Precal Blog

For a large portion of the class period, we worked on some more problems to refine our skills finding the `x`-intercepts (zeros) for cubic functions, working with synthetic substitution rather than long division. We went over two problems that Mr. Bieniek gave us to work out on our own at the end of class yesterday.

The first of these was a typical cubic equation in which we had to find zeros, only without having a zero given to us initially: `f(x)=x^3-3x^2+4x-2`. Guessing and testing made it easy enough to find that 1 was a zero in this equation by simply plugging it in and getting 0. The next step involved synthetic substitution:

`-1| 1  -3  4  -2`

………….`1 -2   2`

…….`1  -2  2   |0|`

The remainder confirms that 1 is a zero. The resulting factor is the quadratic equation `x^2-2x+2=0`. To find the other two zeros in this cubic equation, we plug the values in to the quadratic formula (no calculator–it’s easy to remember with the Pop Goes the Weasel mnemonic) and simplify to get `x=1+i` and `x=1-i`.

The second challenge he gave us was to derive a function when given the zeros `x=2` and `x=4-i`. He gave us the following expression as a hint to start us off: `[x-2][x-(4-i)][x-(4+i)]`. When we went over it today, his first bit of advice was to shift both pairs of inner parentheses to `x-4`. This way, when the second and third sets of brackets were distributed into each other, two of the resulting terms would cancel each other out. Here are the algebraic steps:

`[x-2][(x-4)+i][(x-4)-i]`

`[x-2][(x-4)^2-i^2]`

`[x-2][(x-4)^2+1]`

`[x-2][x^2-8x+17]`

And… well, quite frankly, I don’t think I understand the rest of the work I have written down in my notes for this problem, but I digress. I’ll leave the rest here for y’all to interpret.

`x^3-8x^2+17x`

`  -2x^2+16x-34`

`x^3-10x^2+33x-34`

And somehow I think the answer wound up being `2(x^3-10x^2+33x-34)`.

The constraint we’ve dealt with in all of our problems has been the need for a given zero (or an easily guessed one) to find the others. However, we eliminated this necessity in the remaining class time today as we were introduced to a theorem whose derivation we can’t understand at our current level of mathematical comprehension. We had to take Bieniek’s assertion as truth without proof:

“If there are any rational zeros then they are in this set:

`+-`factors of the constant term`/`factors of the leading coefficient.”

Take the function `p(x)=2x^3+3x^2+2x+3`, for example. According to this theorem, the potential zeros are `1/1`, `1/2`, `3/1`, `3/2`, `-1/1`, `-1/2`, `-3/1`, and `-3/2`. However, these aren’t all necessarily zeros. To find the zeros, you test all of these eight numbers as zeros with synthetic substitution, and if one of them is a zero, it will have a remainder of 0. In this problem’s case, when we began guessing and testing, the number `-3/2` was found to be zero, so we took its quadratic equation–`2x^2+2=0`–found by synthetic substitution, factored it and simplified to find the other two zeros, `x=i` and `x=-i`.

That’s all, folks.

P.S. Sorry, the blog isn’t cooperating with underlining and spacing, so some parts are kind of messy.

We started class today reviewing concepts learned yesterday…

To find any rational zeros we use this hint to make things easier. This hint is called the “Rational Root Theorem”.

= ± factors of the constant/factors of the leading coefficient

for example: f(x)=(x^3)-(8x^2)+(17x)-10…           =±1,2,5,10/1     so your possible options for zeros are 1/1, 2/1, 5/1, or 10/1 or -1/1, -2/1, -5/1, or -10/1        so then we can them to get the zeros.

we know that it has to be a positive zero because if we plug in a negative number into the equation then the answer will always be less than zero… regardless of the value.   we can prove this!

-1^3=-1, -(-8^2)=-64, 17*-1=-17 then add these together and you get a negative number… not a zero!      so we know it has to be positive. so we test 1/1

and it is a zero so instead of doing that for 2/1, 5/1 and 10/1 we can just use the quadratic formula now which is (x^2)-(7x)-10 to find the other two zeros.

sometimes they won’t be rational so we started talking about that a little bit today. but i think we will get more in depth about it tomorrow..

The Big Idea For Today!!!!…. WE AREN’T GOING TO BE GIVEN THE ZEROS ANYMORE

comment if you have any questions!

“I wish Aunt Sally was dead!” – Mr. Bieniek 4/20/2009

Today in Honors Precalculus, we were exposed to Synthetic Substitution. Synthetic Substitution is a shortcut for Polynomial Division. However, it’s not used divide out Factors, but to find Zeros. The main concept behind Synthetic Substitution is the multiply by x, then add the next coefficient property.

So, for the problem `f(x)=x^3-4x^2-3x+18` you would start by putting the coefficients starting from the greatest multiple of x. In this situation the first coefficient will be 1, which is from `x^3` or `1x^3`. The coefficient will be -4, which is from `4x^2`. Just keep on doing this until you end up with the last number of a plus 18. You numbers should be:

1     -4     -3     18

Next, you need to either find a zero of the equation or you’ll be given one. In this situation you are given that one of the zeros is x=-2 which is the next part to put into the synthetic substitution formula.  The -2 goes in front of all the previous numbers, however it’s not part of that same group. If you need a visual reminder feel free to circle or box out the zero. It should look like this.

-2 1     -4     -3     18

Now you are ready to use synthetic substitution. Draw a line underneath those numbers, but give it some room so you can write more numbers underneath. Starting with the 1 you just carry it down. What you are doing is addition so 1+0=1. The 0 is from it being the first addition you do in the problem. Next, you take your zero of the equation, and multiply that and the 1 you just put down. This gives you -2 which you put right underneath your next coefficient.

-2 1     -4     -3     18

0      -2

____________________________

1(-2)

What you’re doing is the mulitply by x and then add the next coefficient property. Just multiply your zero by the coefficent you’re adding until you run out of coefficients. The final product should look like this:

-2 1     -4     -3     18

0     -2     12     -18

__________________

1     -6       9      0

If you get a 0 at the end, that means that the number you multiplied by was indeed a zero of the equation. The numbers you get at the end of this are your coefficients for your next equation to derive the zeros from.

`f(x)=(x+2)(x^2-6x+9)`

Finally, all you have left to do is solve for the zeros. All you have to do is use your quadratic formula for this.

`f(x)=(-6+(-6^2-4(*9)))/2` Which equals -3. Then you solve for the other one by solving for `f(x)=(-6-(-6^2-4(*9)))/2` which is 3.

Your zeros are -2,-3, and 3.

By the way this is a link to the Fundamental Theorem of Algebra:

http://www.cut-the-knot.org/do_you_know/fundamental2.shtml

We discussed this is class, and this website has done a great job of explaining it.

Today we started class with the equation `f(x)=x^3-9x^2-x+105`, we simplified this using the nested form to get `f(x)=((x-9)x-1)x+105`.  This process follows “multiply by x, then add the next coefficient.” Using this method you can factor this equation using a method easier than the long division.  You need to setup the equation like this…

-1| 1     -4     -3     2

You first drop the 1 below the line because there isn’t and x to multiply by, then you add the next coefficient which would be -4.  Then you would multiply that number which is -5,  by the x to get 5.  then you add five and continue the process.  In the end you should get zero. see work below

-1| 1     -4     -3     2

…………..-1      5    -2

……1     -5      2       0

As you can see this makes everything much more simple.  After we went over this, we talked about what would happen if the function only crossed the x-axis once.  We determined that you would end up with two complex numbers.  We then decided that all complex numbers must come in pairs.  After this we went over what the Fundamental Theo rm of Algebra was.  It states an nth degree polynomial function has exactly n zeros in the set of complex numbers, counting multiple zeros.

-Curtis

Today we learned a concept that was completely unlike anything we have ever learned before.  Up to this we have learned about number that are all real, but today we were taught about imaginary numbers.  These are numbers that someone just decided to make up and in order to fully understand them, we went to a very helpful website with John and Betty.  Those kids were great.  Here is the website if you would like to look at it for review or if you were not here.

mathforum.org/johnandbetty

This website starts by talking about biscuits and how the whole biscuits are like real numbers.  Then it goes on to talk about parts of biscuits, which we know as fractions.  John and Betty had never used fractions, so they just made them up.  This is important to remember because it will help you understand the development of imaginary numbers.  After John and Betty talked about fractions, they developed square roots.  In class we got into an interesting discussion about how Pythagoras killed someone for suggesting the idea of square roots because he didn’t believe it.  Just like we accept square roots, we must accept imaginary numbers.

We gave imaginary numbers the value of “i” where `(i*i)=-1`.   Now this was just made up, so we accept it for a value.  What happens if you multiply `5i*3i`?  Well if you separate this out, you get `5*3*i*i` and since `i*i=-1` and `3*5=15` the equation equals -15.  A rule that we developed in class was that if you multiply or divide just imaginary numbers, the answer will be a real number.  However, when you add and subtract them you will get an imaginary number.

The next part of imaginary numbers is complex numbers, which have more to do with the polynomials we are working with.  An example of this type of number is `(-4+i)/(3-2i)`.  In order to get the imaginary number out of the bottom, we multiplied the fraction by a complex conjugate, which in this case would be`(3+2i)/(3+2i)`.  When dealing with the complex imaginary numbers, we can figure out a better way to write ours zeros for our polynomial equations.

One of the non real zeros we have dealt a lot with is `(7+-sqrt(-23))/2`.  We can use i here because we have established that `i*i=-1`, which means `i=sqrt(-1)`.  so we can rewrite our complex number as `7+-i*sqrt(23)`.  This was the biggest part of imaginary numbers, but an interesting side note is that if you reflect a parabola with imaginary zeros, you will get those exact values on the x-axis as real numbers.  Something else to remember is that if the zeros are non real, then there must be two of them if it is quadratic.  Those zeros come in pairs.

Besides learning about imaginary numbers, we also spent the last five minutes of class learning about remainders.  Given the polynomial `3x^3+2x^2-3x+6` and that one of its factors is `(x+1)` we can find the remainder by plugging in the zero -1.  We find the remainder to be 8, which is the exact value we would have found if we had done the long division.  This method saves an enormous amount of time when dealing with these equations and it allows us to develop an equation.  If we use “c” as our variable for the zero then `p(c)=r`, where r is the remainder.  This also means that `(px)/(x-c)` where (x-c) is the factor.

Lastly, we learned one more thing about these tricky polynomials.  If you take the polynomial `3x^3+2x^2-3x+6` and factor out an `x^2` from the first two terms to get `(3x+2)*x^2-3x+6`.  If you then take out an x from the new term and the `3x`, then you get `((3x+2)*x-3)*x+6`.  If you look at the original equation you will find that the coefficients are the numbers in this equation.  This is also a quicker way for solving polynomials.

Today in math class we learned a ton, but i think that the most important thing was that it was EPT’s BIRTHDAY!! We sang happy birthday to him and he loved it!!! So one last time HAPPY 17th BIRTHDAY EPT!

Everyone enjoy your spring break and don’t forget we have a party tomorrow!

1, 2, 3, 4, 5, 1/2, and pi. These are all number’s, but they are also all real. This is not the case for every number though. It’s possible to have imaginary numbers, and one such number is i.

Now we all know that math has to be complicated, so when the question was asked, “What number times itself gets you to -1?” they had to make up a number to do that, since no real number will get you to that. This number that they decided on was i.

John and Betty are two colorful strangers who lack artistic style, but make up for it with mathematical intellect. So join them on an….exciting? adventure to learn all about the number i. The site is http://mathforum.org/johnandbetty/. But seriously, it is very helpful to understand many of the basic concepts of i, and you should go to it if you have any questions.

Now, I will try to do a quick explaination on some of the things behind i. First i^2=-1 so  i=sqare root of-1. This is the main reason that i is imaginary, the fact that it is the number for the square root of a negative. But i doesn’t just work by itself, it can be put together with regular numbers to make a complex number. An example of a complex number would be as simple as 3+2i or 5-3i. These complex numbers would just like a regular number would except for the fact that there is now an i in the statement.

Example: What would be the result of 3i*5i? It would be 15i!   What would be the result of 3i+5i? …8i. Think if the i were just a variable like x, it would all be like normal.

Ok, now one with complex (just one because others are explained on the site. (3-5i)+(4+2i), just combine all your like values to get 7-3i. For all the other basic operations, the idea is the same. What works well, if you are confused with the whole i deal, is to think of it as a variable like x. Then solve it like you would with an x in place of the i, or something like that.

Finally, how would you veiw these complex or imaginary numbers if they are imaginary. Well, they would fit on a regular number line, because there would not be a number for them. So make an imaginary number line. There are good images of it on the site, but the biggest thing about them to remember, is that they go parrallel to the real number line, sort of like an (x,y) graph to show imaginary, real, AND complex numbers.

Well that’s all I’ll talk about for now, but if you just have to know more or have any questions, talk to John and Betty, they have this down.

P.S. sorry for the slightly confusing equations.

<[((____Andrew_))]

Okay so while the title is pretty irrelevant, today’s class was all about blogs and power functions. For this quarter Mr. B has two extended take home assignments, and they’re both blogs, so make ‘em good. The way it works is at the beginning of each class he’ll take a volunteer to do the blog for that class period. Anyhoo, onto power functions.

First, we went through the thought processes involved with square roots. Here are some examples

`sqrt9`: think 3*3=9, so the answer is 3

`sqrtx`: think `x^(1/2) * x^(1/2)=x`, so the answer is `x^(1/2)` 

`3sqrt(x^2)`: think `x^(2/3)*x^(2/3)*x^(2/3)=x^2`, thus the solution would be `x^(2/3)`

So basically… `c*sqrt(x^d)=x^(d/c)` We can use this to help us with power functions (it’s easier to work with an exponent than a square root), which was the second part of the class.  To review, a power function is any function to the form of `f(x)=ax^b`. The shape of the graph is very dependant on b, which can cause 8 different graphs. This is only four, i know, but the shape also changes depending on whether b (or the denominator of b) is odd or even. Instead of forcing any chance reader to memorize what the graphs look like, you can do what I did, and remember a couple simple clues.

If the graph doesn’t go into the third quadrant, the exponent’s even, since multiplying something to an even exponent will never get you a negative number.

If the graph doesn’t go through zero, it means b’s negative, since a negative exponent means a positive inverse exponent and we all know that (1/0) is undefined.

Depending on the growth of the graph, we can know if b is less than one or greater than. If it’s greater than 1 the graph will grow exponentially until it’s basically vertical. If b is less than 1 the graph will show the steepest slope closest to zero and then taper off. Everything gets flipped around if b’s negative, and the best advice I can give you is to imagine the graph inverted over the y=x line and look at it.

Of course, all these rules can go both ways, if it doesn’t go through zero that means its negative but if it does it means it’s positive, etc.