Can someone explain Polynomial Long Division to me?

- anonymous

Can someone explain Polynomial Long Division to me?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

I do very well with the division of a polynomial by a monomial, but once I have to divide by a binomial, I get very confused.

- UsukiDoll

let me find an example somewhere brb

- UsukiDoll

ah suppose we are given this problem and we need to use long division\[\LARGE \frac{x^2-9x-10}{x+1}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- UsukiDoll

so we need to rewrite this starting with the denominator outside and the numerator inside|dw:1437820660785:dw|

- UsukiDoll

so now we have to start finding what variables we need to multiply x+1 with

- UsukiDoll

so how do we produce that x^2 that we need in the numerator?
well we need an x
so \[\LARGE x(x+1) \rightarrow x^2+x\]
after we obtain this result, we need to switch the signs so when we add this to the\[\LARGE x^2-9x\] our \[\LARGE x^2 \] should be gone

- UsukiDoll

so \[\LARGE x^2+x\] becomes
\[\LARGE -x^2-x \]

- UsukiDoll

|dw:1437820910409:dw|

- UsukiDoll

we have used all of our variables up, so the only thing remaining is the number

- UsukiDoll

so based on what we are seeing we have a -10x-10
so there are negatives.
If I choose 10 then we will have 10(x+1) = 10x+10 but we're going to have a problem. Our goal is to cancel -10x-10 and not have a remainder
if I switch the sign of 10x+10
I will have -10x-10 and that's something we don't want.

- UsukiDoll

so I need to choose -10
so that -10 is distributed all over the x+1
-10(x+1) = -10x-10
switching those signs gives mean 10x+10 and that's what I need so I can cancel the -10x-10 in the division.

- UsukiDoll

|dw:1437821086041:dw|

- UsukiDoll

so our final answer is (x+1)(x-10) and 0 remainders
expanding (x+1)(x-10) with FOIL gives back
\[\LARGE x^2-10x+x-10 \rightarrow x^2-9x-10\]

- UsukiDoll

so we have done this correctly

- UsukiDoll

I'm going to give another example where we do have a remainder.

- UsukiDoll

now suppose we are given
\[\LARGE \frac{2x^3-9x^2+15}{2x-5}\]

- UsukiDoll

whoa oh... we don't have an x... we just put a placeholder like 0x

- UsukiDoll

|dw:1437821546765:dw|

- UsukiDoll

our x's in the numerator must increase to decrease and it needs to be present.. in this example the regular x was missing so a placeholder 0x is inserted

- UsukiDoll

so starting with variables only . I notice that there is a 2x^3 and we have 2x-5
if I multiply
\[\LARGE x^2(2x-5) =2x^3-5x^2\]
switching the signs gives me this
\[\LARGE -(2x^3-5x^2) => -2x^3+5x^2\]

- UsukiDoll

|dw:1437821703230:dw|

- UsukiDoll

our new goal is to get rid of the -4x^2
we have already used x^2 so we can get rid of the 2x^3
so now we need an x and a number

- UsukiDoll

so suppose we choose 2x
then \[2x(2x-5) => 4x^2\]... whoops if I change the sign I'll have a negative -4x^2-4x^2 = -8x^2 and that's a big no no
ok we need -2x
\[\LARGE -2x(2x-5) => -4x^2+10x\]
switching signs we have
\[\LARGE -(-4x^2+10x) =4x^2-10x \]

- UsukiDoll

|dw:1437821921260:dw|

- UsukiDoll

so now that we have used up all the variables we are down to numbers
our new goal is to get rid of -10x+15
if we have a negative we need to multiply 2x-5 with -5 , so that when the signs are switched we have a 10x.
\[\LARGE -5(2x-5) = -10x+25\]
switching signs
\[\LARGE -(-10x+25) = 10x-25 \]

- UsukiDoll

|dw:1437822060614:dw|

- UsukiDoll

|dw:1437822081374:dw|

- UsukiDoll

we have a remainder of -10...
we need to write our answer a little bit differently

- UsukiDoll

so we write the
\[\LARGE x^2-2x-5\] first
the remainder \[\LARGE-10\] will be on the numerator and \[\LARGE 2x-5 \] will be on the denominator
so the answer will look like
\[\LARGE x^2-2x-5+\frac{-10}{2x-5}\]

- UsukiDoll

so do my examples make this a bit easier to understand? do you have questions?

- anonymous

I understand more clearly now in terms of dividing a trinomial by a binomial, but once I get to larger polynomials, I run into trouble.

- anonymous

Does the process differ at all once you get to the larger polynomials? @UsukiDoll

- UsukiDoll

hmmmm... the only types of division (for polynomials) is dividing
a trinomial with a monomial and a trinomial with a binomial wait brb I have to do something that needs my attention immediately

- anonymous

Okay, no problem.

- UsukiDoll

alright I'm back.. so is there any specific example that we can go over together?

- UsukiDoll

like a polynomial divided by a trinomial?
the process should be the same... just takes longer.

- UsukiDoll

ah .. here's an example
So suppose we have a longer polynomial divided by a trinomial like
\[\LARGE \frac{x^4-3x^3+2x^2-x+1}{x^2-x+1}\]

- UsukiDoll

a big heads up... this is going to take more than one drawing to do .

- UsukiDoll

|dw:1437823307750:dw|

- UsukiDoll

so now our goal to get rid of the x^4
so we need x^2
\[\LARGE x^2(x^2-x+1) =x^4-x^3+x^2\]
switching signs
\[\LARGE -(x^4-x^3+x^2) = -x^4+x^3-x^2\]

- UsukiDoll

|dw:1437823443863:dw|

- UsukiDoll

so now our new goal is to get rid of the -2x^3
so we need to multiply -2x with x^2-x+1
\[\LARGE -2x(x^2-x+1) = -2x^3+2x^2-2x\]
switching signs we have
\[\LARGE 2x^3-2x^2+2x\]

- UsukiDoll

|dw:1437823607833:dw|

- anonymous

I see, so it's still the same process for the most part. However, I can see how it all aligns when you're dividing polynomials with a dividend and divisor only 1 term in difference, but some of my class problems are 2, sometimes even 3 terms difference.

- UsukiDoll

hmm I would do this one at a time though to avoid errors

- UsukiDoll

anyway we have used up all of our variables and we are left with numbers. since we have a -x^2, we need to multiply -1 with x^2-x+1
and switch signs
\[\LARGE -1(x^2-x+1) = -x^2+x-1\]
switching signs
\[\LARGE x^2-x+1 \]

- UsukiDoll

|dw:1437823911327:dw|

- UsukiDoll

|dw:1437823935681:dw|

- UsukiDoll

so our remainder is 2

- UsukiDoll

\[\LARGE x^2-x+1+\frac{2}{x^2-2x-1}\]

- UsukiDoll

sorry got those switched

- UsukiDoll

\[\LARGE x^2-2x-1+\frac{2}{x^2-x+1}\]

- UsukiDoll

the process is the same. It's just that the bigger the polynomial, the more time consuming and tedious it gets

- alekos

good work, this type of maths is not easy

Looking for something else?

Not the answer you are looking for? Search for more explanations.