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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2each has a common factor of \(k\) so you can "factor it out"

lovinglyhated
 2 years ago
Best ResponseYou've already chosen the best response.0so i keep getting 16 but i'm not sure if thats right

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.0It's not correct. When you take out the common factor k, what is left for each term?

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0remember.. \[7k^2 = 7k \times k\]

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.0You cannot multiply the terms... One example: factor x^2 + 2x In this case, x is the common factor. Take it out and group the rest of the terms, that is x^2 + 2x = x (x) + 2(x) = x ( x+2) Can you try again for your question?

lovinglyhated
 2 years ago
Best ResponseYou've already chosen the best response.0so is k(7k+9) correct
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