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

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

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

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

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

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.0
Let'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?
 one year ago

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

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

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

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

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

whpalmer4 Group TitleBest 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 \[(x3)^2(x^24) =0\]
 one year ago
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