 and 7. Then, we have (2c + 6) (2c +7)

The last step is divide by the number that we multipy in the first step.

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