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

- 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 :(

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- anonymous

Show me what you have done.

- anonymous

How many (2t-3) can you pull out?

- anonymous

I used 2 to factor since it goes in both 4 and -6

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## More answers

- anonymous

1

- anonymous

ok so that is part of it

- 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

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

Do you follow what @psymon did?

- anonymous

yes but why can't we pull out a 2 in the outside since it' can go into 4 and -6?

- 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

Because I rather not do too much at once and worry about mistakes.

- anonymous

and no i didn't follow... this is where im stuck

- anonymous

But Psymon could have done that too.

- anonymous

yes i see

- anonymous

Look you actually only have two terms with several factors each. Hang on..

- anonymous

The first term is 4(2t-3)(8-2t)^3

- anonymous

That is one big term because it is composed of things being multiplied together (called factors)

- anonymous

ok

- anonymous

the factors of the first term are 4, (2t-3), and (8-2t)

- anonymous

right

- anonymous

So what would be the 2nd term? @mathcalculus

- anonymous

-6, (2t-3) and (8-2t)

- anonymous

@Paynesdad

- 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

yes

- anonymous

In other words you factor out 3 and x and leave (2x-5) right?

- anonymous

correct

- anonymous

So now look at your equation ... what things and how many can you factor out of both terms

- anonymous

thats what i was saying... 2 goes in for 4 and -6..

- anonymous

Ok what other factors can you factor out?

- anonymous

i can factor (2t-3) and (8-2t)

- 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

1

- anonymous

right and how many 8-2t?

- anonymous

@Paynesdad right i know i can factor that too but what about the outer numbers. should i factor that firsT?

- anonymous

2

- anonymous

you can it shouldn't matter.

- anonymous

so factor out 2(2t-3)(8-2t)^2 and show me what you are left with

- anonymous

yes it does.... for ex::

##### 1 Attachment

- anonymous

see they factored out using the 4

- anonymous

That doesn't dispute what I was saying it doesn't matter if you factor out the 2 right now or not.

- 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

i want to know why can't i in this problem. which confuses so much.

- 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

@Psymon is choosing to do it at the end...it will get done ... it's just a matter of when

- 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

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

@Psymon got it^_^

- anonymous

that helped out alot. then what?

- anonymous

don't i need to finish?

- anonymous

Show me what you have. so far.

- anonymous

2(2t-3)(8-2t)^2[2(8-2t)-3(2t-3)

- anonymous

That looks good

- anonymous

@Paynesdad

- anonymous

now simplify wh

- anonymous

is in the backets

- anonymous

now simplify what is in the b

- anonymous

*brackets

- anonymous

what about the left side?

- anonymous

add them all up?

- anonymous

no don't just the stuff in the brackets

- anonymous

ook i did

- anonymous

i got [16-4t-6t+9]

- anonymous

Keep going

- anonymous

oh 16-10t+9

- anonymous

keep going

- anonymous

fine

- anonymous

:-)

- anonymous

-10t+7

- anonymous

errrrrr no re-think that last step

- anonymous

there's no where to go. now what about the left side.

- anonymous

oops 25

- anonymous

-10t+25

- anonymous

So now write it out again (yes the whole thing)

- anonymous

2(2t-3)(8-2t)^2[-10t+25]

- 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

let's say im loooking for the critical points.... i would still need to follow to this step right?

- anonymous

yes it is in deed helpful.

- anonymous

indeed*

- anonymous

the 2?

- anonymous

Nope there is something you can pull out of the last factor...

- anonymous

the (8-2t)^2?

- anonymous

no Look ---> (-10t+25)

- anonymous

oh i thought you meant only from looking at the left side.. :?

- anonymous

yes 2 *-5 =-10

- anonymous

There is no side it is just one big thing...

- anonymous

ok

- anonymous

well what are you factoring out of the last factor?

- anonymous

-5

- anonymous

ok so do that and combine it with the 2 in the front and re-write it one last time...promise.

- anonymous

2*-5(2t-3)(8-2t)^2[25]

- anonymous

Whoa whoa whoa

- anonymous

You

- anonymous

You factored out a -5 and had what left in the parantheses

- anonymous

you mean in the brackets t+25

- anonymous

No it was (-10t+25) if you just had this and factored out -5 what would the result be

- anonymous

2t-5

- 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

right

- 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

So the final answer is \[-10(2t-3)(8-2t)^2(2t-5)\]

- anonymous

So

- anonymous

Looking at the final answer ... how many terms are there?

- anonymous

@mathcalculus

- anonymous

3

- anonymous

nope remember terms are things that are separated by + or - and not in parantheses.

- anonymous

1

- anonymous

For example 5x+2y +16x^2 is three terms

- anonymous

right

- anonymous

Right only 1... Now how many factors in that 1 terms?

- anonymous

?

- anonymous

Factors are things that are multiplied together so 6xy has 3 factors.

- anonymous

7(x+1) (x-2) has 3 factors too

- anonymous

this is confusing me more.

- anonymous

3?

- anonymous

yep three things multiplied together... good

- anonymous

ok

- anonymous

for a small problem.. it's so long..:?

- anonymous

Well actually 4...-10, 2t-3, 8-2t, and 2t-5

- anonymous

okay 4.

- anonymous

...I know ... do you want to know why this form of the equation is better?

- anonymous

next

- anonymous

yeah

- anonymous

ok it has to do with this rule...if xy=0 what do you know?

- anonymous

idk

- anonymous

i have to keep going... it's getting too late for one problem:/

- anonymous

is it possible to show me so that i can how...

- anonymous

ask*

- 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

0

- anonymous

1and 0

- anonymous

right! so look at the original equation...if you set that to 0 you cant solve for t

- anonymous

omg obvious......

- anonymous

but if you get it into this form \[-10(2t-3)(8-2t)^2(2t-5)=0\]

- anonymous

it's the factoring i need help with. that's all

- anonymous

whats next after that.

- 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

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

im looking for t right.

- anonymous

they've multiplied -10 to 2t-5

- anonymous

but why? the last one?

- anonymous

not the 1st, nor the 2nd?

- anonymous

Right that is exactly what you had but YOU factored the -10 out...they didn't.

- anonymous

Hang on you are thinking of distirbution

- anonymous

you don't distribute over multiplication !

- anonymous

this is taking forever.. for a small problem ___-

- anonymous

thank you for you help @Paynesdad really appreciate it.

- 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

No problem ... Good luck in tough subject but honestly

- anonymous

you have a better comprehension then 50% of the students I have had. @mathcalculus

- anonymous

*than

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