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Shtirlitz [24]
4 years ago
14

Find the illegal values of c in the multiplication statement c^2-3c-10/c^2+5c-14 times c^2-c-2/c^2-2c-15

Mathematics
1 answer:
Citrus2011 [14]4 years ago
3 0

Answer: c ≠ {-7, -3, 2, 5}

<u>Step-by-step explanation:</u>

  \frac{c^{2} - 3c - 10}{c^{2} + 5c - 14} * \frac{c^{2} - c - 2}{c^{2} - 2c - 15}

= \frac{(c - 5)(c + 2)}{(c + 7)(c - 2)} * \frac{(c - 2)(c + 1)}{(c - 5)(c +3)}

restrictions:

c + 7 ≠ 0           c - 2 ≠ 0          c - 5 ≠ 0          c + 3 ≠ 0

    c ≠ -7                 c ≠ 2               c ≠ 5                c ≠ -3

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