At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

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

ok so that is part of it

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

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)

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 and how many 8-2t?

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.

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

so factor out 2(2t-3)(8-2t)^2 and show me what you have and I will tell you if you are right.

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]

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

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

So

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

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

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*

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.

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.

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

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

*than