Algebra

Polynomial remainder theorem | Polynomial and rational functions | Algebra II | Khan Academy

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Let’s get acquainted with the theorem for dividing polynomials by a remainder (Bezou’s theorem). And, as we’ll see in a moment, at first you will think of it as magic. But in future videos we will prove it and see … like a lot of things in math, when you think well maybe it’s not really magic. What is the theorem for dividing polynomials by a remainder? She tells us that if we start with some polynomial, f (x) … This is a polynomial here. Polynomial. And we share it on (x – a). Then the rest of this long division the polynomial will be f (a). He will be f (a). I know this may seem a bit abstract at the moment. I’m talking about f (x) and (x – a). Let’s do it a little more specifically. Let’s say that f (x) is equal to – I will invent something – let’s say a second degree polynomial. But this will be true for any polynomial. 3x ^ 2 – 4x + 7. Let’s say that a is 1. We will divide this into … we will divide it into (x – 1). In this case a = 1. Let’s do the long division of the polynomial. I encourage you to pause the video. If you are not familiar with the long division of polynomials, I advise you to watch the video for him this video, because I will accept that you know how a long division of polynomials takes place. Sections 3x ^ 2 – 4x + 4 on (x – 1). See how much you will get as a balance and whether this remainder is really f (1). I accept that he tried. Let’s work together. Let’s divide 3x ^ 2 – 4x + 7 on (x – 1). Okay, a little long division of polynomials it is never a bad start to the morning. It’s morning for me. I don’t know about you. Okay, I’m looking at the x member here – the member of the highest degree. And then I’ll start with the highest-ranking member here. How many times x goes into 3x ^ 2? Enters 3 times. 3x by x is 3x ^ 2. I will record 3x here. I will write it over the place for the first degree. 3x by x is 3x ^ 2. 3x by -1 is -3x. And now we want to take that out. This is done in the traditional long division. How much do we get? 3x ^ 2 minus 3x ^ 2, this will be just 0. And to this -4x we will have plus 3x. Negative value of a negative number … -4x + 3x will be -x. I will do this in a new color. This will be -x. And then we can download this 7. A complete analogy with when he first learned about the long division in maybe, I don’t know, third or fourth grade. All I did was multiply 3x by that. You get (3x ^ 2 – 3x) and then I subtracted that from (3x ^ 2 – 4x), to get this here. Or you could say I took it out of this whole polynomial and then I got (-x + 7). How many times (x – 1) contained in (-x + 7)? x is contained in -x … -1 times. -1 over x is -x. -1 over -1 is +1. But then we’ll want to get that out and this will give us the rest. That is -x – (-x). This is the same as -x + x. These will add up to 0. And then you have 7. This will not be 7 + 1. Remember, you have a negative sign, so when you open the brackets, this will be -1. 7 – 1 is 6. Your remainder here is 6. One way to think about it is to say that … In fact, I will keep this for future video. This is our remnant here. And since this is a negotiation of the long polynomial division, you know the rest is when you get something that is of a lower degree. That is, I guess you can call it polynomial of degree 0. This is to a lesser extent than to which you divide, or by (x – 1), by the divisor. This is of a lower degree; this is the rest. This can no longer go into this. According to the theorem for dividing a polynomial by a remainder, if that’s true – and here I just chose a random example. This is not proof, but simple way to concretize what the theorem for dividing a polynomial by a remainder tells us. If the theorem for dividing a polynomial by a remainder is true, it tells us that f (a), in this case, 1, f (1) must be equal to 6. It must be equal to this remainder. Let’s make sure of that. This will be equal to 3 over 1 ^ 2, which will be 3, minus 4 over 1, so this is going to be – 4, plus 7. 3 – 4 is -1, plus 7 – we deserve applause – is really equal to 6. This, at least for this example in particular, seems to support the fact that the theorem for dividing a polynomial by a remainder works. But its application is, for example, if someone says: “What is the remainder if I divide 3x ^ 2 – 4x + 7 of (x – 1) if I\’m interested in the remainder? ” They are not interested in the private. They are only interested in the rest and you can to say that in this case a is 1. I can introduce that. I can calculate f (1) and get 6. I don’t have to do all this. I just have to do this, to find how much is the remainder of 3x ^ 2 – 4x + 7, divided on (x – 1).

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