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whpalmer4
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0um not totally sure

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

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

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

whpalmer4
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0no substitutions necessary. : )

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

ambius
 2 years 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
 2 years 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
 2 years 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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