Results 1 to 6 of 6

Math Help - Problems with factorization in regard to defining a polynomial

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    9

    Problems with factorization in regard to defining a polynomial

    Hi

    Following cordinate-points are given to define a function:

    (˝,o), (-2,0) and (1,-3) with the parable roots being ˝ and -2. Since d>0, I used the formula: f(x)= a(x-r1)(x-r2), where r1= first root and r2= second root.

    So this is what I end up with:

    f(x) = a(x-˝)(x+2) <=> f(x)= a(x^2+ 2x- ˝x- 1) <=> f(x)= ax^2+ 1˝x-1

    Then I isolate a, by inserting the x and y values of the third coordinate-point (1,-3) and I get:

    -3= a(1-0.5)(1+2) <=> -3= a(1+2-0.5-1) <=> -3= a(1.5) <=> -3/1.5 = a <=> a= -2

    So the final definition of the function is:

    f(x) = -2ax^2 + 1˝x - 1

    But in the answer list in my text-book, it says the definition should be:

    f(x)= 2x^2 + 3x- 2

    Can someone please explain to me where I went wrong?

    Thank you in advance.

    Regards
    Lia
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,889
    Thanks
    326
    Awards
    1
    Quote Originally Posted by Lia85 View Post
    Hi

    Following cordinate-points are given to define a function:

    (˝,o), (-2,0) and (1,-3) with the parable roots being ˝ and -2. Since d>0, I used the formula: f(x)= a(x-r1)(x-r2), where r1= first root and r2= second root.

    So this is what I end up with:

    f(x) = a(x-˝)(x+2) <=> f(x)= a(x^2+ 2x- ˝x- 1) <=> f(x)= ax^2+ 1˝x-1
    First, you should probably leave the 1 \frac{1}{2} as 3/2.

    Second in the line above the final entery should be ax^2 + \frac{3}{2}ax - a.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    9
    Hi Dan

    If a is the slope, then why does it appear thrice? The general formula for a quadratic is ax^2 + bx + c. But maybe it was just a typo? Or maybe there's something I'm just not seeing?

    I'll be sure to rephrase 1˝x to 3/2x, but even so, it still doesn't solve my problem and i still end up with the same expression: f(x) = -2ax^2 + 3/2x - 1, which is somewhat of a far cry from what the answer in my textbook is, namely
    f(x)= 2x^2 + 3x- 2

    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    When you used the formula f(x) = a(x-r_1)(x-r_2) the a is out front so you need to expand it out. Indeed, it is not a slope, only straight lines have slope.

    Do you understand that f(x) = -2(x^2+1.5x-1) = (-2 \cdot x^2) + (-2 \cdot 1.5) + (-2 \cdot -1)

    Once simplified that is a factor of -1 out but I see no way to reconcile that (perhaps it's an even function?)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2011
    Posts
    9
    Hi e^(i*pi)

    Quote Originally Posted by e^(i*pi) View Post
    When you used the formula f(x) = a(x-r_1)(x-r_2) the a is out front so you need to expand it out. Indeed, it is not a slope, only straight lines have slope.
    Yes you're right, a in this case is just a coefficient that appears infront of the quadratic term. I just got thrown away by assigning the same arbitrary letter to all of the constants. Sorry for the mix-up.

    Quote Originally Posted by e^(i*pi) View Post
    Do you understand that f(x) = -2(x^2+1.5x-1) = (-2 \cdot x^2) + (-2 \cdot 1.5) + (-2 \cdot -1)
    Yes I understand the point you're trying to make, but I don't know if it is applicable in this case.

    The value of f(x)= -2x^2 + 3/2x - 1 is not conserved if you rephrase the expression to f(x)= (-2*x^2) + (-2*1.5x) + (-2* -1), because the coefficient -2 is only to be multiplied with the quadratic term, as seen in the original expression I ended up with. Had the expression been f(x)= -2(x^2 +1.5x -1), then the above method would be right.

    The only way I can see how you might have arrived at f(x)= -2(x^2 +1.5x -1) is if maybe you extracted it from the formula f(x)= a(x-r1)(x-r2). If so could you please show me how, step by step?

    I apologise in advance for my limited math-abilities, I just want to understand, that's all.

    Thanks for the contributions
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2011
    Posts
    9

    I got it now...

    Having to do it the other way around, i.e. breaking down a predefined functions into factors, made me understand what both of you were trying to say. Thanks for the help... And yeah I guess the problem is solved, well except the negative sign infront of a... Oh well maybe it's an error in the textbook.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Factorization of the polynomial
    Posted in the Algebra Forum
    Replies: 5
    Last Post: August 1st 2011, 09:27 PM
  2. Factorization of the polynomial
    Posted in the Algebra Forum
    Replies: 5
    Last Post: December 1st 2010, 07:13 PM
  3. Polynomial Unique Factorization
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: August 9th 2010, 05:20 AM
  4. Polynomial factorization
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 22nd 2010, 10:03 AM
  5. Replies: 3
    Last Post: June 26th 2008, 11:26 PM

Search Tags


/mathhelpforum @mathhelpforum