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

Lesson 22: Synthetic division of polynomials

# Why synthetic division works

Sal explains why synthetic division gives you the same result as traditional algebraic long division. Created by Sal Khan.

## Want to join the conversation?

• Sal said there is a way to use synthetic division even if there is a coefficient in front of x, but I haven't found any videos so far. How would you do it then? factor out the coefficient?
• I would just divide everything by the coefficient, so that you get x + a while maintaining the equation.
• Why does algebraic long division work?
• It is actually fairly simple, and just requires you to take a step back and ask yourself what the problem you are facing actually means instead of getting confused by all of the variables. It is the same basic principal of dividing whole numbers just taken to the next level. Basically what is happening is you are multiplying the denominator X times so that it is equal to the numerator. In normal division, 42/6 is saying "How many times do you multiply 6 to get 42?" Its the exact same concept. Except in this case it is slightly a bit abstract, it is "How many times do you multiply your denominator to equal your numerator?"

These examples you would normally do by factoring out values, but just bear with me here and ill try to help you understand that they are actually the same thing and the same process. Starting out simple, if you have the expression 4x / 4. You don't think of it this way because your brain is doing it subconciously, you just realize that 4 over 4 is 1 with an x left over. You could write it as 1x, but the answer is just x. Now thinking in terms of what the "/" part of division(a fraction) actually means, "How many times do you multiply 4 to get "4x." The answer is "1x," which just simplifies to "x."

Now lets add another term to the numerator, say you have the expression (8x + 8y) / 4 setup as a fraction. Your brain automatically goes, "Hey, 8x over 4 is just 2x." You can then rewrite the 8x/4 as just 2x, you simplified the fraction to a whole number. Lets not simplify the 8y just yet. You then end up to the expression "2 + 8y/4". In long division, 8y/4 is the remainder, what was left over in the original division you didn't do. If you set it up long division wise now as 8x + 8y divided by 4 you see this work itself out. 4 goes into 8x 2x times (4 * 2x = 8x.) You can then take the 8x out of the numerator as its equivalent whole number: 2x. You then look at what is remaining in the fraction (hence, the remainder) and say "How many times does my denominator go into this term?" 4 goes into 8y 2y times. Again, taking the 8y out of the numerator. If you are still lost, while looking at the fraction, always think of it like it is written in plain english... "How many times does my denominator go into my numerator?" or "How many times does 4 go into 8y?"

This will keep going on and on as you add more terms. It is always the same question though, and you can always think of it as "How many times do I need to multiply the denominator so that I can subtract a term out of the numerator?" (starting with the highest degree term first.) Keep going until you can no longer remove any terms out of the numerator without adding any extra terms. I hope this helped, I watched Sal's videos too and learned how to do it very easily, and I too was curious as to exactly what was going on. You just really need to sit down and solve problems of increasing complexity. Start with a whole number division problem, ask yourself "Why am I doing this, and what is it doing?" Then setup numerator with a variable, and ask yourself that same question. It will always be the exact same reason no matter if you have whole numbers or a variable raised to the 100th power.
• Will there ever be any skills practice concerning the division of polynomials?
• What if the answer doesn't have remainder?
• If it doesn't have a remainder, this is a beautiful thing! Just as when you divide 18 by 3 you get a nice "6" as opposed to 19 divided by 3. You still get the "6" but you have a little left over...1/3.
• is there a video about the fundamental theorem of algebra?
• I think if there is a number in front of x you just factor it out of the denominator, and then ignore it and perform synthetic division normally. Then divide your answer by the number that you factored out.
• Not quite:
(2x^3+4x^2-5x+6)/(2x -3) = (1/2 (2x^3+4x^2-5x+6))/(1/2(2x -3)) is true.
However, for remainder, you need to multiply 2 back.

(2x^3+4x^2-5x+6)/(2x -3)
= (x^2 + 7/2 x + 11/4) + (14 1/4)/ (2x -3) ----------(1)
(1/2 (2x^3+4x^2-5x+6))/(1/2(2x -3))
= (x^2 + 7/2 x + 11/4) + (7 1/8)/(1/2(2x -3)) ---(2)
if we multiply (2x -3) on both side of (1) we get:
(2x^3+4x^2-5x+6) = (2x -3)(x^2 + 7/2 x + 11/4) + (14 1/4)
if we multiply (2x-3) on both side of (2), (1/2 cancel out) we get
(2x^3+4x^2-5x+6) = (2x -3)(x^2 + 7/2 x + 11/4) + (7 1/8)/(1/2)
i.e.
(2x^3+4x^2-5x+6) = (2x -3)(x^2 + 7/2 x + 11/4) + (7 1/8)*2
That is why you need to multiply 2 back on the remainder.
• If synthetic division works for polynomials, would it also work for integers?
• No synthetic division will not work for integers. You can test this by taking a three digit number, like 224, and dividing it(for this example 2). 200....20....4. Next you can divide it by the negative of 2, -2. If you follow the synthetic division process, you will not reach the correct conclusion. Thus synthetic division does not work for integers.
• Would synthetic division work with numbers like 2x or x^2? Or would it work only with x?
• no it only works with X. the coefficient and exponent must both be implied 1's. this is because, like sal just explained, since you can asume that your dividing by an X you can simplify the problem by only multiplying 1 term.
• When do you use synthetic division?