Math Is Fun Forum

Thats good to know.

I hereby proclaim myself superior to the writers of my mathbook and everyone involved in making it.

brilliant work, guys!

Logarithms rock!

One thing I noted is if x^n = n, then x is then x is the nth root of n. (you can prove this by raising both sides to the 1/nth power. But I don't think that will help here.

This thread needs something. Its needs.....something to do with math!

y = arcsin (ax/b)

dy/dx = a/sqrt( b - a^2 x^2 )

y = arctan (ax/b)

dy/dx = ab/(a^2 x^2 + b^2)

YAY!

(stands back to look at the thread) Much better!

I was already there for quite a while. Must have spent and hour on the problem. lol.

Hopefully I'll be able to do this without drawing lots of pictures.

The remaindor theorem.

Lets talk about remainders for a sec. Suppose you had 13 cookies and four kids. Being the wise parent that you are, you know one kid getting one extra cookie then the rest would start World War III. So you want each kid to get an equal number of cookies. But 13 will not divide evenly into 4. You consider cutting the remainders into 4th's with a knife, but they are crunchy cookies and would probably just crumbe. You could try to break them in fourths by hand but if they don't break 100% evenly, the war will start anyway. You decide to eat the remainder(s) yourself.  If we use integer division which we all learned a gazillion years ago, finding how many cookies you get to eat is easy!  13/4, the answer is 3 with a remaindor of 1. (aw... you only get one cookie! >:-( ).

Now lets look at this again. What does this remainder represent? Well, if we separate the cookies into 4 groups of 3, 1 will be left The remainder. We had a 13 cookies in all and thus its obvious that 3 * 4 + 1 = 13.

But lets look at this yet again. We divided 13 by 4, got an answer of 3 with a remainder of 1. 

13/4 = 3, Remainder: 1 

Thus quotient * divisor + remaindor = dividend.

I use the term quotient loosely here. Technically its not a real quotient. Its an integer quotient.

When we learned fractions, we learned to put the remainder on top of the divisor to get an exact answer with no remainders. For situations when an object can be divided into parts of the whole.

Notice, with synthetic division we do the same thing. In the previous example, we got a remainder of 133. We wanted an exact answer so we placed 133 over (x + 2) to get an exact answer.

In algebra, we learned more clearly the direct connection between multiplication and division.

if x/y = 5 then x = 5 * y

Likewise, if: 

(5x^4 - 6x^3 + 2x + 9)/ (x + 2)  =  5x^3 - 16x^2 + 32x -62 + 133/(x+2)

then:   (5x^4 - 6x^3 + 2x + 9) = (x + 2) (5x^3 - 16x^2 + 32x -62 + 133/(x+2))

But the expression on the left is the original polynomial thus we have rewritten the polynomail with a factor of (x + 2)

Remember how I said 133 was the remainder? We cut divided it by (x +2) to even it out, but what if we didn't?

If divide 13/4 and put the remainder over the divisor we get a remaindor of 1/4. But what if we didn't?

Remember earlier? 13/4 = 3 remainder 1. Thus 3 * 4 + 1 = 13.

Likewise (5x^4 - 6x^3 + 2x + 9)/ (x + 2) = 5x^3 - 16x^2 + 32x -62  Remainder 133

Thus (5x^3 - 16x^2 + 32x -62 )(x + 2) + 133 =  (5x^4 - 6x^3 + 2x + 9).

But the expression on the left is the original expression. Therefore we've just rewritten it in a different form.

But lets take a look now, if we have the expression, (5x^3 - 16x^2 + 32x -62 )(x + 2) + 133, what value would it have if x = -2? well this would bring the the (x +2) factor to zero, and the whole expression in the parenthesis would have a coefficient of zero, and thus a value of zero. So what would be left? 133. The remainder.

Now lets think about what just happened. We divided the original expression by x + b using synthetic division and hence put -b in the box. The synthetic division allowed us to rewrite the expression into a form with x + b as a factor, and a remainder. (x+b)m + remainder = original expression.
When x has a value of -b, m has a coefficient of zero and no value and thus the expression has the value of the remainder.

Thefore, if we use synthetic division to devide a polynomail by x + b, then if we use -b as a value for x, the expression will have a value of the remaindor of synthetic division. But when we use synthetic division to divide by x + b, we put -b in the box.

THEREFORE! If you use synthetic division to divide a polynomial by placing b in the little box, the remaindor will be the value of the expression evaluated at b! This is the remaindor theorem.

I drew up a quick diagram to illustrate using synthetic division to quickly evaluate f(x) at x values of 3, -2, and 5.

Yep yep. But like I said, it is not imediatly apparent that it can be factored into that form. If it was, then we could solve the problem right away. It leaves no remainder and thus is a zero of the expression or function when the factor equals 0.

Hmmm... solids of rotation? My book teaches that. Disks, washers, shells. I haven't gotten to any yet but I think I'll be learning them within a week.

Calculus III? w00t! :-D Talk about t3h r0xx0r! Too bad Saxon books only go up to calculus one.

In my oppinion computer science is something every mathematician should take. It allows you to truely harness its power!

Ah the power of math!

Uh oh... I feel it coming again..... ugh.... hold on... keep it together man... keep it to-....oh no!

wigout();

OH MY FREAKIN GOSH! MATH ROXXORS T3H B1G ONEZ!1111111