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we can multiply each one out and see what works, or perhaps we can work smarter
you see how the last term is \(2y\) right?
That's my problem, I get so confused multiplying everything out :( and yeah
that gets rid if b and d do you see why?
Yeah, that's actually a really cool loophole
\[ (5x + y)(2x + 3)\] last term will be \(3y\) so that is no good
same here \[ (5x + 3)(2x + y)\]
I figured it out, it's A
so we can try the first one \[ (3x + 2)(5x + y)\]
yeah that looks good
Yayyyyy I got it! :)
you are quite welcome got more?
A few, yes hold on
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
ok we can do this
you see how it only touches where \(x=-1\)?
Yeah, and where x=1
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
so it has to zeros, one at \(x=-1\) and one at \(x=1\) because it crosses at \(x=1\) one factor is \((x-1)\)
because it touches at \(x=-1\) then the factor looks like \((x+1)^2\)
lol "has two zeros" i meant
final job is to multiply \[(x+1)^2(x-1)\] to see what we get i would cheat!!
looks like \[x^3+x^2-x-1\] is a winner is that a choice?
Yup :) Wolfram Alpha has been helping me a lot this unit lol
might as well use it instead of making an algebra mistake i do!
Haha! I have just one more question and them I'm done lol
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
The twos are squares
Wait nvm I found the answer haha
I got an 88% :D
then time for a tune https://www.youtube.com/watch?v=smdNetoPFV0
I LOVE THAT SONG!!!!!
I love that whole band, actually!
lol thought so !!
How'd you know? XD
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?
i can be here at about 9 or 9:30 earlier probably not
tag me if you like, i will try to help if i can!
my bed time too
That's a good time for me :) alrighty, thanks so much for helping me! good night c: