anonymous
  • anonymous
⇒ 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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Show me what you have done.
anonymous
  • anonymous
How many (2t-3) can you pull out?
anonymous
  • anonymous
I used 2 to factor since it goes in both 4 and -6

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

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