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Math Help - need some help with polynomials

  1. #1
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    need some help with polynomials

    heres the questions i have to answer can someone help me out with the answers:

    v^2-8v+7
    w^2-2w-48
    6t^2-36t+48
    3x^7+6x^6-9x^5
    a^6+7a^5b-8a^4b^2
    x^2+9x+18
    x^2-x-2
    r^2+14r+45
    t^2+2t-35
    s^2+2s-35


    sorry they arents polynominals they are trinomials i need to factor them
    Last edited by sdtronrud; December 8th 2009 at 10:15 AM.
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  2. #2
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    Are you saying you can't do ANY of them?
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  3. #3
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    ya i am having a hard time figuring these out
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  4. #4
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    Are they equal to something or should they be factorised?
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  5. #5
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    i have to complete the factoring
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  6. #6
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    Quote Originally Posted by sdtronrud View Post
    i have to complete the factoring
    I'll take the first as an example but bear in mind nobody will give you all the answers, moreso if you don't show any attempt of having done them yourself

    v^2-8v+7

    As the number infront of v is 1 we know that we have (v+p)(v+q) where p and q are constants

    We also know that pq must equal 7. As 7 is prime only +7 and +1 or -7 and -1 will work. To figure out which p+q must add up to the middle number infront of the v (which is -8). As -7+(-1) = -8 we have the answer

    v^2-8v+7 = (v-7)(v-1)
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  7. #7
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    i am still very confused i just dont get these damn nomials, please break it down even more.
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  8. #8
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    Sounds like you need to do a lesson on them, look around there is plenty of websites around
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  9. #9
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    ok like in this example:

    ex. 6x^2 +19x -20
    1) multiply the first coefficient by the constant
    (6)(-20)= -120
    2) rewrite it to x^2 +19x -120 (notice that 6 is no longer there)
    3) factor the new trinomial (changes to x^2 +bx +c)
    (x-5)(x+24)
    4) add the first coefficient back to the equation (in front of x)
    (6x-5)(6x+24)
    5) find the common factor of the numbers in each bracket
    ex. for the first bracket (6 and 5) there is no common factor so
    it's left as is
    for the second bracket (6 and 24) the common factor is 6
    then divide each bracket by the common factor
    6) result is (6x-5)(x+4)


    i understand how to get from point 1-2 but what i dont get is in #3 where the hell did
    the (x-5)(x+24) come from?
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  10. #10
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    Hi...

    I didn't really check the previous steps but since you said that you understand how to get from 1-2, I'm just going to explain #3. You said that you didn't know where (x-5)(x+24) came from...

    Going back to the previous polynomial, you came up with x^2 + 19x-120. Remember, when we want to factor this out we want to find two numbers whose sum is 19 and whose product is -120. This can take some time depending on the numbers. You can factor 120 and pair the numbers accordingly to see which ones better fit the scenario. Or, you can use the quadratic formula. The only thing is that you will come up with 5 and -24 with the quadratic formula and to turn it into factored form, you would just have to
    set x=5 and solve for x and set x=-24 and solve for x, which will result in
    (x-5)(x+24)....

    I hope this helps....if not I'll try to explain better (I am a newbie after all).
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  11. #11
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    can u put it in more easier terms still extremely confused
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  12. #12
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    I assumed you were talking about crystal's explanation, so I'll go from there.
    Your polynomial appears to be

    x^2 + 19x -120. So, instead of thinking about it in terms of factoring out the 120 and pairing up terms, let's look at the quadratic formula. I'm going to assume that you know what the quadratic formula is, because I'm not sure how to type it in here and make it look good.

    But in our problem we see that a = 1, b = 19, and c = -120. By plugging these values into the quadratic formula you will get two roots. Those roots should be x = 5 or -24. However, we want this in factored form.

    x= 5 gives x-5=0
    x=-24 gives x+24=0. So our factors are (x-5)(x+24).

    The factoring 120 is like a trial and error.
    1 x 120
    2 x 60
    3 x 40
    4 x 30
    5 x 24
    6 x 20
    8 x 15
    10 x 12
    12 x 10....we can stop here because the numbers are just going to repeat, just in the opposite order. But aside from that, what crystal said was true. I'll give you an example from the list....

    Take 1 and 120. If we add these numbers together we get 121, which is not equal to 19 and if we multiply them together we won't get -120. Let's say -1 and -120. Once again, if we add these together, we get -121 and multiplying would give 120.
    1 x -120 = -120, but 1+(-120) = -119.
    -1 x 120 = -120, but -1+ 120 = 119.

    These are the kinds of quick calculations you would have to do if you were thinking about taking this route.

    Skipping to the answer. take 5 and 24.
    -5 + 24 = 19 and (-5) x (24) = -120....so this we see works. Remember to check your answer, just multiply the factors together and you should come up with the original polynomial. Hope this isn't too wordy hehehe. Good luck!
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  13. #13
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    Quote Originally Posted by sdtronrud View Post
    ex. 6x^2 +19x - 20
    .........
    6) result is (6x-5)(x+4)
    Do you at least realise that (6x-5) times (x+4) = 6x^2 + 19x - 20 ?
    The multiplication:
    6x - 5
    x + 4
    =====
    6x^2 - 5x : multiply top by x
    24x - 20 : multiply top by 4
    Now add 'em up:
    6x^2 - 5x + 24x - 20 = 6x^2 + 19x - 20 ; get that?

    In other words, factoring is coming up with WHAT WAS MULTIPLIED.
    The "how" takes practice.
    It's usually easier to use the quadratic formula.
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