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What have you done so far?

nothing

Well, do you know how to find the factors of each term?

cf is 4

That is a start. Pull a factor of 4 from each term and see what you have left.

4, 8, 5

No, I mean write the new expression after you factor out a 4

4(4s^3-8s-5)

Not quite. You're missing a s^5, but otherwise that's good.

Now what other factors do some of your terms have in common?

Ok, so factor those out from the terms that have it

not sure how

show me step by step plz and then explain

How did you factor out the 4?

4(4s^3-8s-5)

Right, but I mean what did you do?

Also that's not right

divide each term by 4

You keep forgetting the 4s^5

then what would it be

4(s^5 +4s^3 - 8s - 5)

But as I said, you can factor an s from the first 3 terms inside the parens.

Or rather, like you said.

And you would do it the same way. Put an s out front, and divide each term by the s you took away.

Just don't take one away from the term that doesn't have an s to start with.

s(^5+4^3-8-5)

No. What is \[\frac{s^5}{s}\]

idk im really bad with fractions

Lets start now.
What is \[\frac{32}{2}\]

16

Ok, and if I said
\[\frac{2^5}{2}\]
?

rawr!

a^5

What is \(2^5\)

32

Interesting. And what was \[\frac{32}{2} \] again?

16/1

Right.. Just curious, but what is 2^4?

rawr! rawr!

16

That's interesting.
\[\frac{2^5}{2} = 2^4\]

Does this work with other numbers?
\[3^3 = 27\]
\[\frac{27}{3} =\ ?\]

And 9 is \(3^2\)

Lets see if we can figure out why this might be happening.

What is \(a^2\)

Or better still, \(a^3\)

a*a*a

a^2

Right.
So now, coming back to the original question..
What is \[\frac{s^5}{s}\]

no.

then just s

What was \[\frac{a^3}{a}\] again?

a*a*a

No, you just said it 6 mins ago. just scroll up

a^2

And do you remember why?

because we cancel out 1

So now. What is \[\frac{s^5}{s}\]
Use the same argument as before

s^4

yes, precisely.

So if we divide s^5 by s, we get s^4

And if we divide s^3 by s?

s^2

And if we divide s by s?

s/s = s?

What is 2/2

What is 5/5

And w/w?

Why was 5/5 = 1?

s/s=1

why?

s goes into s 1 time

that's a good reason.

So now.

thats why i said s i thought instead of putting 1 you put the variable

If we need to factor an s from s^5 +4s^3 - 8s what will our expression be?

s^4+s^2-8

not quite.

But very close!

i hate this math

\[s^5 + 4s^3 - 8s\]

factor out an s, does that effect the numbers out in front?

no

so where'd the 4 go when you factored?

s(s^4+4s^2-8-5)?

Where'd that 5 come from?

the original equation

jumped the gun a bit

That 5 didn't have an s

So we cannot factor a s from it

no

yes what happens to the 5

What is the original equation (after we factored out the 4)

4(s^5+4s^3-8s-5)

Right. So now take all those terms with an s, and replace them with the factored form you found.

s(s^4+4s^2-8)-5

Where'd the 4 out front go?

4(s^4+4s^2-8)-5

[4(s^5+4s^3-8s]-5)

hrm. I think there's a vocab problem

what is a term?

each number 4s^5

close, but not quite

I realize this has been a very lengthy process, but bear with me for a min

im really bad at math

What is your goal?

it just doesnt make sense to me

I know that, but it's not for the reasons you think.

So lets fill a few holes quickly

do you mind?

I promise it will make math less hard

And I assume that you have a reason for taking the math classes you are taking

and don't want to give up on those goals

Believe me. You can learn this.

im majoring in acct

That's awsome!

i drop out in 9th grade got my GED and somehow passed the math test

12yrs ago i dropped out and got GED 1yr ago

Sounds like my story ;)

i had my son dropped out and raised him and had my daughter and dont want to wait anymore

That's good!

yeah so how do i make this easier

So lets learn some math. =)

Starting with vocabulary

Vocabulary is important, because that way when I say something you can know exactly what I mean.

I was trying to.

But you weren't understanding what I was saying

So vocabulary.

First off, we have a sum.
This is a sum:
a + b + c

Very simple I know

right

This sum has how many terms?

A term is just the things you are summing together.

How many terms are in the sum:
a+b+c
?

And how many terms are in the sum
a*b + c +d*e

How many different things are we adding together?

a*b+c
c+d*e
so 2?

(a*b) + (c) + (d*e)

so 3

Yes. There are 3 things being added together. Those things are called terms.

Now this is a product:
a*b

right

What are the factors of this product?

a
b

So how many factors are there?

Ok now
5 - 3a + 8c
Is that a product, or a sum?

difference of a sum

I'd rather say it was a sum with one negative term

How many terms are in this sum?

How many different things are being added (and subtracted) together?

5-3a
3a+8c

Why do you keep breaking it up with repeated parts?

Is 5 - 3 + 2 = 5 - 3 and 3+2 ?

no

So what do you mean
5-3a
3a + 8c
?

5 - 3 + 2
Has how many terms?

(5) - (3) + (2)

Or if you prefer
(5) + (-3) + (2)

How many things are being added together?

yes

5 - 4 + 2 + 6

How many terms?

so each number is a term

no, each thing you are adding together as part of the sum is a term

a + b + c
Has no numbers, but still has 3 terms

so 4a+6b+8c is 3 or 6 terms

3 terms.

But!
4*3c + 7b + 4*12*7a
Also has 3 terms

What is the second term of the above expression?

7b

right.

Ok, now going back to products

7b is a product

you know that 7b is the same as 7*b yes?

I think you're being trolled...

yes or 7(b)

Yes, exactly

Ok, so how many factors are in the product 7b?

good

Ok, so what is \(7bc^2\) a sum, or a product?

product

Ok, here's a trickier one.
4(a + 3)
Sum or a product?

sum

If I tell you that b = a+3
Wouldn't that expression be 4b ?

is 4b a sum or a product?

product

If 4b is a product, then 4(a+3) must be a product, cause they're basically the same expression

ok

So if 4b is a product, what are its factors?

3,1

No. The factors of 4b are 4, and b

oh yeah sry

So if
b = a+3
What are the factors of the product
4(a+3)

a
3

What are the factors of 4b again?

4
b

4
a
4
3

No.

4(a+3) is
4 times (a+3)
What two things are being multiplied!!?

4
(a+3)

yes

The product 4(a+3) has 2 factors. They are 4 and (a+3)

One of those factors is a sum. Which one is a sum?

(a+3)

What are the terms of the sum (a+3)

a
3

Correct

Lets try this one

(x+1)(x-2)
A product? or a sum?

(x+1) sum
(x-2) difference

no.

It is a product.
(x+1)(x-2)
is
(x+1) TIMES (x-2)
It is two things multiplied together.

oh ok i get it

Each of those two things by themselves is a sum, but when you multiply them together it's a product.

What is
4(s^5 + 4s^3 - 8s -5)
A product? or a sum?

product

yay!

What are the factors of this product?

4
s^5+4s^3-8s-5

yes!

Now, the second factor has many terms with s in them

I want you to find each term with a factor of s

s^5
4s^3
8s

s+4^3-8

i just dont understand what you mean

4 + 16 + 12
Factor a 4 from each of those terms.

4(1+4+3)

yes! perfect

4a + 3a + 2a
Factor an a from each of those terms

a(4+3+2)^3

What is the ^3 for?

Why didn't you do the same thing you did for the 4?

i thought for the a's

It wasn't 4(1+4+3)^3

Just put an 'a' out front, and then divide each term by a.

ok

So what is it?

s(^5 +4^3-8)

No. We went over this before.
s^5 divided by s is?

s^4

s(s^4+4s^2-8)

Yes

[4(s^4+4s^3-8s-5]

a) that is not the original equation
b) the 4 out front is not one of the terms you took an s from

If you do not understand the instructions why don't you ask to clarify?

s(s^4+4s^2-8)
this is where i factored the s

just guessing is [4+4][2-8] the answer you are looking for?

4([s^5+4s^3-8s]-5)

yes!

yes

Bah!

You didn't check my work ;p
I made a mistake

It should be s(s^4 + 4s^2 - 8)

I had a 3 on the power of the middle term.

4([s^4+4s^2-8]-5)

where'd the s out in front go?

4s([s^4+4s^2-8]-5)

you changed something outside the brackets

so then explain where im suppose to put the s that was in front

Exactly where it was in the factored version.
s(s^4 + 4s^2 -8)

4(s[s^4+4s^2-8]-5)

Delete what's in the brackets. Take that whole expression, and put it in the brackets instead.

Sure, that is at least correct.

If you want, you can switch the brackets to parenthesis.

is that the right answer

almost. We still have more we can factor.

\[4(s (s^4 + 4s^2 - 8) - 5)\]
There are two terms that have an \(s^2\)

4(s[s^4 + 4s^2 -8)]-5)
is this what it should look like with the brackets

yes, except for the ) after the 8

sry forgot to take that out

Which two terms in this expression have an s still?

s^4
s^2

s^4 and 4s^2

The 4 is part of the term

yes forgot the 4

Ok, so you can factor an s^2 from each of those terms
s^4 + 4s^2 = s^2( ? + ? )

show me

How did you factor out the 4, or the s, or anything else you've factored so far?

2(s^2+4

I can only assume you made some typos there, but you meant to have
s^2(s^2 + 4)

yes ty

Ok, so replace those two terms (ONLY) with the newly factored version of the expression

The two terms in the big expression we are working on.

4(s[s^2+4-8]-5)

where did the s^2 out in front go?

And the parenthases?

s^4 + 4s^3 = s^2(s^2 + 4)
So take this [s^4 + 4s^3]
and put in this [s^2(s^2 + 4)]

Err I miss typed that exponent again, but you know what I mean.

s^4 + 4s^2 = s^2(s^2 + 4)
Take out [ s^4 + 4s^2 ]
Put in [ s^2(s^2 + 4) ]

4(s[s^2(s^2 + 4) -8]-5)

yes. exactly. the end.

There are no terms that have any common factors, so we cannot factor anything else out anywhere.

4(s[s^2(s^2 + 4) -8]-5) this is the final answer

Yes. Now I'm going to go do my math homework and go to bed before I have to get up in 5 hours.

At a minimum the instructions should have been to factor using the greatest common factor.

then at least you would know that the factorization you wanted would include all those terms.

and also you learned a lot more about factoring from the extra practice.

yes i did ty for everything