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Class 10 (Foundation)
Course: Class 10 (Foundation) > Unit 3
Lesson 3: Division of polynomialsDividing polynomials: long division
Sal divides (x^2-3x+2) by (x-2) and then checks the solution. Created by Sal Khan and Monterey Institute for Technology and Education.
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- Is there a video on synthetic division.(22 votes)
- Sal has made some videos about it now:
http://www.khanacademy.org/math/algebra/polynomials/v/synthetic-division(36 votes)
- Hi, I'm stuck on this question from school...
When P(x) = x^4 + ax^2 + bx + 2 is divided by x^2 +1 the remainder is -x+1. Find the values of a and b.
Thank you!(7 votes)- If x⁴ + ax² + bx + 2 divided by (x²+1) leaves a remainder of -x + 1, then:
(x²+1) divides x⁴ + ax² + bx + 2 - (-x +1) = x⁴ + ax² + (b + 1) + 1 exactly.
Since (x²+1) = (x + i)(x - i) this tells us (x - i) also divides x⁴ + ax² + (b + 1)x + 1 and, by the Polynomial Remainder Theorem, i is a zero.
Substituting x = i in to x⁴ + ax² + (b + 1)x + 1 = 0 gives:
1 - a + (b + 1)i + 1 = 0
And by comparing real and imaginary parts we get a = 2, and b = -1(19 votes)
- How do you divide a monomial by a polynomial? Thank you!!(5 votes)
- With the monomial in the numerator, your only option is to factor the denominator and cancel out any common factors shared with the numerator.(15 votes)
- What do you call a 9 side polygon?(5 votes)
- As Ellie said, it's a nonagon.
Alternatively, especially for polygons with a LOT of sides, you can just call them a n-gon, with n being the number of sides. So a less common way to call a nonagon is a 9-gon, while a 129 sided polygon would just normally be called a 129-gon.(4 votes)
- Could you use (x^2/x-2) - (3x/x-2) + (2/x-2). I can't seem to find a way to divide a monomial over a polynomial. Is it possible?(5 votes)
- Well, they have the same denominator, so you have (x^2 -3x +2)/(x-2). Then just model what they do in this video.
OR, you can notice that (x^2 -3x +2) = (x-2)(x-1), so:
(x^2 -3x +2)/(x-2) = (x-2)(x-1)/(x-2) =x-1 (where x != 2)(3 votes)
- What if you have a problem where the denominator has variables raised to higher exponents than the numerator has? I see a lot of those in calculus.(5 votes)
- Then there's no such possible division. There's no integer number that multiplied by the denominator would result in the numerator. Depends on what part of calculus you're studying, we would have to see how the function behaves and apply limits or other concept proper from calculus. But is not so related to this.(1 vote)
- I kinda feel i am the only one here lol, all the comments are from 5 to 10 years!(4 votes)
- You are seeing the top ranked questions and answers. This is the default sort order. To see the newer ones, select the "recent" option to see the more recent questions and answers.(3 votes)
- Atit was -3x minus 2x, which would be -5x. But when he multiplied by -1 it turned into -3 plus 2x. Is that a mistake or does it work somehow? This confuses me... 1:13(2 votes)
- I love the Keepers of the Lost Cities too! (Sorry for how off topic this is...)(5 votes)
- this stuff kills my brain i'm just sayin'!
(Polynomial long division)
(y 3 - y 2 + y + 3) ÷ (y + 1)
GOD PLEASE HELP MEEEEEE!(4 votes) - Can you show how to divide polynomials when all terms has an exponent? =((4 votes)
- that you should ask directly to Sal himself or one of the workers of this site(0 votes)
Video transcript
Divide x squared minus 3x
plus 2 divided by x minus 2. So we're going to
divide this into that. And we can do this really
the same way that you first learned long division. So we have x minus
2 being divided into x squared minus 3x plus 2. Another way we
could have written the same exact expression is
x squared minus 3x plus 2, all of that over x minus 2. That, that, and that are
all equivalent expressions. Now, to do this type
of long division-- we can call it algebraic
long division-- you want to look at the highest
degree term on the x minus 2 and the highest degree term on
the x squared minus 3x plus 2. And here's the x, and
here's the x squared. x goes into x squared
how many times? Or x squared divided
by x is what? Well, that's just equal to x. So x goes into x
squared x times. And I'm going to write
it in this column right here above all of the x terms. And then we want to
multiply x times x minus 2. That gives us-- x
times x is x squared. x times negative
2 is negative 2x. And just like you first
learned in long division, you want to subtract
this from that. But that's completely the
same as adding the opposite, or multiplying each of
these terms by negative 1 and then adding. So let's multiply
that times negative 1. And negative 2x times
negative 1 is positive 2x. And now let's add. x squared minus x
squared-- those cancel out. Negative 3x plus 2x--
that is negative x. And then we can bring
down this 2 over here. So it's negative x plus 2 left
over, when we only go x times. So then we say, can x minus
2 go into negative x plus 2? Well, x goes into negative
x negative one times. You can look at it right here. Negative x divided
by x is negative 1. These guys cancel out. Those guys cancel out. So negative 1 times x minus
2-- you have negative 1 times x, which is negative x. Negative 1 times
negative 2 is positive 2. And we want to subtract
this from that, just like you do
in long division. But that's the same thing
as adding the opposite, or multiplying each of
these terms by negative 1 and then adding. So negative x times
negative 1 is positive x. Positive 2 times
negative 1 is negative 2. These guys cancel
out, add up to 0. These guys add up to 0. We have no remainder. So we got this as being
equal to x minus 1. And we can verify it. If we multiply x minus 1 times
x minus 2, we should get this. So let's actually do that. So let's multiply x
minus 1 times x minus 2. So let's multiply negative
2 times negative 1. That gives us positive 2. Negative 2 times x--
that's negative 2x. Let's multiply x
times negative 1. That is negative x. And then x times x is x squared. And then add all the like terms. x squared, negative 2x minus
x-- that's negative 3x. And then 2 plus
nothing is just 2. And so we got that
polynomial again.