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03.16.2019: One Lesson of Math - Polynomials, 1/5: The Remainder Theorem, Part 1


Today's soundtrack is Reg Downey: You're Not Alone (Not Quickly Broken), a thoughtful album with worshipful lyrics.


This morning, I'm learning how to solve polynomial division equations by using both polynomial long division and synthetic division, the latter of which I touched on in the past.


By solving polynomial division equations through the two above methods, I was able to confirm the Remainder Theorem: "When a polynomial P(x) is divided by x-b, the remainder is P(b)." So if we need to quickly find the remainder of a division equation, we can substitute b into P(x) to find our answer. If the coefficient of x-b is any value besides one, we use a modified version of the equation: When a polynomial P(x) is divided by ax-b, the remainder is P(-b/a), which can be found by substituting -b/a for x.


By investigating the relationship between a divisor and the equation we are dividing, I found out that "x-b is a factor of the polynomial (P)x, and b is a root of the polyomial P(x)" (contentconnections.ca).


And with that...suddenly, this whole thing has clicked! The real value of the Remainder Theorem is that we can quickly determine whether or not a polynomial is a factor of another polynomial by finding whether or not it produces a remainder.


That's all for today! Tomorrow, I'll go through some examples.

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