how do you factor this: -6x^4-12x^3+18x?
ok, so what's something they all have in common?
6
great, and...
and X?
yep, so you could all divide them by 6x.-6x^2 would be better, to get rid of the negatives. just an idea. so you would then have: -6x^2(x^2+2x-18)
i decided to divide it by -6^2, because we want our first value to be x^2, not x^3. we must also remember that the -6x^2 didn't just poof away. so we need to keep that. next we're going to factor for x^2+2x-18.
ok
do you think you can handle it from there, or do you want me to keep on going?
so when we factor x^2 + 2x - 18 the -6x^2 stays to the side?
do u mind helpin me with the next step?
so far i have (for factoring x^2 +2x-18) x(x )(x )....do u have to find 2 numbers that equal 18?
okay, im back sorry
yes and yes. you can just ignore the -6x^2 for now, and add it in the end. and you do need to find two numbers that equal -18, not just 18.
so first start of by list multiples of -18
ok cool... -6 x 3 = -18
so now i have x (x - 6) (x + 3)
ok, so now multiply it out and see if you get x^2 +2x-18
wait! i just noticed a huge mistake!
ok
we forgot to divide 18 by -6. so you need to factor for x^2+2x-3
i hope i didnt confuse you!
so now find factors for -3. this should be easier
we could use -1 and 3: (x-1)(x+3)
sry my computer froze ...im lost now
so i shouldnt of used -18?
when we multiply that out, we get : x^2+2x-3. which means this is right. remember how we divided -6x^2? well this is when we put it back. so the final factored answer is: -6x^2(x^2+2x-3)
but where did the (x^2+2x-3) come from?
nevermind...I figured it out! lol ...i strolled back up in the convo
thank u so much...i just have 1 more quick question if u dnt mind
ok, remeber when we divided -6x^2? we forgot to divide that on -18x.
ok, lol. sorry for confusing you! i confused myself too. my cp kept freezing
my comp was freezing too!
Um...I think you factored -6x^2 out of 18x.
when factoring 27^3 - 48x ... being that there is no common numbers between the both of them...what would be the 1st step?
27^3 or 27x^3
im sry 27x^3
well then they can both be divided by x
ohh! wait a sec. what grade are you in?
so then i would have x(x )(x )...and then do i find #'s for 48?
im in 12th grade
have you learned about completing the square?
whoa! im teaching a 12 grader. sorry. im a freshie_
it doesn't sound familiar ... im really bad in math, its like another language to me!
lol well u did a good job helpin me on the 1st 1 =)
no prob, glad to help anytime
so the answer is : -6x^2(x^2+2x-3) for the first one
yea i got that 1
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