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A few areas of algebra I don't understand

  1. Sep 27, 2008 #1
    As I mentioned in another post I'm trying to keep up with my math class and since I didn't finish high school theres some things I have to learn on my own. Heres a few things that I didn't figure out in the class. This is in chronological order, things are gradually getting more complex so I'd better keep up now if I'm gonna have any chance of doing well when it gets hard. We're still in the warmup period.

    1.) Unsquaring short equations
    Q.) (2A + B)sq A.) 4Asq + Bsq + 4ab
    I would have just answered it 4Asq + Bsq I have no idea where the 4ab comes from.


    2.) Turning equations with x to the power of 3 into quadratic equations
    I'm fairly lost here. Heres an example question.
    Q.) Xcubed + 5Xsq - 12X - 16 divided by X + 1
    I know how to do the division but I'm wondering what its all about. The aim seems to be to turn this cubed equation into a simpler quadratic equation. To do this can I just divide any of its factors into it?

    3.) Even harder cubed equations. The teacher started throwing the coefficient "K" into the equations. Example:
    Q.) 2Xcubed - Xsq + KX - 12 divided by 2X + 1
    I have no idea what to do with this K

    4.) Finding the coefficients of a cubed equation instead of X. Example
    Q.) Find A and B, factors are X - 1 and X - 2. Equation = Xcubed + AXsq - 7X + B

    I'm having trouble keeping up with the class because when I take time to go over one thing he moves onto another and I gradually fall behind more and more so I have to figure out as much of my notes as I can when I go home. One problem is I don't know the language of math so I have trouble searching for help on certain things I'm having trouble with. I'm glad theres forums like this around.
     
    Last edited: Sep 27, 2008
  2. jcsd
  3. Sep 27, 2008 #2

    cepheid

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    1) This is a common misconception. Here's the explanation:

    Anything squared is merely multiplied by itself (by definition). Hence:

    [tex] (2a + b)^2 = (2a + b) \cdot (2a + b) [/tex]

    Now, an algebraic expression with two terms is called a binomial. The question is, how do we multiply two binomials together? The answer comes from the fact that multiplication is distributive: *each* term in the first binomial has to be multiplied by every term in the second binomial (i.e. it gets distributed through), and vice versa. So the expression becomes:

    [tex] = (2a \cdot 2a) + (2a \cdot b) + (b \cdot 2a) + (b \cdot b) [/tex]

    Note that the two middle terms (which are called the "cross" terms because they involve multiplying the *unlike* terms from each binomial) are identical and can be added together to give you 2ab + 2ab = 4ab. This is true generally and gives you a bit of a shortcut for squaring a binomial. Anyway, so our result is:

    [tex] = 4a^2 + 4ab + b^2 [/tex]
     
    Last edited: Sep 27, 2008
  4. Sep 27, 2008 #3
    Ah that explains it thanks. I was looking at it like it was 2a^ + b^.
     
  5. Sep 27, 2008 #4

    cepheid

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    k is just a constant, unlike x, which is a variable. As the name suggests, the value of a variable can change, whereas you can think of a constant has having some fixed value. So having a k there is no different than having a "2" or a "3" or any other number for that matter. He's just simply not specifying what the value of k is. It's some constant.

    Very simple example:

    I have 2x + kx. How can I "perform addition" in order to rewrite this? Well, the first term says that I have 2 times x, i.e. I have two copies of "x". The second terms says that I have "k" copies of "x", whatever k happens to be. So how many copies of x do I have in total? Answer: Just add together the copies to get a total number of (k + 2) copies

    So we can rewrite 2x + kx as (2 + k)x or as (k + 2)x. (The order in which addition is performed doesn't matter. We say that addition is "commutative").

    This is an example of "collecting" *like* terms. Both terms in this example are terms involving x multiplied by some coefficient. Since both terms contain x, they can be collected together, which amounts to keep track of the total number of "x's" you have in your equation.
     
  6. Sep 27, 2008 #5

    cepheid

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    I don't know what you mean. The aim is not to magically turn the cubic expression into a quadratic one. The aim is to perform the long division i.e. to figure out an algebraic expression for the quotient of these two polynomials -- this expression will truly be their quotient no matter what value x has.

    Since you are dividing a cubic by a linear polynomial, the result will be quadratic in *this* case.
     
  7. Sep 27, 2008 #6
    Just keep at it! Ask for additional help from here and your teacher where you need it.

    Be careful about your terminology. We do not call this "unsquaring".

    In algebra, we have a number of rules associated with arithmetic operations. Here is a quick search result on Google listing some of the rules:

    http://www.themathpage.com/alg/rules-of-algebra.htm

    Some of them are redundant, but you should be familiar with most of them.

    In this instance, you have made a mistake by "inventing" your own rule: [tex](a + b)^2 = a^2 + b^2[/tex]. This statement isn't true! It's ok, it happens to everyone from time to time.

    We know that squaring an expression is the same as multiplying it to itself.
    [tex](a + b)^2 = (a + b) (a + b)[/tex]

    How do we multiply the rest of this out? We use something called the FOIL rule: http://en.wikipedia.org/wiki/FOIL_method

    The poster above me explained roughly how foil works. FOIL is your best friend in algebra. Memorize it and learn to love it =-)





    Again, please be careful with terminology. We don't "turn cubic equations into quadratic equations". The process you are describing is called "factorization". We want to "factorize" [tex]x^3 + 5x^2 - 12x - 16[/tex].

    You should be familiar with prime factorization: that for every integer, you can break it down into prime components.

    6 = 2 * 3
    12 = 2 * 2 * 3
    121 = 11 * 11
    17 = 17

    Each set of numbers on the right hand side is a prime number. In certain cases (namely, for prime numbers such as 17), there is no way to break the number down further.

    For polynomials, we can do something very similar:

    (x^2 +2x + 1) = (x + 1)(x + 1)
    (x^2 - 1) = (x + 1)(x - 1)
    (x^2 + 1) = (x^2 + 1)

    Note that the first two can be factorized into a pair of polynomials. The last one cannot, and is something very much like a prime number.

    For many problems, you might be able to factor by trial and error. Make a guess what the factors might be, and use FOIL to see if you're right. Doing this got me a B in highschool algebra ;-)

    There is a general technique called synthetic division. It's basically long division for polynomials. I never learned it in high school, but I really wish I did now!




    I'm not sure what to do with this either! It looks like the final answer will have a K in it somewhere. You should ask your teacher about this question.

    I'm not sure one this one either. Go in during your teacher's office hours. It's worded a little ambiguously.

    Good luck!
     
  8. Sep 29, 2008 #7
    For problems two and three there is a hint in what you are dividing the cubic equation by to help you factor it.

    x^3+5x^2-12x-16/x+1

    x+1 can be factored out of the denominator

    (x+1)(ax^2+bx+c)/(x+1)

    x+1 cancels out and you are left with just the quadratic to solve.

    Whenever you are dividing one ploynomial by another first try to factor the polynomial that is in the denominator out of the numerator, this will save you the hassle of having to try and divide two polynomials. Keep in mind it won't work out everytime like it did in this one, so don't always expect it. Just try it first everytime to see.
     
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