If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder

- anonymous

If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder

- schrodinger

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- UsukiDoll

\[x^4+4x^3+2x^2+x+4\] by \[x^2+3x\]
have you made any attempts to this problem?

- anonymous

x4+4x3+2x2+x+4byx2+3x
=x4+4x3+2x2+x+4bx2y+3x
Combine Like Terms:
=x4+4x3+2x2+x+4bx2y+3x
=(4bx2y)+(x4)+(4x3)+(2x2)+(x+3x)
=4bx2y+x4+4x3+2x2+4x
Answer:
=4bx2y+x4+4x3+2x2+4x
right?

- UsukiDoll

whoa man.. direct answers are against openstudy's code of conduct

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## More answers

- UsukiDoll

and this is long division... I'm not doing that until the official poster responds

- anonymous

i have made attempts and none of the answer are there

- UsukiDoll

so you had tried the long division method to this problem?

- anonymous

thats not a direct answer i said it step by step and just to let you know i did this on the other post and they said that they under stood it completely now

- anonymous

yes sir i have. would you like to see the answer choices

- UsukiDoll

yeah I like to see the choices

- anonymous

A) x2 + 3x
B) 4x + 4
C) x2 + x – 2
D) 4
E) 0
That is all of the answer choices

- UsukiDoll

ok... so this problem is asking to divide... so we need long division.
|dw:1435149580706:dw|

- UsukiDoll

wow.. this is long.. so first of all we need to get rid of the x^4

- anonymous

yea it sucks i'm in Algebra 2.

- UsukiDoll

I've learned this in Algebra 1 O_O
anyway , are you aware of the exponential rules?

- anonymous

Don't you combine them? If its not that then I don't know about it

- UsukiDoll

one of the exponential rules is \[n^{a+b}=n^{a}n^{b}\]
let n = x and a = 2
\[x^{2+b}\] what does b need to equal in order to get \[x^4 \]?

- anonymous

2 if i'm not mistaking

- UsukiDoll

yes because \[x^{2+2} =x^2x^2\]

- UsukiDoll

so we need an x^2

- UsukiDoll

|dw:1435149984739:dw|

- anonymous

Okay. Inside of the sqrt there is a 2x^2

- UsukiDoll

we have to go slowly.. getting rid of the x^4 and x^3 is a great idea at this point

- anonymous

Alright so would we divide x^2+3x to x^4

- UsukiDoll

we need to multiply (x^2+3x) by x^2

- UsukiDoll

|dw:1435150150946:dw|

- anonymous

Okay so its going to be x^4+3x^3

- UsukiDoll

yeah...one problem we need to switch the signs.

- UsukiDoll

then we can use addition.

- UsukiDoll

let's just distribute the -
-(x^4+3x^3) = -x^4-3x^3

- UsukiDoll

|dw:1435150318360:dw|

- UsukiDoll

oh the plus sign is invisible for long division. I'll just put it in () |dw:1435150362975:dw|

- UsukiDoll

so what's x^4-x^4 and what's 4x^3-3x^3 ?

- anonymous

Okay so if we are trying to change the sign wouldn't it be x^8 and x^6

- UsukiDoll

whoa no...

- UsukiDoll

you only use the exponent rule to figure out what's missing. Like what I did earlier. with the x^4 = x^{2+b}

- anonymous

ok. sorry im not much of a math person. I love history

- anonymous

Oh Okay.

- UsukiDoll

|dw:1435150555870:dw|
let's try this again
what's x^4-x^4 and 4x^3-3x^3 ?

- anonymous

x and x. right.

- anonymous

If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder? Here is the full question

- UsukiDoll

......
let's make it easier
I'm going to factor out the
x^3
x^3(4-3)

- UsukiDoll

well we need to do long division...this is the right method

- UsukiDoll

or synthetic division which is faster but I forgot

- UsukiDoll

x^4(1-1)
x^3(4-3)
try solving those two ^

- anonymous

You forgot how to do synthetic division

- UsukiDoll

I've practiced more with long division...it stuck in my head.

- anonymous

Oh okay, and i got x^4-x^4 and 4x^3-3x^3

- UsukiDoll

wow @_@!

- UsukiDoll

try again...
x^4-x^4 is the same as x^4(1-1)
what's 1-1 ?

- anonymous

sorry i suck at math.

- anonymous

1-1=0

- UsukiDoll

yes and what's x^4(0) also known as what's x^4 multiplied by 0

- anonymous

0

- anonymous

cause anything that is multiplyed by 0 is 0

- UsukiDoll

yes! so the x^4's are gone!
SImilarly
what's x^3(4-3) ? what;s 4-3?!

- UsukiDoll

what's 4-3?

- anonymous

1

- UsukiDoll

yes and x^3 times 1 is?

- anonymous

x^3

- UsukiDoll

yay!

- anonymous

lol im getting it slowly but surely

- UsukiDoll

|dw:1435151077510:dw| so after we solve that we bring the rest of the terms down

- UsukiDoll

now our new goal is to get rid of the x^3

- UsukiDoll

remember the exponent rule?

- anonymous

na+b=nanb
let n = x and a = 2
x2+b
what does b need to equal in order to get
x4
that one

- UsukiDoll

\[x^{2+b} =x^2x^b\]

- UsukiDoll

yup now what value of b is needed to produce x^3?

- anonymous

1

- UsukiDoll

yes.

- anonymous

ok so it would be x^3-x^3

- anonymous

which would get rid of each other

- UsukiDoll

|dw:1435151215690:dw|

- UsukiDoll

now this time
x(x^2+3x) and switch signs after distributing

- anonymous

ok so it is 4x^3

- UsukiDoll

|dw:1435151316916:dw|

- UsukiDoll

?!?!?!! |dw:1435151357797:dw|

- anonymous

my bad. x^3-3x^2 Right.

- UsukiDoll

something is off.. are you sure the x^2 part of your original question is correct?

- anonymous

yes it is ill post the whole question in here

- anonymous

If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder?

- UsukiDoll

let's try continuing and see what happens..

- anonymous

Alright

- UsukiDoll

|dw:1435151602784:dw|

- anonymous

so the x^3 would cancel and it would be -x^2

- UsukiDoll

I see the question is asking for the remainder XD. we're not done xD we're doing it right heheeh
so what's x^3-x^3 and 2x^2-3x^2 ... yes x^3 is gone and we will have -x^2

- anonymous

so i did that part right

- UsukiDoll

|dw:1435151714443:dw|

- UsukiDoll

the last part is a digit, but it has to be negative

- UsukiDoll

because when switching the sign it will be positive.. and I have a negative -x^2 right at me

- anonymous

what do yo mean by that

- UsukiDoll

we've reached the end of the problem.. after all the exponents are used up, it's down to digits. like 1 2 3 4 ...

- UsukiDoll

there can only be 1 that will produce x^2

- anonymous

ok so the only one that can be right is 4 then. cause i have 0 and 4 as a answer

- UsukiDoll

0 is no remainder.. I doubt that will happen

- anonymous

A) x2 + 3x
B) 4x + 4
C) x2 + x – 2
D) 4
E) 0
Those are the answer

- UsukiDoll

|dw:1435151939146:dw|

- UsukiDoll

it's not going to be 0 . trust me.. I did the final step in my head.

- UsukiDoll

quick question what's -1(x^2) ?

- anonymous

-x^2 right

- UsukiDoll

yeah.. noticed that I needed the -1 to produce the -x^2-3x but I need to switch signs because we already have a -x^2 present!

- anonymous

ok so it becomes positive

- UsukiDoll

|dw:1435152046028:dw|

- UsukiDoll

yup you just need to compute x+3x and 4+0 that's your remainder

- anonymous

SO it would be 4

- UsukiDoll

oh I saw what happened I confused regular division with polynomial division. In polynomial division we can have bigger or smaller than the original numbers... can't have that in regular division xD (that was for the x^2 remark)
you're missing x+3x!!!!!!!!

- UsukiDoll

x(1+3) what's 1+3 ?

- anonymous

4

- UsukiDoll

yeah and 4(x) is ?

- anonymous

4x

- UsukiDoll

yeah with 4x and 4 combined together we have 4+4x

- UsukiDoll

|dw:1435152315128:dw|

- anonymous

Ok so thats the remainder

- UsukiDoll

anything left over on the bottom is the remainder.. what aloud did was way off.

- UsukiDoll

so that's done... yay x)

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