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Math Help - very confused on factoring

  1. #1
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    Exclamation very confused on factoring

    I'm so so confused on how to factoring problems that have trinomials..

    just as an example i have 3x^2 + 7x +2
    i'm jst getting really confused and i need someone to explain it to me in SIMPLE words, my teacher is really confusing

    so basically with the problem i gave above i need to factor it completely, and then pick one of the factors and match it with a column thats adjacent to it..but that's what i'll do after someone explains HOW to factor!!

    please help
    (just for some more examples
    3x^2 - 16x + 5
    7x^2 - 9x + 2
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ncbabe View Post
    just as an example i have 3x^2 + 7x +2
    This is called the "ac method."

    Multiply the coefficient of the quadratic term by the constant term. In this case
    3 \cdot 2 = 6

    Now write all the possible factors of 6 in pairs:
    1, 6
    2, 3 <-- We don't need to do 3, 2 and 6, 1. They are considered repeats.
    -1, -6
    -2, -3

    Now add each of these factors together:
    1 + 6 = 7
    2 + 3 = 5
    -1 + -6 = -7
    -2 + -3 = -5

    Now compare this list to the coefficient of the linear term. In this case we have a 7. Thus we want to choose the 1, 6 pair. (Note: If no such pair exists, we cannot factor the trinomial.)

    So:
    3x^2 + 7x +2

    Write the linear term as a sum using the pair we found above: 7x = 1x + 6x
    3x^2 + x + 6x +2

    = (3x^2 + x) + (6x + 2)

    Now factor what you can from the parenthesis. The first pair of terms has a common x and the second pair of terms has a common 2:
    = x(3x + 1) + 2(3x + 1)

    Now note that each term has a common 3x + 1. Factor this out to the right:
    = (x + 2)(3x + 1)

    As a check you should always multiply this out:
    (x + 2)(3x + 1) = x \cdot 3x + 2 \cdot 3x + x \cdot 1 + 2 \cdot 1

    = 3x^2 + 6x + x + 2 = 3x^2 + 7x + 2

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ncbabe View Post
    3x^2 - 16x + 5
    I'll do one more.

    3x^2 - 16x + 5

    3 \cdot 5 = 15

    Factors of 15:
    1, 15 --> 1 + 15 = 16
    3, 5 --> 3 + 5 = 8
    -1, - 15 --> -1 + -15 = -16 <-------------
    -3, -5 --> -3 + -5 = -8

    So we choose the -1, -15 pair.

    Thus
    3x^2 - 16x + 5

    = 3x^2 - 1x + -15x + 5

    = (3x^2 - x) + (-15x + 5)

    The first pair of terms has a common x and the second pair of terms has a common 5. It will be useful to factor a -5 from both terms; try to keep that linear coefficient positive on this one. And watch the signs!

    = x(3x - 1) + -5(3x - 1)

    Now factor the common 3x - 1:
    = (x + -5)(3x - 1) = (x - 5)(3x - 1)

    I'll leave it to you to check that this is factored properly.

    -Dan
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  4. #4
    Junior Member Dergyll's Avatar
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    Hey dan

    Is there a better way to factore ^3s and ^4s...etc (trinomials and up) other than using synthetic division and/or dividing?

    Thanks
    Derg
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Dergyll View Post
    Hey dan

    Is there a better way to factore ^3s and ^4s...etc (trinomials and up) other than using synthetic division and/or dividing?

    Thanks
    Derg
    i think you mean third degree polynomials and up, and the answer is no. there are formulas for third and fourth degree polynomials, but they are a mess and are often not worth the hassle. I don't think there is any formula for fifth degree polynomials and up. Synthetic division and long division are definately the easiest ways to tackle these problems, unless you can spot factors right away, which is sometimes possible
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Jhevon View Post
    i think you mean third degree polynomials and up, and the answer is no. there are formulas for third and fourth degree polynomials, but they are a mess and are often not worth the hassle. I don't think there is any formula for fifth degree polynomials and up. Synthetic division and long division are definately the easiest ways to tackle these problems, unless you can spot factors right away, which is sometimes possible
    It has been proven impossible (to write a formula) for 5th degree polynomials and up.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    It has been proven impossible (to write a formula) for 5th degree polynomials and up.
    I thought so, but i wasn't absolutely sure, so i said i think
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    Quote Originally Posted by ecMathGeek View Post
    It has been proven impossible (to write a formula) for 5th degree polynomials and up.
    I am supprised how many people know this famous result from Galois theory. Because it is by no means close to elementary.

    Where did you here it from?
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  9. #9
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I am supprised how many people know this famous result from Galois theory. Because it is by no means close to elementary.

    Where did you here it from?
    The proof has never been shown to me (I doubt I would understand it).

    I actually heard this from a few different teachers. I think the first to tell me this was a College Algebra teacher back in High School.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I am supprised how many people know this famous result from Galois theory. Because it is by no means close to elementary.

    Where did you here it from?
    i actually never heard about it until college

    it was around the time i was doing calc 3, i heard some students talking about it
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