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anonymous
 one year ago
Can someone explain Polynomial Long Division to me?
anonymous
 one year ago
Can someone explain Polynomial Long Division to me?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.5let me find an example somewhere brb

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5ah suppose we are given this problem and we need to use long division\[\LARGE \frac{x^29x10}{x+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so we need to rewrite this starting with the denominator outside and the numerator insidedw:1437820660785:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so now we have to start finding what variables we need to multiply x+1 with

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so 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\[\LARGE x^29x\] our \[\LARGE x^2 \] should be gone

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so \[\LARGE x^2+x\] becomes \[\LARGE x^2x \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437820910409:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5we have used all of our variables up, so the only thing remaining is the number

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so based on what we are seeing we have a 10x10 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 10x10 and not have a remainder if I switch the sign of 10x+10 I will have 10x10 and that's something we don't want.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so I need to choose 10 so that 10 is distributed all over the x+1 10(x+1) = 10x10 switching those signs gives mean 10x+10 and that's what I need so I can cancel the 10x10 in the division.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437821086041:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so our final answer is (x+1)(x10) and 0 remainders expanding (x+1)(x10) with FOIL gives back \[\LARGE x^210x+x10 \rightarrow x^29x10\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so we have done this correctly

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5I'm going to give another example where we do have a remainder.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5now suppose we are given \[\LARGE \frac{2x^39x^2+15}{2x5}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5whoa oh... we don't have an x... we just put a placeholder like 0x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437821546765:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5our 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
 one year ago
Best ResponseYou've already chosen the best response.5so starting with variables only . I notice that there is a 2x^3 and we have 2x5 if I multiply \[\LARGE x^2(2x5) =2x^35x^2\] switching the signs gives me this \[\LARGE (2x^35x^2) => 2x^3+5x^2\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437821703230:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5our 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
 one year ago
Best ResponseYou've already chosen the best response.5so suppose we choose 2x then \[2x(2x5) => 4x^2\]... whoops if I change the sign I'll have a negative 4x^24x^2 = 8x^2 and that's a big no no ok we need 2x \[\LARGE 2x(2x5) => 4x^2+10x\] switching signs we have \[\LARGE (4x^2+10x) =4x^210x \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437821921260:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so 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 2x5 with 5 , so that when the signs are switched we have a 10x. \[\LARGE 5(2x5) = 10x+25\] switching signs \[\LARGE (10x+25) = 10x25 \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437822060614:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437822081374:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5we have a remainder of 10... we need to write our answer a little bit differently

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so we write the \[\LARGE x^22x5\] first the remainder \[\LARGE10\] will be on the numerator and \[\LARGE 2x5 \] will be on the denominator so the answer will look like \[\LARGE x^22x5+\frac{10}{2x5}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so do my examples make this a bit easier to understand? do you have questions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Does the process differ at all once you get to the larger polynomials? @UsukiDoll

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5hmmmm... 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5alright I'm back.. so is there any specific example that we can go over together?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5like a polynomial divided by a trinomial? the process should be the same... just takes longer.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5ah .. here's an example So suppose we have a longer polynomial divided by a trinomial like \[\LARGE \frac{x^43x^3+2x^2x+1}{x^2x+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5a big heads up... this is going to take more than one drawing to do .

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437823307750:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so now our goal to get rid of the x^4 so we need x^2 \[\LARGE x^2(x^2x+1) =x^4x^3+x^2\] switching signs \[\LARGE (x^4x^3+x^2) = x^4+x^3x^2\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437823443863:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so now our new goal is to get rid of the 2x^3 so we need to multiply 2x with x^2x+1 \[\LARGE 2x(x^2x+1) = 2x^3+2x^22x\] switching signs we have \[\LARGE 2x^32x^2+2x\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437823607833:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.5hmm I would do this one at a time though to avoid errors

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5anyway 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^2x+1 and switch signs \[\LARGE 1(x^2x+1) = x^2+x1\] switching signs \[\LARGE x^2x+1 \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437823911327:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437823935681:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5so our remainder is 2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5\[\LARGE x^2x+1+\frac{2}{x^22x1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5sorry got those switched

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5\[\LARGE x^22x1+\frac{2}{x^2x+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.5the process is the same. It's just that the bigger the polynomial, the more time consuming and tedious it gets

alekos
 one year ago
Best ResponseYou've already chosen the best response.0good work, this type of maths is not easy
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