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Course: Algebra (all content) > Unit 10
Lesson 22: Synthetic division of polynomialsDividing polynomials: synthetic division
Sal divides (2x^5-x^3+3x^2-2x+7) by (x-3) using synthetic division. Created by Sal Khan.
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What if there isn't a remainder?(68 votes)
IF there isnt a remainder, than there wouldnt be a number at the end, it would cancel out(35 votes)
I'm trying to find a video of yours to help me figure out how to do these math problems that involve synthetic division but they also factoring. She told us, " use synthetic division to show that X is a solution of the third degree polynomial equation and use the result to factor completely. List all real solutions". I'm not real sure how to do all this, i've never been very good at factoring anyway... help?(27 votes)
Let's say you have a 3rd degree polynomial p(x). If R is a root of p(x), the monomial (x - R) divides p(x) and there is no remainder.
After you've divided p(x) with (x - R) and thus proven that R is a root, you should have a quadratic equation which you can probably factor on your own.
Example: x = 1 is a solution to p(x) = 4x^3 - 8x^2 - 20x + 24 = 0. Factor completely and list all real solutions.
Step 1: Divide p(x) with (x - 1): (4x^3 - 8x^2 - 20x + 24) / (x - 1) = 4x^2 - 4x - 24.
There's no remainder, so x = 1 is indeed a root of p(x).
Step 2. Factor what we got in step 1: 4x^2 - 4x - 24. You can factor it by solving its roots with the quadratic formula, or whichever way you want to do it.
The roots of 4x^2 - 4x - 24 are x = -2 and x = 3, so it factors to 4(x + 2)(x - 3).
So the original polynomial factors to 4(x + 2)(x - 3)(x - 1) and its real roots are x = -2, x = 1 and x=3.(36 votes)
How did the first term of 2x^5 go to 2x^4?(15 votes)- you do 2x^5/x and you get 2x^4 you basically subtract an x from the highest exponent so if you had 5x^3+18x^2+7x-6 / x+3 you would take 5x^3/x and get 5x^2 as your answer(17 votes)
He says there is a different way to do it if you are not dividing by x + or - a number. Are there Videos on this?(14 votes)- If the coefficient is greater than 1, I think you would just take that into account by multiplying 1/a (the coefficient of x that isn't 1) to the values you obtain by performing synthetic division.
:)(4 votes)
Does anyone know how/who synthetic divison was thought of/who though of it? And also, where did the "operation" symbol come from?(5 votes)
The "operation symbol" is not actually an operator, it's just a template to organize the calculations for this particular operation, similar to long division or multiplication by hand. You could think of other "operators", but this one is conventional. As for your other questions, Google is your friend :)(10 votes)
- Is there a video on how to do synthetic division if the number being divided by starts with x squared?(9 votes)
What about in the case of the polynomial having more than 1 variable. In one question, there was a polynomial; 9a^2 - 30ab + 25b^2 Divided by 3a - 5b, What do I do then?(5 votes)
it would be easier to do polynomial long division because this synthetic division only works when the denominator / devisor is x+number or x-number, with a coefficient of 1. at1:21, sal says this.
with a coefficient greater or less than one connected to the x of denominator, you would divide the denominator and polynomial with that coefficient and solve using your new polynomial and denominator (when the coefficient of x is 1), in other words multiplying 1/coefficient(1 vote)
- When we finally write the answer as 17x^2 + 54x + 160 + 487/x-3 , why do we write 487/x-3 and not just 487?(2 votes)
The 487 is the remainder. We treat it the same as we do when we divide something like 17/5 and get an answer of 3 2/5. The remainder is shown as a fraction.
Hope this helps.(3 votes)
So does this mean that every time you use synthetic division, there will be a remainder?(2 votes)- No, like Joel said for long divide synthetically, and you WILL not get a remainder.(2 votes)
At1:35Sal says something about there being a different process if the x-3 was 3x-3 or 5x^2-3, what is that called? Is it still synthetic division?(2 votes)- Well you could technically use 3x-3 for synthetic division because if you set that expression equal to zero, then you get 3x-3=0. Then add 3 to both sides, 3x=3. After that divide both sides by 3 to get the coefficient off the x term, x=1. But for denominator expressions where you can't do what I just did, you would need to use long division. I can't really explain how to do it because I'm bad at it myself, but if you go to the lesson before this one, Long division of polynomials, you can figure it out. Sal uses the denominator x-2, but you can do long division of polynomials with an x coefficient greater than 1 too. Hope this helps!(2 votes)
1:21Video transcript
Let's do another synthetic
division example. And in another
video, we actually have the why this works relative
to algebraic long division. But here it's going
to be another just, let's go through
the process of it just so that you get
comfortable with it. And now is a good chance to
give it a shot, to actually try to simplify this
rational expression. So let's think about
this step by step. So the first thing
I want to do is write all of the coefficients
of the numerator. So I have a 2. Oh, I have to be careful here. Because the 2 is the
coefficient for x to the fifth, I have no x to the fourth term. Let me start over. So I have the 2 from
2x to the fifth. And then I have no
x to the fourth. So it's really 0x to the fourth. So I'll put a 0
as the coefficient for the x to the fourth term. And then I have a negative
1 times x to the third. And then I have a positive
3 times x squared. Negative 2 times x. And then I have a constant
term, or zero degree term, of 7. I just have a positive 7. And now let me just draw my
little funky synthetic division operator-looking symbol. And remember, the type of
synthetic division we're doing, it only applies when we are
dividing by an x plus or minus something. There's a slightly
different process you would have to do if it
was 3x or if was negative 1x or if it was 5x squared. This only works when we have
x plus or minus something. In this case we have x minus 3. So we have the negative 3 here. And the process
we show-- there's other ways of doing it-- is
you take the negative of this. So the negative of
negative 3 is positive 3. And now we're ready to perform
our synthetic division. So we'll bring down
this 2 and then multiply the 2 times the 3. 2 time 3 gives us 6. 0 plus 6 is 6. And then we multiply that times
the 3, and we get positive 18. Negative 1 plus 18 is 17. Multiply that times the 3. 17 times 3 is 51. 3 plus 51 is 54. Multiply that times 3. The numbers are getting
kind of large now. So that's going to be what? 50 times 3 is 150. 4 times 3 is 12. So this is going to be 162. Negative 2 plus 162 is 160. And then finally, 160
times 3 is going to be 480. And you add 480 to
7, and you get 487. And you can think of it, I only
have one term or one number to the left-hand side
of this bar here. Or I'm just doing the standard,
traditional x plus or minus something version of synthetic
division, I should say. So I can separate
this out, and now I've essentially gotten my answer. And it looks like voodoo,
and it kind of is voodoo. And that's why I
don't like to do it, because you're just
memorizing an algorithm. But there are other
videos why we explain why. And it can be fast
and convenient and paper saving very often,
like you see right here. But then we have
our final answer. It's going to be-- and
let me work backwards. So I'll start with
our remainder. So our remainder is 487. And it's going to be
487 over x minus 3. And so this is
our constant term. And so you're going to have plus
160 plus 487 over x minus 3. Now this is our x term. So it's going to be
54x plus all of this. This is going to be
our x squared term. So this is going to be 17x
squared plus 54x plus 160 and all of that. Then this is going to
be x to the third term. So this is going to be 6x to
the third plus all of that. And then finally, this is
our x to the fourth term-- 2x to the fourth. And let me erase this. So then I have my x
to the fourth term. So it is 2x to the fourth. And we are done. This thing simplifies
to this right over here. And I encourage you to verify it
with traditional algebraic long division.