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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

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