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### Course: Algebra (all content)>Unit 10

Lesson 23: Practice dividing polynomials with remainders

# Divide polynomials by x (with remainders)

Learn how to simplify complex expressions by dividing polynomials by 'x'. Discover how to break down the numerator, distribute the division, and use exponent properties. Master the art of simplifying fractions and handling negative exponents. Sal demonstrates by dividing (18x^4-3x^2+6x-4) by 6x. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• Shouldn't the two fractions at the end of the video have been added together to make a single fraction?
• Bobby The 1/2 times x has an x to the positive one power and the 2/3 x has a negative one power there for they do not qualify as like terms. Because the first x is the same as X/1 and the second is 1/x. Its the same as the difference between 100 and 1/100 :D Hope this helps
• Could you use synthetic division to solve this problem?
If yes, how?
• No, you cannot. In order to divide polynomials using synthetic division, the denominator (the number(s) on the bottom of the fraction) must satisfy two rules:
1 - Be a linear expression, in other words, each term must either be a constant or the product of a constant and a single variable to the power of 1.
2 - The leading coefficient (first number) must be a 1.
For example, you can use synthetic division to divide a polynomial by (x + 2) or (x – 6), but you cannot use synthetic division to divide by 6x, or (2x + 3) or (3x^2 – x + 3).
• At , Sal said, "anything divided by anything is just one." I think what he meant to say was "anything divided by itself is just one." :)
• Yeah,
Sal was referring to the first anything when he said anything the second time : )
• At the End of the video, he said we can write it as x^-1 power, but by the way of the polynomials it is not a polynomial because every degree in the expression must have a non-negative integer, so what's with that Am I wrong or I am missing something.
• Dividing 2 polynomials doesn't guarantee that your result will be a polynomial. This is just like if we divide 2 whole numbers, our result may or may not ge a whole number. For example: 17 divided by 5 = 3 2/5 or 3.4. The result is not a whole number. We get a rational number (one involving a fraction).

Sal's result is a rational expression (one that involves a fraction).

Hope this helps.
• what if instead of 6x on the bottom as the denominator its negative 6x how would you distribute it when doing the division
• Then you would simplify your equation by distributing the negative from the denominator (in your case the 6x) throughout its numerator.
So using Mr. Khan's example of the numerator being: 18x^4-3x^2+6x-4... and dividing it by its new denominator of -6x (your question), then you would multiply both the numerator and the denominator by -1, which would simplify the equation for you by making it a negative numerator divided by a positive denominator like this:
Numerator... -1(18x^4-3x^2+6x-4)= -18x^4+3x^2-6x+4.
& Denominator... -1(-6x)= +6x.

Hope that helped!
• At to , could the `-½x` also be written as `-x/2`?
• yes they can be because they are similar expressions
(1 vote)
• I heard in some videos when the power of x is raised to negative exponent it is not consider as polynomial or is this just some convention?
• That is correct, it is not a polynomial. By definition, the exponents of all the x's must be a nonnegative integer. If the exponent is negative or not an integer, then it is not a polynomial.

There are some more advanced definitions of "polynomial" that professional mathematicians work with, but for this level of study we can define a polynomial as a function that can be written (that doesn't mean it is currently written this way, but only that it is possible) as the sum of a finite number of terms that are composed of a constant multiplied by a variable raised to a nonnegative integer power. Note that some of these terms can have the constant multiplied by the variable raised to the zeroth power which would be equal to just the constant.
• when divide a polynomial why do we get a reminder that is 1 power less x^n than the divisor?
• At , Sal writes it as x^1 but it's just a remainder.
• Do you add or subtract the exponents?
• When multiply numbers with the same base, you add their exponents. When dividing numbers with the same base, you subtract the exponent of the denominator from the exponent of the numerator.
• While dividing the polynomial with the fifth degree power,2x^5-x^3 +3x^2-2x ÷ x-3, you added 0 because there was no x^4 but in the 18x^4 -3x^2+6x-4 ÷6x you didn't add 0 where X^3 was missing why?
• In the first example, the division was being done by more than one term, so that required long division.
The divisor in the second example was only a single term, so that could divide into each term of the given polynomial without resorting to long division.
(Be careful with parentheses. it is (x-3) doing the division in the first example, not just x.)

## Video transcript

Simplify the expression 18x to the fourth minus 3x squared plus 6x minus 4, all of that over 6x. So there's a couple ways to think about them. They're all really equivalent. You can really just view this up here as being the exact same thing as 18x to the fourth over 6x plus negative 3x squared over 6x, or you could say minus 3x squared over 6x, plus 6x over 6x, minus 4 over 6x. Now, there's a couple of ways to think about it. One is I just kind of decomposed this numerator up here. If I just had a bunch of stuff, a plus b plus c over d, that's clearly equal to a/d plus b/d plus c/d. Or maybe not so clearly, but hopefully that helps clarify it. Another way to think about it is kind of like you're distributing the division. If I divide a whole expression by something, that's equivalent to dividing each of the terms by that something. The other way to think about it is that we're multiplying this entire expression. So this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you-- they're are all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here, we could just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the-- well, they don't tell us. But if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the 4 minus 1 power, or x to the third power. Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1/2. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just-- well, I'll write it. I could write a 1 here. And let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. You could view it two ways-- anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1, which is x to the 0, which is also equal to 1. Either way, you knew how to do this before you even learned that exponent property. Because x divided by x is 1, and then assuming x does not equal to 0. And we kind of have to assume x doesn't equal 0 in this whole thing. Otherwise, we would be dividing by 0. And then finally, we have 4 over 6x. And there's a couple of ways to think about it. So the simplest way is negative 4 over 6 is the same thing as negative 2/3. Just simplified that fraction. And we're multiplying that times 1/x. So we can view this 4 times 1/x. Another way to think about it is you could have viewed this 4 as being multiplied by x to the 0 power, and this being x to the first power. And then when you tried to simplify it using your exponent properties, you would have-- well, that would be x to the 0 minus 1 power, which is x to the negative 1 power. So we could have written an x to the negative 1 here, but x to the negative 1 is the exact same thing as 1 over x. And so let's just write our answer completely simplified. So it's going to be 3x to the third minus 1/2 x plus 1-- because this thing right here is just 1-- and then minus 2 times 1 in the numerator over 3 times x in the denominator. And we are done. Or we could write this. Depending on what you consider more simplified, this last term right here could also be written in minus 2/3 x to the negative 1. But if you don't want a negative exponent, you could write it like that.