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

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

Mathematics
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What have you done so far?
nothing
Well, do you know how to find the factors of each term?

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Other answers:

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?
s
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
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.
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? \[3^3 = 27\] \[\frac{27}{3} =\ ?\]
9
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
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?
a^2
Right. So now, coming back to the original question.. What is \[\frac{s^5}{s}\]
5
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 = s?
What is 2/2
1
What is 5/5
1
And w/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
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?
yes what happens to the 5
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?
?
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
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.
[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
I want you to know. You can be very good at math. You have the skills. You are a very hard worker clearly.
What is your goal?
it just doesnt make sense to me
I know that, but it's not for the reasons you think.
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.
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 ;)
Except I graduated, but not with terribly good grades. Then screwed around for 15 years before finally going back to school
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.
can we finish the problem we were working on first so i can close my homework this is the last problem
I was trying to.
But you weren't understanding what I was saying
k
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 ?
3
And how many terms are in the sum a*b + c +d*e
4
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?
2
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?
2
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?
2
(5) - (3) + (2)
Or if you prefer (5) + (-3) + (2)
How many things are being added together?
3
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.
k
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?
2
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
Ok, in the expression (the product) 4(a+3) What are the factors (things being multiplied together to form the product)?
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
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.
s+4^3-8
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.
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
Now take the terms in the original equation that you factored that s from, and put square brackets [] around them
[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?
4s^5+16s^3-32s-20 this is original equation where do i put the brackets i got s from all except the -20
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.
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
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?
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
4([s^5+4s^3-8s]-5)
yes!
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?
yes
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.
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.
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
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.
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

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