Use Synthetic Division and NEVER Use Polynomial Long Division EVER again!
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- Опубликовано: 5 фев 2025
- Struggling with polynomial long division? Well in this video I will explain how to use synthetic division to divide polynomials so that you never have to use polynomial long division EVER again!
Polynomial long division is a long, tedious, and difficult process that many of my students find confusing. Fortunately, the synthetic division process for dividing polynomials makes things fast, accurate, and fun!
In this video I will walk you through how to apply the synthetic division algorithm to divide a polynomial of degree 3 by a binomial of degree 1. Note that this example results in a remainder of zero.
There are many different ways to go about dividing polynomials. Be sure to learn both synthetic division and polynomial long division so that you can master the topic of polynomial functions!
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Very informative ❤
Hope this helps make dividing polynomials easier for you! Just watch out for remainders that aren’t zero and divisors that have coefficients that aren’t 1. These will be covered in future videos!
What about dividing by x^2 + 2 or some other polynomial with degree >1
It can definitely be done with SD. Although it is a much more complex process than what is shown here. In fact, I would even argue an example of that caliber would be better suited to PLD.
I'll keep all the tools of the math-trade, thank you.
Definitely. PLD is actually really helpful in more advanced problems (when your division is of exponent 2, for example)
This doesn't work in all cases, does it? For instance, I tried dividing x^3 - 7x^2 - 2x + 6 by 2x + 2, and it only gives the correct answer in the long division method. I'm assuming this method would work as long as the coefficient of x of the divisor is 1?
This is a really good question and something I want to explore in another video. If you have a divisor with a leading coefficient that is NOT 1, you will want to first divide each coefficient in your table by the coefficient that is NOT 1. But, in doing this, you need to remember that your remainder will need to be multiplied at the end for it to work out properly. I know that's confusing in text. Hopefully I can explore it in a future video!
Does synthetic division still work when the coefficient of the divisor is NOT 1?