Here's the question you clicked on:
Grazes
Find the term that is a constant in the binomial expansion of...
\[(x ^{3}-\frac{ 2 }{ x ^{2} })^{5}\]
\[\left(\begin{matrix}5 \\ U\end{matrix}\right)(x ^{3})^{n-U}(-\frac{ 2 }{ x ^{2} })^{U}\]
Perfect. Now, were is \((x^{3})^{n-U}\cdot\left(\dfrac{1}{x^{2}}\right)^U = x^{0}\)
\[(-\frac{ x ^{2} }{ 2 })^{-U}\]?
nvm. I got it. Is -80 right?
We can worry about the constants later. We just need to know which termn it is for now. \((x^{3})^{5}\cdot (x^{-2})^{0} = x^{15}\) \((x^{3})^{4}\cdot (x^{-2})^{1} = x^{10}\) \((x^{3})^{3}\cdot (x^{-2})^{2} = x^{5}\) \((x^{3})^{2}\cdot (x^{-2})^{3} = x^{0}\) \((x^{3})^{1}\cdot (x^{-2})^{4} = x^{-5}\) \((x^{3})^{0}\cdot (x^{-2})^{5} = x^{-10}\)
But set 3(5-U)-2U to zero and solve for U. Then you can plug it back in and get -80 o-o
Fair enough. It's not perfectly clear to me what "find the term" means. Does it mean completely define it (-80) or just figure out which one it is (Term #4)? Excellent work!
It means to find the term that is a constant, rather than a coefficient and a variable(s)
That doesn't clear it up at all. If you find a rock under a flower, do you necessarily know what color the rock is? What if it's dark out? It may seem a little silly, but exceptional clarity is a bit more difficult than allowing various assumptions. Let's see how it goes in this household. Mom: Steve, will you help me find my car keys? Steve: Sure. Where have you looked? M: In the kitchen and the bedroom. S: Okay, I'll go check the living room. {a few moments pass} S: Found them! M: Where were they? S: They're under the couch. {a few moments pass} M: Are you bringing the keys to me? S: You said "find" them. I found them. They're still under the couch. Maybe Steve was just being funny, but that doesn't mean Mom was being particularly clear.