We’re studying polynomials and got introduced to the remainder theorem. I’ve been having a really hard time wrapping my head around the logic of it. Like why do we have to find the zero of the divisor and how does inputting that Into the dividend work?? In our books, they gave a really confusing explanation of how we use it to divide the dividend with a divisor which is a linear equation of x-a or something like that. Would appreciate some help!
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I’m assuming, from context, that you’re talking about the [polynomial remainder theorem](https://en.wikipedia.org/wiki/Polynomial_remainder_theorem) (there are other theorems by that name).
So, first, let’s set some context.
You’re dividing some polynomial *f* by another polynomial *g*. In general, f might not be divisible by g, so this results in a quotient q and a remainder r. Of course, all of these are polynomials and, at the level you’re operating at, functions, but it’s helpful to think of them as objects in their own right.
When we say “f divided by g has quotient q and remainder r”, what we *mean* is f = g*q + r. This might not be obvious, but try it with regular old integers. When I say that 17 divided by 3 has quotient 5 and remainder 2, I mean 17 = 3 * 5 + 2. That is, *remainders and quotients are really statements about multiplication and addition*, and that’s going to matter in a second.
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In the remainder theorem, we’re interested in a particular kind of polynomial *g* where g(x) = x – a. (I’m going to use lowercase *a* here, since we’re using *r* for the remainder; many books use capital Q and R for the quotient and remainder and *r* for the constant here.) For example, we might be dividing by the polynomial g(x) = x – 3, or the polynomial g(x) = x + 7 (where we think of this as “x – (-7)”). Such a polynomial – hopefully obviously – has a root at *a*, since g(a) = a – a = 0.
The remainder, by definition always has a lower degree (lower biggest power of x) than the thing we’re dividing by. Since we’re dividing by a polynomial of degree 1, the remainder, in this case, is necessarily just a constant (that is, a polynomial of degree zero). That is, r(x) = r is just a number. It might be 14, it might be -28, it might be pi/7, but whatever it is, it doesn’t have x or any powers of x in it.
The remainder theorem tells us that **f(a) = r**. That is, if we plug in the root of the polynomial we’re dividing by to the original polynomial, we get the remainder. This is weird, and it’s at the heart of your question.
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But let’s rephrase. Remember, back in the first section, we said that “f divided by g has quotient q and remainder r” really *meant* f = q*g + r.
Let’s try plugging *a* into both sides of this equation. Since these are equal *functions*, they have to be equal for every value we can plug in, so f = q*g + r implies that f(a) = q(a) * g(a) + r(a) (that is, if we have equal functions, we can plug equal values into them and still get an equation of numbers). With me so far?
But wait a minute. g(x) was just x-a, which has a root at *a*. So g(a) is just 0! Then f(a) = q(a) * **0** + r(a). And as usual when multiplying by zero, this makes the first term vanish. So f(a) = r(a). And since r is just a constant, it doesn’t depend on *a* at all – it’s just *r*.
Therefore, f(a) = r, which is what we said in the first place.
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The key idea here is that when you plug in anything that’s a root of what you’re dividing by, it ends up creating a zero in the equation f = q*g + r. This applies even if g(x) isn’t just of the form x – a, but in practice that’s often the form you’re most interested in.
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