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Algebra (all content)
Course: Algebra (all content)ย >ย Unit 10
Lesson 23: Practice dividing polynomials with remainders- Divide polynomials by x (with remainders)
- Divide polynomials by monomials (with remainders)
- Divide polynomials by monomials (with remainders)
- Dividing polynomials with remainders
- Divide polynomials by linear expressions
- Divide polynomials with remainders
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Divide polynomials by monomials (with remainders)
CCSS.Math:
Sal divides (7x^6+x^3+2x+1) by X^2, and writes the solution as q(x)+r(x)/x^2, where the degree of the remainder, r(x), is less than the degree of x^2. Created by Sal Khan.
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Can someone tell me the parts of a polynomial like : degree (etc)(3 votes)
Degree is the highest exponent a variable is raised to in a given function. Take the function:
f(x) = x^3 + x^2 + x + 5
3 is the degree.
The variable is x (self-explanatory).
The constant term is 5 (basically the coefficient of the x^0 term).(5 votes)
im having trouble I need help(4 votes)
In this video professor Sal is explaining how to divide polynomials by monomials with reminders i.e.,
(4x^2+2x^2+x+1)/x^2 [my own example]
for this {(4x^2/x^2),(2x^2/x^2),[(x+1)/x^2]}
so 4x+2x+(x+1)/x^2 is the answer
"[(x+1)/x^2]}" this is because as in question they have asked to answer in the format - "q(x)+r(x)/x^2, where the degree of the remainder, r(x), is less than the degree of x^2"
Hope it helps(2 votes)
How are you getting 7xto the 6th power to 7x to the 4th power? I'm not understanding this at all...(2 votes)- Review your properties of exponents: https://www.khanacademy.org/math/pre-algebra/exponents-radicals/exponent-properties/v/exponent-rules-part-1
(x^7)/ (x^5) = x^(7-5) = x^2(6 votes)
- I looked in multiple places to find polynomial division with binomial's and remainders, but I couldn't find it anywhere, could someone please direct me to where to find this if it exists, and if not could someone tell me in full description what to do? I would be very grateful,
Many Thanks.(2 votes) - Why can't we use long division to solve this problem? And when do I know to use long division, or (this type) of division? Thanks in advance!(1 vote)
This question was answered before:why not use long division?
"Since this is a monomial we are dividing by, it is straightforward enough to do directly. It certainly is possible to do it long division style."(3 votes)
Isn't the remainder of the division supposed to be 2x-1 instead of 2x+1?
Why did sal write 2x+1/x^2 instead of 2x-1/x^2(1 vote)- No. The remainder is indeed 2x+1. You get to that remainder as Sal did, or you can perform a long division to be sure. But in both cases the remainder is 2x+1.(2 votes)
- x square + 3x square+3x +1 / x + pi. remainder equals to?(1 vote)
- 4pi^2-3pi+1. And the quotient will be 4x-4pi+3.
Or you made a mistake and actually meant for the first term in the numerator to be cubed. In that case, you get the remainder -pi^3+3pi^2-3pi+1 and the quotient x^2-xpi+3x+pi^2-3pi+3. If you look carefully at the remainder, it can be simplified to -(pi-1)^3
On both cases, as far as performing a division to simplify the expression goes, you are better of by just leaving the original expression untouched.
Also a note:
I reached both results above by performing a long division and double checking my results. The process may seem daunting at first, but you really just need to look at pi as if it was just another variable. Essentially performing long division as if you were dividing polynomials with xy terms. You just look at pi as being your y.(2 votes)
How would you do a problem on the exercises though? The vide is a bit different.(1 vote)- x^2-3x-10/x+2 Can you please show me the steps for long division(1 vote)
- At1:30why cant it be 2/x? if it is 2x/(x)(x), cant an x cancel out?
Thanks for the help(1 vote)- Why do you call it division of the exponents. when you subtract them?(1 vote)
1:30Video transcript
The quotient of two
polynomials-- a of x and b of x-- can be
written in the form a of x over b of x is equal to
q of x plus r of x over b of x-- where q of x and r
of x are polynomials and the degree of r of x is
less than the degree of b of x. Write the quotient
7x to the sixth plus x to the third plus 2x plus
1 over x squared in this form. Well, this one is
pretty straightforward because we're
dividing by x squared. So you could literally
view this as 7x to the sixth
divided by x squared plus x to the third divided
by x squared plus 2x divided by x squared plus 1
divided by x squared. So we could just do
this term by term. What's 7x to the sixth
divided by x squared? Well, x to the sixth divided by
x squared is x to the fourth. So it's going to be 7x
to the fourth power. And then, same thing
right over here. Plus x to the third
divided by x squared. Well, that's just going to be x. So plus x. And then, we're going to
have 2x divided by x squared. But remember, we
want to write it in a form of r of x over
b of x-- where r of x has a lower degree than b of x. Well, 2x has a lower
degree than x squared. Here this is degree 1. This is degree 2. So you could write it as
plus 2x over x squared. Like that. And then, you could write
plus 1 over x squared. So you could do this--
plus 1 over x squared. So you could write it like that. But that's not exactly
the form that they want. They want us to write it q
of x-- and you could view that as 7x to the fourth plus x. And then, they want plus
r of x over b of x So plus some polynomial over
x squared in this case. So instead of writing it as 2x
over x squared plus 1 over x squared, we could just write
it as 2x plus 1 over x squared. So one way to think about it. So let me just put
some parentheses here so that it interprets
my typing correctly. So notice, this part
of the polynomial, these terms have an equal or
higher degree than x squared. So I just divided those. 7x to the sixth divided by x
squared is 7x to the fourth. x to the third divided
by x squared is x. And then, once I got two terms
that had a lower degree than x squared, I just left on there. I just said plus whatever 2x
plus 1 divided by x squared is. And that's the form that they
wanted us to write it in. We'll check our answer. And we got it right.