I got stuck on this question. Use long division to solve

Mathematics

1 answer:

Phantasy [73]3 years ago

8 0

Answer:

5x² +19x +76 +310/(x-4)

Step-by-step explanation:

The process is straightforward. Find the quotient term, multiply it by the divisor and subtract from the dividend to get the new dividend. Repeat until the dividend is a constant (lower-degree than the divisor).

The tricky part with this one is realizing that there is no x-term in the original dividend, so that term needs to be added with a 0 coefficient. The rather large remainder is also unexpected, but that's the way this problem unfolds.

Unlike numerical long division, polynomial long division is simplified by the fact that the quotient term is the ratio of the highest-degree terms of the dividend and divisor. Here, the first quotient term is (5x^3)/(x) = 5x^2.

You might be interested in

Let f be a continuous function defined on the interval [0,1] such that f(0) = f(1). Show that there exists a number 0 <= a &l

Leni [432]

Explanation:

In order to prove that affirmation, we define the function g over the interval [0, 1/2] with the formula $g(x) = f(x+1/2)-f(x) .$

If we evaluate g at the endpoints we have

g(0) = f(1/2)-f(0) = f(1/2) - f(1) (because f(0) = f(1))

g(1/2) = f(1) - f(1/2) = -g(0)

Since g(1/2) = -g(0), we have one chance out of three

g(0) > 0 and g(1/2) < 0
g(0) < 0 and g(1/2) > 0
g(0) = g(1/2) = 0

We will prove that g has a zero on [0,1/2]. If g(0) = 0, then it is trivial. If g(0) ≠ 0, then we are in one of the first two cases, and therefore g(0) * g(1/2) < 0. Since f is continuous, so is g. Bolzano's Theorem assures that there exists c in (0,1/2) such that g(c) = 0. This proves that g has at least one zero on [0,1/2].

Let c be a 0 of g, then we have

$0 = g(c) = f(c+1/2)-f(c)$