Factor
4sˆ5+16s³-32s-20
4s^5 is suppose to be 4s exponent5 but i dont know how to do it

- anonymous

Factor
4sˆ5+16s³-32s-20
4s^5 is suppose to be 4s exponent5 but i dont know how to do it

- Stacey Warren - Expert brainly.com

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- anonymous

What have you done so far?

- anonymous

nothing

- anonymous

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

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## More answers

- anonymous

cf is 4

- anonymous

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

- anonymous

4, 8, 5

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

s

- anonymous

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

- anonymous

not sure how

- anonymous

show me step by step plz and then explain

- anonymous

How did you factor out the 4?

- anonymous

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

- anonymous

Right, but I mean what did you do?

- anonymous

Also that's not right

- anonymous

divide each term by 4

- anonymous

You keep forgetting the 4s^5

- anonymous

then what would it be

- anonymous

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

- anonymous

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

- anonymous

Or rather, like you said.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

idk im really bad with fractions

- anonymous

Then you should practice! =)
Everything depends on everything else. If you practice your fractions until you are not bad with them, factoring will be a very fast process.

- anonymous

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

- anonymous

16

- anonymous

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

- anonymous

rawr!

- anonymous

a^5

- anonymous

What is \(2^5\)

- anonymous

32

- anonymous

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

- anonymous

16/1

- anonymous

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

- anonymous

rawr! rawr!

- anonymous

16

- anonymous

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

- anonymous

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

- anonymous

9

- anonymous

And 9 is \(3^2\)

- anonymous

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

- anonymous

What is \(a^2\)

- anonymous

Or better still, \(a^3\)

- anonymous

a*a*a

- anonymous

Ok, now if we have
\[\frac{a\cdot a\cdot a}{a}\]
We can cancel one of the a's on top with the a on the bottom.
So what would that equal?

- anonymous

a^2

- anonymous

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

- anonymous

5

- anonymous

no.

- anonymous

then just s

- anonymous

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

- anonymous

a*a*a

- anonymous

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

- anonymous

a^2

- anonymous

And do you remember why?

- anonymous

because we cancel out 1

- anonymous

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

- anonymous

s^4

- anonymous

yes, precisely.

- anonymous

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

- anonymous

And if we divide s^3 by s?

- anonymous

s^2

- anonymous

And if we divide s by s?

- anonymous

s

- anonymous

s/s = s?

- anonymous

What is 2/2

- anonymous

1

- anonymous

What is 5/5

- anonymous

1

- anonymous

And w/w?

- anonymous

w

- anonymous

Why was 5/5 = 1?

- anonymous

s/s=1

- anonymous

why?

- anonymous

s goes into s 1 time

- anonymous

that's a good reason.

- anonymous

So now.

- anonymous

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

- anonymous

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

- anonymous

s^4+s^2-8

- anonymous

not quite.

- anonymous

But very close!

- anonymous

i hate this math

- anonymous

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

- anonymous

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

- anonymous

no

- anonymous

so where'd the 4 go when you factored?

- anonymous

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

- anonymous

Where'd that 5 come from?

- anonymous

the original equation

- anonymous

jumped the gun a bit

- anonymous

That 5 didn't have an s

- anonymous

So we cannot factor a s from it

- anonymous

no

- anonymous

Our original expression is
\[4( s^5 + 4s^3 -8s -5)\]
And you said that
\[s^5 + 4s^3 -8s = s(s^4 + 4s^2 -8)\]
Right?

- anonymous

yes what happens to the 5

- anonymous

We leave it alone. We are going to replace
\[s^5 + 4s^3 -8s\]
in the original expression with
\[s(s^4 + 4s^2 -8)\]
What does it look like?

- anonymous

?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Where'd the 4 out front go?

- anonymous

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

- anonymous

No. Take a second. Look at the original equation
\[4(s^5+4s^3-8s-5)\]
I want you to retype this equation with a [ and a ] around the terms with an s.

- anonymous

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

- anonymous

hrm. I think there's a vocab problem

- anonymous

what is a term?

- anonymous

each number 4s^5

- anonymous

close, but not quite

- anonymous

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

- anonymous

im really bad at math

- anonymous

I want you to know. You can be very good at math. You have the skills. You are a very hard worker clearly.

- anonymous

What is your goal?

- anonymous

it just doesnt make sense to me

- anonymous

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

- anonymous

Somewhere a long time ago, you were learning math. And there were other classes, and due dates, or other life events, and somewhere along the way you missed something here and there. Not a lot, just a little bit. But it was a little bit here, and a little bit there, and slowly over time these little bits add up. They are not hard things to learn. But now you are expected to know them and they are holes in your knowlege. And these holes are making it 100 times harder to learn this new material.

- anonymous

So lets fill a few holes quickly

- anonymous

do you mind?

- anonymous

I promise it will make math less hard

- anonymous

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

- anonymous

and don't want to give up on those goals

- anonymous

Believe me. You can learn this.

- anonymous

im majoring in acct

- anonymous

That's awsome!

- anonymous

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

- anonymous

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

- anonymous

Sounds like my story ;)

- anonymous

Except I graduated, but not with terribly good grades. Then screwed around for 15 years before finally going back to school

- anonymous

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

- anonymous

That's good!

- anonymous

yeah so how do i make this easier

- anonymous

So lets learn some math. =)

- anonymous

Starting with vocabulary

- anonymous

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

- anonymous

can we finish the problem we were working on first so i can close my homework this is the last problem

- anonymous

I was trying to.

- anonymous

But you weren't understanding what I was saying

- anonymous

k

- anonymous

So vocabulary.

- anonymous

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

- anonymous

Very simple I know

- anonymous

right

- anonymous

This sum has how many terms?

- anonymous

A term is just the things you are summing together.

- anonymous

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

- anonymous

3

- anonymous

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

- anonymous

4

- anonymous

How many different things are we adding together?

- anonymous

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

- anonymous

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

- anonymous

so 3

- anonymous

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

- anonymous

Now this is a product:
a*b

- anonymous

right

- anonymous

What are the factors of this product?

- anonymous

a
b

- anonymous

So how many factors are there?

- anonymous

2

- anonymous

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

- anonymous

difference of a sum

- anonymous

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

- anonymous

How many terms are in this sum?

- anonymous

2

- anonymous

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

- anonymous

5-3a
3a+8c

- anonymous

Why do you keep breaking it up with repeated parts?

- anonymous

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

- anonymous

no

- anonymous

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

- anonymous

5 - 3 + 2
Has how many terms?

- anonymous

2

- anonymous

(5) - (3) + (2)

- anonymous

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

- anonymous

How many things are being added together?

- anonymous

3

- anonymous

yes

- anonymous

5 - 4 + 2 + 6

- anonymous

How many terms?

- anonymous

so each number is a term

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

3 terms.

- anonymous

k

- anonymous

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

- anonymous

What is the second term of the above expression?

- anonymous

7b

- anonymous

right.

- anonymous

Ok, now going back to products

- anonymous

7b is a product

- anonymous

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

- anonymous

I think you're being trolled...

- anonymous

yes or 7(b)

- anonymous

Yes, exactly

- anonymous

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

- anonymous

2

- anonymous

good

- anonymous

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

- anonymous

product

- anonymous

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

- anonymous

sum

- anonymous

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

- anonymous

is 4b a sum or a product?

- anonymous

product

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

3,1

- anonymous

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

- anonymous

oh yeah sry

- anonymous

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

- anonymous

a
3

- anonymous

What are the factors of 4b again?

- anonymous

4
b

- anonymous

Ok, in the expression (the product)
4(a+3)
What are the factors (things being multiplied together to form the product)?

- anonymous

4
a
4
3

- anonymous

No.

- anonymous

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

- anonymous

4
(a+3)

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

(a+3)

- anonymous

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

- anonymous

a
3

- anonymous

Correct

- anonymous

Lets try this one

- anonymous

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

- anonymous

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

- anonymous

no.

- anonymous

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

- anonymous

oh ok i get it

- anonymous

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

- anonymous

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

- anonymous

product

- anonymous

yay!

- anonymous

What are the factors of this product?

- anonymous

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

- anonymous

yes!

- anonymous

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

- anonymous

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

- anonymous

s^5
4s^3
8s

- anonymous

Yes, now. Factor an s out of each of those terms, this will make a product where one factor is an s, and the other factor is the sum of the terms that you took the s from.

- anonymous

s+4^3-8

- anonymous

I'm not sure where to begin there. You were doing well up until that point, so something about my instructions must not have made sense.

- anonymous

i just dont understand what you mean

- anonymous

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

- anonymous

4(1+4+3)

- anonymous

yes! perfect

- anonymous

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

- anonymous

a(4+3+2)^3

- anonymous

What is the ^3 for?

- anonymous

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

- anonymous

i thought for the a's

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

So what is it?

- anonymous

s(^5 +4^3-8)

- anonymous

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

- anonymous

s^4

- anonymous

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

- anonymous

Yes

- anonymous

Now take the terms in the original equation that you factored that s from, and put square brackets [] around them

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

4s^5+16s^3-32s-20
this is original equation where do i put the brackets i got s from all except the -20

- anonymous

That is a very well thought out question. Thank you.
I would like you to go to the form of the expression where we factored out the 4 already. Then place the brackets such that they surround ONLY the terms (from the sum) that you took the s from, and no others.

- anonymous

Realize that you haven't actually factored the s out yet. I just want you do denote which terms you will be taking the s from by placing them in the brackets

- anonymous

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

- anonymous

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

- anonymous

Ok
This is the expression I want you to work with:
\[4(s^5 + 4s^3 - 8s -5)\]
Put square brackets around the terms that have s in them

- anonymous

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

- anonymous

yes!

- anonymous

Now, do you agree that
s^5+4s^3-8s = s(s^4 + 4s^3 - 8)
And that s^5+4s^3-8s is what is inside the brackets?

- anonymous

yes

- anonymous

Good. Then delete what is inside the brackets and replace it with the version where the s is factored out. You should not change anything not inside the brackets.

- anonymous

Bah!

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

where'd the s out in front go?

- anonymous

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

- anonymous

you changed something outside the brackets

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Sure, that is at least correct.

- anonymous

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

- anonymous

is that the right answer

- anonymous

almost. We still have more we can factor.

- anonymous

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

- anonymous

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

- anonymous

yes, except for the ) after the 8

- anonymous

sry forgot to take that out

- anonymous

Which two terms in this expression have an s still?

- anonymous

s^4
s^2

- anonymous

s^4 and 4s^2

- anonymous

The 4 is part of the term

- anonymous

yes forgot the 4

- anonymous

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

- anonymous

show me

- anonymous

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

- anonymous

2(s^2+4

- anonymous

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

- anonymous

yes ty

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

where did the s^2 out in front go?

- anonymous

And the parenthases?

- anonymous

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)]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes. exactly. the end.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

i typed in the answer we came up with and it is wrong the answer is 4(s^5+4s^3-8s-5) i had that 2 and 1/2 hrs ago

- anonymous

That's the problem with having a computer grade. The form we came up with is correct. You can factor it a lot of different ways. But "Factor" doesn't make it clear which bits to factor, or how far to factor.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes i did ty for everything

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