anonymous
  • anonymous
Can someone explain Polynomial Long Division to me?
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • 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
  • UsukiDoll
let me find an example somewhere brb
UsukiDoll
  • 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
  • UsukiDoll
so we need to rewrite this starting with the denominator outside and the numerator inside|dw:1437820660785:dw|
UsukiDoll
  • UsukiDoll
so now we have to start finding what variables we need to multiply x+1 with
UsukiDoll
  • 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
  • UsukiDoll
so \[\LARGE x^2+x\] becomes \[\LARGE -x^2-x \]
UsukiDoll
  • UsukiDoll
|dw:1437820910409:dw|
UsukiDoll
  • UsukiDoll
we have used all of our variables up, so the only thing remaining is the number
UsukiDoll
  • 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
  • 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
  • UsukiDoll
|dw:1437821086041:dw|
UsukiDoll
  • 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
  • UsukiDoll
so we have done this correctly
UsukiDoll
  • UsukiDoll
I'm going to give another example where we do have a remainder.
UsukiDoll
  • UsukiDoll
now suppose we are given \[\LARGE \frac{2x^3-9x^2+15}{2x-5}\]
UsukiDoll
  • UsukiDoll
whoa oh... we don't have an x... we just put a placeholder like 0x
UsukiDoll
  • UsukiDoll
|dw:1437821546765:dw|
UsukiDoll
  • 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
  • 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
  • UsukiDoll
|dw:1437821703230:dw|
UsukiDoll
  • 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
  • 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
  • UsukiDoll
|dw:1437821921260:dw|
UsukiDoll
  • 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
  • UsukiDoll
|dw:1437822060614:dw|
UsukiDoll
  • UsukiDoll
|dw:1437822081374:dw|
UsukiDoll
  • UsukiDoll
we have a remainder of -10... we need to write our answer a little bit differently
UsukiDoll
  • 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
  • UsukiDoll
so do my examples make this a bit easier to understand? do you have questions?
anonymous
  • 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
  • anonymous
Does the process differ at all once you get to the larger polynomials? @UsukiDoll
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
  • anonymous
Okay, no problem.
UsukiDoll
  • UsukiDoll
alright I'm back.. so is there any specific example that we can go over together?
UsukiDoll
  • UsukiDoll
like a polynomial divided by a trinomial? the process should be the same... just takes longer.
UsukiDoll
  • 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
  • UsukiDoll
a big heads up... this is going to take more than one drawing to do .
UsukiDoll
  • UsukiDoll
|dw:1437823307750:dw|
UsukiDoll
  • 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
  • UsukiDoll
|dw:1437823443863:dw|
UsukiDoll
  • 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
  • UsukiDoll
|dw:1437823607833:dw|
anonymous
  • 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
  • UsukiDoll
hmm I would do this one at a time though to avoid errors
UsukiDoll
  • 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
  • UsukiDoll
|dw:1437823911327:dw|
UsukiDoll
  • UsukiDoll
|dw:1437823935681:dw|
UsukiDoll
  • UsukiDoll
so our remainder is 2
UsukiDoll
  • UsukiDoll
\[\LARGE x^2-x+1+\frac{2}{x^2-2x-1}\]
UsukiDoll
  • UsukiDoll
sorry got those switched
UsukiDoll
  • UsukiDoll
\[\LARGE x^2-2x-1+\frac{2}{x^2-x+1}\]
UsukiDoll
  • UsukiDoll
the process is the same. It's just that the bigger the polynomial, the more time consuming and tedious it gets
alekos
  • 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.