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whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Can you rearrange that so it is a product? If you can, then set each component of the product = 0 and solve for x.

farmergirl411
 one year ago
Best ResponseYou've already chosen the best response.0um not totally sure

farmergirl411
 one year ago
Best ResponseYou've already chosen the best response.0can you solve it another way I am totally lost with this problem

farmergirl411
 one year ago
Best ResponseYou've already chosen the best response.0can you help please

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0What do you get if you multiply \[(x^24)(x3)\]?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Let's make a substitution to make it easier to see what is going on. Instead of \[x^2(x3)^2  4(x3)^2 = 0\]We'll let u = (x3) and write\[x^2u^24u^2 = 0\]Can you factor that at all?

ambius
 one year ago
Best ResponseYou've already chosen the best response.0We know that: \[= (\sqrt{x^2}x \sqrt{x^2}3)^2  (\sqrt{4}x3 \sqrt{4})^2\]

ambius
 one year ago
Best ResponseYou've already chosen the best response.0no substitutions necessary. : )

ambius
 one year ago
Best ResponseYou've already chosen the best response.0\[0 = (x^2  3x)^2  (2x  6)^2\]

ambius
 one year ago
Best ResponseYou've already chosen the best response.0\[(x^23x)^2 = (2x6)^2\] then root both sides, simplify, and solve for x.

valentinekid
 one year ago
Best ResponseYou've already chosen the best response.0@whpalmer4 had the right idea. Factor (x3)^2 out: \[(x^24)(x3)^2=0\]

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0@ambius no substitution was *necessary*, but people often have trouble spotting that they can factor out a binomial rather than a single term, until they've done it a few times. Spotting that one can factor out a single term is something that many do more easily. It's easy to see that \[x^2u^2  4u^2 =0\]can be factored to \[u^2(x^24)=0\]and then when you plug \(u = (x3)\) back in, you've got \[(x3)^2(x^24) = 0\]Yes, someone more experienced at factoring of this sort will notice directly that \[x^2(x3)^2  4(x3)^2 = 0\]factors to \[(x3)^2(x^24) =0\]
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