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anonymous
 one year ago
Part A: Factor 3x2y2 − 2xy2 − 8y2. Show your work. (4 points)
Part B: Factor x2 + 10x + 25. Show your work. (3 points)
Part C: Factor x2 − 36. Show your work. (3 points)
anonymous
 one year ago
Part A: Factor 3x2y2 − 2xy2 − 8y2. Show your work. (4 points) Part B: Factor x2 + 10x + 25. Show your work. (3 points) Part C: Factor x2 − 36. Show your work. (3 points)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i need steps i have answers! Part A.) y^2(3x+4)(x2) Part B.) (x5)^2 Part C.) (x+6)(x6)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry I cant help u on this but I love the name :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yw and good luck on ur question!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 3x^2y^2 − 2xy^2 − 8y^2\]What are we doing? :U Factoring this thing?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 3x^2\color{orangered}{y^2} − 2x\color{orangered}{y^2} − 8\color{orangered}{y^2}\]Hmm there isn't a whole lot that we can do with this first one. Notice that the last term has no x's, so we can't pull out any factor out any amount of x. The last two terms have a 2 in them (8 is 4 times 2) but the first term does not, so we can't pull a 2 out. You can do something with the y's though, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oh I guess we can factor the x's after that though :3 hah

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \color{orangered}{y^2}(3x^2−2x−8)\]So you ok with that first step? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Um (y2)(3x2+−2x+−8) =(y2)(3x2)+(y2)(−2x)+(y2)(−8) =3x2y2−2xy2−8y2 =3x2y2−2xy2−8y2

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2You just undid the first step :) Not sure why hehe

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am confused on this..

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440970256523:dwSo we're taking y^2 out of each term.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2And from there, we're going to factor our x's. That first step giving you trouble though? :d

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes, that's the answer. But we're not through the steps quite yet :) lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Since the x^2 has a coefficient, we'll have to try factoring by grouping for this one:\[\large\rm 3x^22x8\qquad\to\qquad ax^2+bx+c\]Multiplying a and c gives us 24, ya? So we need factors of 24 that will add to 2. Example: \(\large\rm 12 * 2 = 24\) But \(\large\rm 12 + 2\ne2\) Hmm so 12 and 2 are NOT the factors we're looking for.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\rm 3\cdot8=24\) But \(\large\rm 3+8\ne2\) Hmm looks like those factors don't work either :[

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Any other ones you can try? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2No no, we want numbers that multiply to 24.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.23 and 8 didn't work. 12 and 2 didn't work. Any other combinations we can try? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(12) *2= 24 \] OR \[12*(2)=24\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2No we already tried 12 and 2. Gotta try some other factors of 24

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\rm 6\cdot4=24\) Ok those work for multiplication. Let's try adding them. \(\large\rm 6+4=2\) Hmm I got a 2 instead of a 2, that simply means I Put the negative on the wrong number. So the factors we want are 6 and 4. \(\large\rm 6\cdot4=24\) \(\large\rm 6+4=2\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 3x^2\color{orangered}{−2x}−8\]So what we want to do is, split up this 2x into 6x and 4x.\[\large\rm 3x^2\color{orangered}{−6x+4x}−8\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2And from there, we can do some grouping.\[\large\rm (3x^2−6x)+(4x−8)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So uh, factor some stuff :O what can you pull out of each bracket?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2factor some stuff out of this: (3x^26x)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm (3x^2−6x)+(4x−8)\]\[\large\rm 3x(x−2)+(4x−8)\]Good good good :O Do something similar with the other set of brackets.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Mmm k, so you get a 4 from the other brackets, ya?\[\large\rm 3x(x−2)+(4x−8)\]\[\large\rm 3x(x−2)+4(x−2)\]And you go a step further and pull the (x2) from each term,\[\large\rm (x−2)(3x+4)\]Mmm k good, and ya, you get that. With the y^2 still in front.\[\large\rm y^2(3x+4)(x2)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2For Part C recall your difference of squares formula:\[\large\rm a^2b^2=(ab)(a+b)\]Notice that in the formula, both numbers are squared on the left side of the equation. So in order to apply this rule to our equation,\[\large\rm x^236\]We'll first need to rewrite 36 as a square somehow.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2ok great!\[\large\rm x^26^2\]Good. Now you can apply the formula:\[\large\rm x^26^2=(x6)(x+6)\]Nice easy one step problem. It's really just about remembering your formula for that one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you write like Part A Part B And Part C I am really confused what is what >_< lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2We skipped Part B, maybe that's the confusion. I gotta go make a sammich, so I was trying to rush through it >.< lol sorry The big long explanation was Part A. And then these last two comments I posted were Part C.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I really need part B too!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2you _ you and your ways lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2ok sec :3 lemme read it again

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm x^2 + 10x + 25\]So this is our typical factoring problem. Don't have a good grasp on these types yet? :o You look for `factors of 25` that will `sum to 10`. Lemme try a one and see what happens. \(\large\rm 25\cdot1=25\) \(\large\rm 25+1=26\) hmm, nope. they didn't add to 10.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Any other factors we can try?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2`factors of 25` is what you're looking for. \(\large\rm 25\cdot(15)\ne25\) Those are not factors of 25.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Example: If I was looking for factors of 24, i would list them like this maybe: 8 * 3 = 24 6 * 4 = 24 12 * 2 = 24 24 * 1 = 24 So those are all of the combinations I would want to try.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.225 ends in a 5, so it's 5 times something

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\rm 5\cdot5=25\) \(\large\rm 5+5=10\) Ah there we go! That's the one we were looking for.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(x5)(x5) or (x5)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(x5)(x5) or (x5)^2\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So for part B, you can either do factoring by grouping again:\[\large\rm x^2 + 10x + 25\]\[\large\rm x^2 + 5x+5x + 25\]\[\large\rm x(x + 5)+5(x + 5)\]\[\large\rm (x+5)(x + 5)\]\[\large\rm (x+5)^2\]Or just remember your shortcut for factoring. Yes. but not 5, +5 silly :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2The solution you listed at the start is incorrect. That's strange :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am confused whats part a and wahts part c?
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