anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i need steps i have answers! Part A.) y^2(3x+4)(x-2) Part B.) (x-5)^2 Part C.) (x+6)(x-6)
anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
sorry I cant help u on this but I love the name :)

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anonymous
  • anonymous
ty :}
anonymous
  • anonymous
yw and good luck on ur question!
anonymous
  • anonymous
ty
anonymous
  • anonymous
yw
zepdrix
  • zepdrix
\[\large\rm 3x^2y^2 − 2xy^2 − 8y^2\]What are we doing? :U Factoring this thing?
zepdrix
  • zepdrix
\[\large\rm 3x^2\color{orangered}{y^2} − 2x\color{orangered}{y^2} − 8\color{orangered}{y^2}\]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?
anonymous
  • anonymous
Yes.
zepdrix
  • zepdrix
Oh I guess we can factor the x's after that though :3 hah
zepdrix
  • zepdrix
\[\large\rm \color{orangered}{y^2}(3x^2−2x−8)\]So you ok with that first step? :o
anonymous
  • anonymous
Um (y2)(3x2+−2x+−8) =(y2)(3x2)+(y2)(−2x)+(y2)(−8) =3x2y2−2xy2−8y2 =3x2y2−2xy2−8y2
zepdrix
  • zepdrix
You just un-did the first step :) Not sure why hehe
anonymous
  • anonymous
Opps >_<
anonymous
  • anonymous
I am confused on this..
zepdrix
  • zepdrix
|dw:1440970256523:dw|So we're taking y^2 out of each term.
zepdrix
  • zepdrix
And from there, we're going to factor our x's. That first step giving you trouble though? :d
anonymous
  • anonymous
y^2(3x+4)(x-2)
zepdrix
  • zepdrix
Yes, that's the answer. But we're not through the steps quite yet :) lol
anonymous
  • anonymous
Ok lol :]
zepdrix
  • zepdrix
Since the x^2 has a coefficient, we'll have to try factoring by grouping for this one:\[\large\rm 3x^2-2x-8\qquad\to\qquad ax^2+bx+c\]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 \(\large\rm -12 + 2\ne-2\) Hmm so -12 and 2 are NOT the factors we're looking for.
zepdrix
  • zepdrix
\(\large\rm 3\cdot-8=-24\) But \(\large\rm 3+-8\ne-2\) Hmm looks like those factors don't work either :[
zepdrix
  • zepdrix
Any other ones you can try? :)
anonymous
  • anonymous
Um... 2*-8=(-16)
anonymous
  • anonymous
\[2\times-8= -16\]
zepdrix
  • zepdrix
No no, we want numbers that multiply to -24.
anonymous
  • anonymous
OH OK.
zepdrix
  • zepdrix
3 and 8 didn't work. 12 and 2 didn't work. Any other combinations we can try? :o
anonymous
  • anonymous
\[(-12) *2= -24 \] OR \[12*(-2)=-24\]
zepdrix
  • zepdrix
No we already tried 12 and 2. Gotta try some other factors of 24
anonymous
  • anonymous
k
anonymous
  • anonymous
\[6•4\]
zepdrix
  • zepdrix
\(\large\rm 6\cdot-4=-24\) 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. \(\large\rm -6\cdot4=-24\) \(\large\rm -6+4=-2\)
anonymous
  • anonymous
^0^
zepdrix
  • zepdrix
\[\large\rm 3x^2\color{orangered}{−2x}−8\]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
  • zepdrix
And from there, we can do some grouping.\[\large\rm (3x^2−6x)+(4x−8)\]
anonymous
  • anonymous
=)
zepdrix
  • zepdrix
So uh, factor some stuff :O what can you pull out of each bracket?
anonymous
  • anonymous
3x2−2x−8
anonymous
  • anonymous
\[3x^2−2x−8\]
zepdrix
  • zepdrix
...?
zepdrix
  • zepdrix
factor some stuff out of this: (3x^2-6x)
anonymous
  • anonymous
3x(x−2)
zepdrix
  • zepdrix
\[\large\rm (3x^2−6x)+(4x−8)\]\[\large\rm 3x(x−2)+(4x−8)\]Good good good :O Do something similar with the other set of brackets.
anonymous
  • anonymous
\[(3x+4)(x−2)\]
zepdrix
  • zepdrix
Mmm 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 (x-2) from each term,\[\large\rm (x−2)(3x+4)\]Mmm k good, and ya, you get that. With the y^2 still in front.\[\large\rm y^2(3x+4)(x-2)\]
anonymous
  • anonymous
k
zepdrix
  • zepdrix
For Part C recall your difference of squares formula:\[\large\rm a^2-b^2=(a-b)(a+b)\]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,\[\large\rm x^2-36\]We'll first need to rewrite 36 as a square somehow.
anonymous
  • anonymous
ok maybe \[6^2\]
zepdrix
  • zepdrix
ok great!\[\large\rm x^2-6^2\]Good. Now you can apply the formula:\[\large\rm x^2-6^2=(x-6)(x+6)\]Nice easy one step problem. It's really just about remembering your formula for that one.
anonymous
  • anonymous
Can you write like Part A Part B And Part C I am really confused what is what >_< lol
zepdrix
  • zepdrix
We 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
  • anonymous
I really need part B too!
zepdrix
  • zepdrix
you -_- you and your ways lol
zepdrix
  • zepdrix
ok sec :3 lemme read it again
anonymous
  • anonymous
Thank yyou!
zepdrix
  • zepdrix
\[\large\rm x^2 + 10x + 25\]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
  • zepdrix
Any other factors we can try?
anonymous
  • anonymous
\[25+(-15)\]
zepdrix
  • zepdrix
`factors of 25` is what you're looking for. \(\large\rm 25\cdot(-15)\ne25\) Those are not factors of 25.
zepdrix
  • zepdrix
Example: 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.
anonymous
  • anonymous
There is not one.
zepdrix
  • zepdrix
25 ends in a 5, so it's 5 times something
anonymous
  • anonymous
Yes 5*5...
zepdrix
  • zepdrix
\(\large\rm 5\cdot5=25\) \(\large\rm 5+5=10\) Ah there we go! That's the one we were looking for.
anonymous
  • anonymous
Is that part B?
anonymous
  • anonymous
(x-5)(x-5) or (x-5)^2
anonymous
  • anonymous
\[(x-5)(x-5) or (x-5)^2\]
zepdrix
  • zepdrix
So 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
  • zepdrix
The solution you listed at the start is incorrect. That's strange :P
anonymous
  • anonymous
I am confused whats part a and wahts part c?
anonymous
  • anonymous
*whats

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