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⇒ s′(t)=4(2t−3)(8−2t)^3−6(2t−3)^2(8−2t)^2 how can i factor this? I keep factoring incorrectly :(

Mathematics
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Show me what you have done.
How many (2t-3) can you pull out?
I used 2 to factor since it goes in both 4 and -6

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Other answers:

1
ok so that is part of it
\[4(2t-3)(8-2t)^{3}-6(2t-3)^{2}(8-2t)^{2}\] \[(2t-3)(8-2t)^{2}[4(8-2t)-6(2t-3)]\] I'll stop here as to not go all the way, but maybe you can see what I was doing?
The problem is im not sure how to factor these anymore. i did a similar problem like this.. only they factored out an outer number and here: we factor one of the insides number. im so confused.
Do you follow what @psymon did?
yes but why can't we pull out a 2 in the outside since it' can go into 4 and -6?
ok I have dealt with this before ... it is tough to see. There really are only 2 terms in your original equation...do you see that?
Because I rather not do too much at once and worry about mistakes.
and no i didn't follow... this is where im stuck
But Psymon could have done that too.
yes i see
Look you actually only have two terms with several factors each. Hang on..
The first term is 4(2t-3)(8-2t)^3
That is one big term because it is composed of things being multiplied together (called factors)
ok
the factors of the first term are 4, (2t-3), and (8-2t)
right
So what would be the 2nd term? @mathcalculus
-6, (2t-3) and (8-2t)
Good so now think of factoring something easy... like 6x^2-15x ... you factor out the common factors and leave whatever is left ... right?
yes
In other words you factor out 3 and x and leave (2x-5) right?
correct
So now look at your equation ... what things and how many can you factor out of both terms
thats what i was saying... 2 goes in for 4 and -6..
Ok what other factors can you factor out?
i can factor (2t-3) and (8-2t)
Right how many 2t-3 can you factor out remember you can only factor out as many as the least number in either of the two terms.
1
right and how many 8-2t?
@Paynesdad right i know i can factor that too but what about the outer numbers. should i factor that firsT?
2
you can it shouldn't matter.
so factor out 2(2t-3)(8-2t)^2 and show me what you are left with
yes it does.... for ex::
see they factored out using the 4
That doesn't dispute what I was saying it doesn't matter if you factor out the 2 right now or not.
|dw:1375832092218:dw| This is just how I might go about doing these on paper. Looks like you're getting closer to understanding though, @mathcalculus ^_^
i want to know why can't i in this problem. which confuses so much.
You can factor out the 2...You can do it now or you can do it in the end ... either way you will end up doing it. I, personally, would do it in the beginning.,,
@Psymon is choosing to do it at the end...it will get done ... it's just a matter of when
so factor out 2(2t-3)(8-2t)^2 and show me what you have and I will tell you if you are right.
I'm personally too focused on the parenthesis terms and not making mistakes that I kinda just leave the coefficients til the end. Its dividing out those groups that worry me the most :P gtg now, though, good luck @mathcalculus
@Psymon got it^_^
that helped out alot. then what?
don't i need to finish?
Show me what you have. so far.
2(2t-3)(8-2t)^2[2(8-2t)-3(2t-3)
That looks good
now simplify wh
is in the backets
now simplify what is in the b
*brackets
what about the left side?
add them all up?
no don't just the stuff in the brackets
ook i did
i got [16-4t-6t+9]
Keep going
oh 16-10t+9
keep going
fine
:-)
-10t+7
errrrrr no re-think that last step
there's no where to go. now what about the left side.
oops 25
-10t+25
So now write it out again (yes the whole thing)
2(2t-3)(8-2t)^2[-10t+25]
Good now is there anything else that can be factored out now that you combined that last factor (notice you didn't multiply out that factor before just INSIDE of the parantheses of that last factor
let's say im loooking for the critical points.... i would still need to follow to this step right?
yes it is in deed helpful.
indeed*
the 2?
Nope there is something you can pull out of the last factor...
the (8-2t)^2?
no Look ---> (-10t+25)
oh i thought you meant only from looking at the left side.. :?
yes 2 *-5 =-10
There is no side it is just one big thing...
ok
well what are you factoring out of the last factor?
-5
ok so do that and combine it with the 2 in the front and re-write it one last time...promise.
2*-5(2t-3)(8-2t)^2[25]
Whoa whoa whoa
You
You factored out a -5 and had what left in the parantheses
you mean in the brackets t+25
No it was (-10t+25) if you just had this and factored out -5 what would the result be
2t-5
right welll you can't just throw that out...So now you should have all together...\[2(2t-3)(8-2t)^2(-5)(2t-5)\] right?
right
Now normally we combine our any numerical factors so we would bring the -5 to the front and multiply it by 2 and get -10
So the final answer is \[-10(2t-3)(8-2t)^2(2t-5)\]
So
Looking at the final answer ... how many terms are there?
3
nope remember terms are things that are separated by + or - and not in parantheses.
1
For example 5x+2y +16x^2 is three terms
right
Right only 1... Now how many factors in that 1 terms?
?
Factors are things that are multiplied together so 6xy has 3 factors.
7(x+1) (x-2) has 3 factors too
this is confusing me more.
3?
yep three things multiplied together... good
ok
for a small problem.. it's so long..:?
Well actually 4...-10, 2t-3, 8-2t, and 2t-5
okay 4.
...I know ... do you want to know why this form of the equation is better?
next
yeah
ok it has to do with this rule...if xy=0 what do you know?
idk
i have to keep going... it's getting too late for one problem:/
is it possible to show me so that i can how...
ask*
well If I am thinking of two numbers and I tell you that they multiply to 0 can you tell me what one of the numbers I am thinking of?
0
1and 0
right! so look at the original equation...if you set that to 0 you cant solve for t
omg obvious......
but if you get it into this form \[-10(2t-3)(8-2t)^2(2t-5)=0\]
it's the factoring i need help with. that's all
whats next after that.
ok I just wanted you to see how it helps you in actual calc problems but I understand...next? no next ...\[-10(2t-3)(8-2t)^2(2t-5)\] is completely factored you did it.!
are you sure? look how simple this person did this problem ds(t) / dt = 4 (2t - 3) (8 - 2t)^3 - 6 (2t - 3)^2 (8 - 2t)^2 = 0 ==> factoring: (2t - 3) (8 - 2t)^2 (4 (8 - 2t) - 6 (2t -3)) = 0 ==> (2t -3) (8 - 2t)^2 (50 - 20t) = 0 so the three critical points are: t = 3/2, 4, 5/2.
im looking for t right.
they've multiplied -10 to 2t-5
but why? the last one?
not the 1st, nor the 2nd?
Right that is exactly what you had but YOU factored the -10 out...they didn't.
Hang on you are thinking of distirbution
you don't distribute over multiplication !
this is taking forever.. for a small problem ___-
thank you for you help @Paynesdad really appreciate it.
5(79)(3)(2)...you can muliply the 5 times any one of those numbers but not all of them ...I would ppick the last one but we could multiply any of them just not all of them
No problem ... Good luck in tough subject but honestly
you have a better comprehension then 50% of the students I have had. @mathcalculus
*than

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