Divide using long division. Write the result as dividend = (divisor)(quotient) + remainder. (x^3

Divide using long division. Write the result as dividend =...

Key Concepts

Polynomial Long Division

A systematic method for dividing one polynomial by another, closely resembling numerical long division. This technique is used to obtain a quotient and a remainder when the dividend does not evenly divide by the divisor.

Dividend

The polynomial that is being divided. It is the initial expression from which parts are successively subtracted during the division process.

Quotient

The result obtained from the division process. It represents the polynomial that, when multiplied by the divisor and added to the remainder, reconstructs the original dividend.

Remainder

The final leftover polynomial after completing the division, having a lower degree than the divisor. It is the part of the dividend that cannot be evenly divided by the divisor within the domain of polynomials.

Divisor

The polynomial by which the dividend is divided. Its degree generally influences how the division is carried out, and it appears as the subtractor in the iterative steps of the long division.

Transcript

00:01 The problem we're going to solve today is this problem here on the left.

00:03 They are two polynomials and we're going to divide them and actually have them labeled so you guys don't get confused on which is which.

00:11 And the way that we're going to solve this is by using long division.

00:15 So remember, there is synthetic division and then there's long division and we're going to use long division.

00:21 And this is how essentially you'd be setting up a long division problem.

00:26 Now, the goal of this video is to format it in this form.

00:31 Right here.

00:35 We have our dividend, we have our divisor, and our quotient, if you guys don't know, will be the actual answer that we get and at the very end we just attach the remainder.

00:47 So let's get started.

00:48 So i've already shown you how to set up the problem under the section of method of solving.

00:54 And this is the actual problem set up.

00:58 So we have x minus 2 on the outside, which is our divisor, and on the inside we have dividend.

01:06 Now right off the bat, our first step is going to be to find our multiplier.

01:16 And the way to find our multiplier is that we take the first term of the dividend, which in this case is x cubed.

01:25 And we actually divide that by our first term in our divisor, which is x.

01:32 If we solve this problem, we get x squared.

01:38 And you guys can see that the x on the bottom cancels one of the x is on the top.

01:43 Now, the whole point of a multiplier is so that we find a term that multiplies by our divisor in order to get the first two terms of our dividend.

01:55 So just watch this.

01:57 Now that we have found our multiplier, that becomes part of our solution, or in other words, our quotient.

02:05 Once we have our multiplier, we can move on to the next step.

02:08 So step two is to multiply.

02:16 Our multiplier by our divisor.

02:25 And we're going to do that to both terms of the divisor.

02:29 And just to clear up some confusion, it's the same thing as when we're dividing two numbers.

02:35 So let's say we're dividing 5 into 255.

02:38 We know that in order to solve this problem, we need to multiply 5 by 5 in order to get this first number, 25, and then we subtract them.

02:49 We get 0 and 0, we bring down the 5, and now we have to find our multiplier, and in this case our multiplier would be 1.

02:58 5 by 1 is equal to 5.

03:01 And boom, we get a reminder of 0 in this case.

03:03 But you can see how this relates to this problem.

03:06 What we're going to do is now do that same process, but now with polynomials.

03:11 So now we can start off with x squared multiplied by x, which is equal to x cubed, and x squared times negative 2, which is equal to negative 2, x.

03:20 Squared.

03:23 Now our next step will be to subtract, but unlike the example here that i provided, we only have one number, but the problem that we're actually solving has, you know, has two terms...

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