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Polynomial Division

  1. Nov 19, 2004 #1
    I have been trying two problems for the longest time, and no one is able to help me I am so stuck, I know how to divide polynomials using long division and synthetic division using simpler numbers but I just cant do these two questions and my course is online, so I can't even ask my teacher :cry: plz help me!
    The question is

    -6x^4+2x^2-8x+1 / 2x^2-3 using long division and synthetic division

    For long division I tried putting in place value zeroes but it still didnt work, and in the synthetic division I didnt even know what the divisor would be, lol Im in big trouble plz show me these two questions step by step thank you :smile:
     
  2. jcsd
  3. Nov 19, 2004 #2
    In synthetic division, the first thing you'll want to do is to find the monomial multiplier for [tex]2x^2[/tex] that will give [tex]-6x^4[/tex] ([tex]-3x^2[/tex]). Next, check to see whether that multiplying the binomial [tex]2x^2 - 3[/tex] by [tex]-3x^2[/tex] is enough to complete the division:
    [tex]-3x^2 \cdot (2x^2 - 3) = -6x^4 + 9x^2[/tex]
    It's not enough, so try to divide the remaining reduced degree polynomial:
    [tex](-6x^4+2x^2-8x+1) - (-6x^4 + 9x^2) = 11x^2 - 8x + 1[/tex]
    in the same manner.
    Your quotient will be [tex]-3x^2 + [/tex] whatever you get from this division. :smile:
     
    Last edited: Nov 19, 2004
  4. Nov 19, 2004 #3
    Have you considered instead of dividing by 2(x^2)-3 just dividing by [tex]x-\sqrt \frac{3}{2}[/tex] and [tex]x+\sqrt \frac {3}{2}[/tex]? Stick with what you know if you only know binomials of degree 1.
     
  5. Nov 19, 2004 #4
    I still dont understand

    Ok for long division I keep getting a number (-7) that can not be divided by my divisor of 2x^2, I dont know how to show you how far I have gotten
    I made 2x^2-3 --> 2x^2+0x-3 and divided this into -6x^4+0x^3+2x^2-8x+1 these zeroes are the ones I was talking about before that I added but it still didnt help me in dividing. well I then saw how many times does 2x^2 go into -6x^4 and got the answer -3x^2 I multiplied this by the whole divisor 2x^2+0x-3 and got -6x^4+0x^3+9x^2, now when i subtract this from what is being divided 2-9 = -7 then I get stuck because the divisor can only go into 7x^2 if it is 3.5. I was thinking that maybe me adding a place value 0 in the divisor was wrong, but I only did that because a x value was missing. I still need help, I need to actually see this question done in the division format if possible, I didnt understand the two replies given to me.

    For the synthetic division with the same numbers I need it in the synthetic format in order to understand so like below if the number was (x+3) then there would be an -3 outside of the "L" shape see my number is 2x^2-3 I dont know from this what I am supposed to write on the outside of this (L) because of the square and the coefficient 2. I hope someone understands what I am saying, and plz help me this thing is driving me crazy :frown: Thanks for any replies :smile:

    l 6 2 8 1
    -3 l__________
     
  6. Nov 22, 2004 #5
    Plz help if u know

    Can somone plz help me????
     
  7. Nov 22, 2004 #6
    Well, I see one of your problems. You have 6x^4 and 2x^2... but you need an x^3 in between. Did you teacher tell you that, if you are missing a term in the sequence.. that you put a 0 in there?

    So, when you use sythetic division, it your list would be -6 0 2 -8 1
     
  8. Nov 22, 2004 #7
    yes my teacher told me

    Yes my teacher did tell me to add zeros and i did but it still didnt work, can any one help me out more and show me how to do this plz
     
  9. Nov 25, 2004 #8

    BobG

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    Science Advisor
    Homework Helper

    For long division, you got down to

    -7x^2 -8x+1

    and weren't sure what to do because that meant you -7 divided 2. In normal long division, you pick the smallest number (abs value) that will be equal to or less (abs value) than the number you divided by.

    In other words, you have -3 on the top, and divide back through. For this problem, you'll wind up with a remainder. Whatever polynomial you wind up with on the top, just add the fraction (remainder/divisor).

    Technically, you could choose -4, as well, and the answer would wind up being equivalent for virtually all values of x. (Except if you graphed the answers, the one that used -4 would have a nasty little 'squiggle' on the left side).
     
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