lalaland_lauren
  • lalaland_lauren
Determine the factors of 15x2 + 3xy + 10x + 2y. (3x + 2)(5x + y) (5x + y)(2x + 3) (3x + y)(5x + 2) (5x + 3)(2x + y)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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lalaland_lauren
  • lalaland_lauren
@satellite73
lalaland_lauren
  • lalaland_lauren
@vmac2122
misty1212
  • misty1212
HI!!

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More answers

lalaland_lauren
  • lalaland_lauren
Hello :)
misty1212
  • misty1212
we can multiply each one out and see what works, or perhaps we can work smarter
misty1212
  • misty1212
you see how the last term is \(2y\) right?
lalaland_lauren
  • lalaland_lauren
That's my problem, I get so confused multiplying everything out :( and yeah
misty1212
  • misty1212
that gets rid if b and d do you see why?
lalaland_lauren
  • lalaland_lauren
Yeah, that's actually a really cool loophole
misty1212
  • misty1212
\[ (5x + y)(2x + 3)\] last term will be \(3y\) so that is no good
misty1212
  • misty1212
same here \[ (5x + 3)(2x + y)\]
lalaland_lauren
  • lalaland_lauren
I figured it out, it's A
misty1212
  • misty1212
so we can try the first one \[ (3x + 2)(5x + y)\]
misty1212
  • misty1212
yeah that looks good
lalaland_lauren
  • lalaland_lauren
Yayyyyy I got it! :)
misty1212
  • misty1212
\[\huge \color\magenta\heartsuit\]
lalaland_lauren
  • lalaland_lauren
Thank youuuu!
misty1212
  • misty1212
you are quite welcome got more?
lalaland_lauren
  • lalaland_lauren
A few, yes hold on
misty1212
  • misty1212
kk
lalaland_lauren
  • lalaland_lauren
Graph of a cubic polynomial that falls to the left and rises to the right with graph touching x axis at negative 1 and x intercept at 1 Which of the following functions best represents the graph? f(x) = x3 + x2 − 4x − 4 f(x) = x3 + x2 − x − 1 f(x) = x3 + 3x2 − 4x − 12 f(x) = x3 + 2x2 − 2x − 1
lalaland_lauren
  • lalaland_lauren
1 Attachment
misty1212
  • misty1212
ok we can do this
misty1212
  • misty1212
you see how it only touches where \(x=-1\)?
lalaland_lauren
  • lalaland_lauren
Yeah, and where x=1
misty1212
  • misty1212
actually there is a difference, where \(x=1\) it CROSSES the axis, at \(x=-1\) it comes up, touches it, but does not cross, goes back down
misty1212
  • misty1212
so it has to zeros, one at \(x=-1\) and one at \(x=1\) because it crosses at \(x=1\) one factor is \((x-1)\)
misty1212
  • misty1212
because it touches at \(x=-1\) then the factor looks like \((x+1)^2\)
misty1212
  • misty1212
lol "has two zeros" i meant
misty1212
  • misty1212
final job is to multiply \[(x+1)^2(x-1)\] to see what we get i would cheat!!
misty1212
  • misty1212
http://www.wolframalpha.com/input/?i=%28x%2B1%29%5E2%28x-1%29
misty1212
  • misty1212
looks like \[x^3+x^2-x-1\] is a winner is that a choice?
lalaland_lauren
  • lalaland_lauren
Yup :) Wolfram Alpha has been helping me a lot this unit lol
misty1212
  • misty1212
might as well use it instead of making an algebra mistake i do!
lalaland_lauren
  • lalaland_lauren
Haha! I have just one more question and them I'm done lol
lalaland_lauren
  • lalaland_lauren
Which shows 602 − 402 being evaluated using the difference of squares method? 602 − 402 = (3600 + 1600)(3600 − 1600) = 10,400,000 602 − 402 = 3600 − 1600 = 2,000 602 − 402 = (60 − 40)2 = 202 = 400 602 − 402 = (60 + 40)(60 − 40) = (100)(20) = 2,000
lalaland_lauren
  • lalaland_lauren
The twos are squares
misty1212
  • misty1212
\[60^2-40^2\]?
lalaland_lauren
  • lalaland_lauren
Wait nvm I found the answer haha
misty1212
  • misty1212
D right?
lalaland_lauren
  • lalaland_lauren
Yup xD
lalaland_lauren
  • lalaland_lauren
I got an 88% :D
misty1212
  • misty1212
that it?
misty1212
  • misty1212
then time for a tune https://www.youtube.com/watch?v=smdNetoPFV0
lalaland_lauren
  • lalaland_lauren
I LOVE THAT SONG!!!!!
lalaland_lauren
  • lalaland_lauren
I love that whole band, actually!
misty1212
  • misty1212
lol thought so !!
lalaland_lauren
  • lalaland_lauren
How'd you know? XD
misty1212
  • misty1212
american psychic
lalaland_lauren
  • lalaland_lauren
*clapping emoji*
misty1212
  • misty1212
\[\color\red\heartsuit\]
lalaland_lauren
  • lalaland_lauren
Unfortunately I have to go to bed now :( I'm extremely tired and I'm pretty sure I have to get up early to keep working on assignments. Will you be on in the morning?
misty1212
  • misty1212
i can be here at about 9 or 9:30 earlier probably not
misty1212
  • misty1212
tag me if you like, i will try to help if i can!
misty1212
  • misty1212
my bed time too
lalaland_lauren
  • lalaland_lauren
That's a good time for me :) alrighty, thanks so much for helping me! good night c:
misty1212
  • misty1212
gnight c:

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