Synthetic Division Question?

Yes, you have f(x)/(x+1) = 5x^2-4x+5-(12/x+1). You did that correctly.

Now multiply by x+1 and you get it into the form you are asked for:

f(x) = (x+1)(5x^2 - 4x + 5) - 12

q(x) is 5x^2 - 4x + 5

Remainder is -12

You got the quotient q(x) = 5x^2 - 4x + 5 when you did the synthetic division, and -12 was the remainder.

I think you are overthinking things. You did it exactly right, but are a bit confused about what it all means.

It's just like dividing numbers.

If you divide 103 by 4, you get 25, remainder of 3.

You could also say 103/4 = 25 + 3/4.

Or you could say 103 = 25x4 + 3. There, 25 is like q(x), and 3 is the remainder.

Maybe you are confused since q(x) is only PART of the whole quotient, since there is a remainder. So calling it the quotient is a bit inaccurate.

See http://mathworld.wolfram.com/PolynomialQuotient.ht... which defines "Polynomial Quotient" as the q(x) part, dropping the remainder.

What's going on is that you are trying to work with just polynomials, and 12/(x+1) is not a polynomial, so how are you going to express the division of one polynomial by another?

Imagine you are working with integers, and don't know what a fraction is. So you can't always divide perfectly. 100/4 = 25 makes sense, but what is 103/4? You get 25, but then you have 3 left over. So you might say the "integer quotient" is 25. Some programming languages might do exactly that if you try to divide 103 by 4 and assign it to an integer (or you might have to specifically ask for "integer division," perhaps by using \ instead of /).

You have the same problem when working with polynomials. You want the answer to be expressed as just polynomials. So you get f(x) = (x+1)q(x) + r, where r, or perhaps r(x), will have a smaller degree than the divisor. So while -12/(x+1) is completely correct, you left the realm of polynomials and came up with a rational function. All the question asks is that you not do that, and express the answer as a Polynomial Quotient as defined at the site I gave you the link to, plus a polynomial remainder (a constant -12 being a polynomial of degree 0, -12x^0).

By the way, as others have noted, -12 is going to end up being f(-1). That's because when you plug in -1 into f(x) = (x+1)(5x^2 - 4x + 5) - 12, the (x+1) term will be 0 and you will be left with -12. That may or may not be relevant to what you are studying. It's definitely something you should be aware of. But if you are dividing by a larger degree polynomial, say (x^2 + x + 1), then the remainder isn't easily expressible as some value of f(x), and will most probably be a polynomial of degree 1 (I haven't checked). In that situation, you will need to use polynomial long division, not simple synthetic division. (When I was in school, all that was classified as synthetic division, but when dividing by x-k, it can be done more simply.)

I hope that helps.

It sounds to me like this is leading up to the fact that the remainder term r will be equal to f(k) when dividing by (x - k).

The division algorithm (or "division theorem") guarantees that you can write:

f(x) = q(x)(x - k) + r

The proof often begins with the phrase "Express f(x) in the form...". The division algorithm says that such a form exists and is unique.

The rest of the proof is simple. Given the above representation then:

f(k) = q(x)(k - k) + r = 0 + r = r

That's the only context I can think of. You certainly can't write down such an easy formula for the coefficients of q in terms of k and the coefficients of f.

f(x)/(x-k) = q(x)+ (r/(x-k))

hence

(x) = (x-k)q(x)+r