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Math Help - Simple Polynomial Multiplication problem

  1. #1
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    Simple Polynomial Multiplication problem

    If you ask me to simplify (x+3)(x+5), I'd do it easily by multiplying both sides.

    (x+3)(x+5)=
    x*x+5x+3x+3*5=
    x2+8x+15

    But if you ask:

    Explain: x2+8x+15 = (x+3)(x+5)

    I'd be dumbfounded.
    How do I "see" that it is so more logically other than remembering what a simple polynomial multiplication looks like? help on this is MUCH required as I'll be tested both ways, I will definitely fail if I can't wrap my head around this one. PLEASE help me see it! Thanks in advance, any thought-provoking comments appreciated.
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  2. #2
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    Quote Originally Posted by kappa View Post
    If you ask me to simplify (x+3)(x+5), I'd do it easily by multiplying both sides.

    (x+3)(x+5)=
    x*x+5x+3x+3*5=
    x2+8x+15

    But if you ask:

    Explain: x2+8x+15 = (x+3)(x+5)

    I'd be dumbfounded.
    How do I "see" that it is so more logically other than remembering what a simple polynomial multiplication looks like? help on this is MUCH required as I'll be tested both ways, I will definitely fail if I can't wrap my head around this one. PLEASE help me see it! Thanks in advance, any thought-provoking comments appreciated.
    For ax^2 + bx+c = 0

    When there is no coefficient on x^2 then it's easier (ie: a=1). Let p and q be the numbers we need to find. Because of FOIL (the method you used to expand) the two last terms need to multiply to make c and add to make b:

    pq = c
    p+q = b

    c and b are given in the quadratic to be factorised and so you can solve for p and q, with practice it gets easier especially since p and q must be factors of c.

    It is also possible that some do not factorise into rational numbers, you can check if your equation is one by checking the discriminant. If b^2-4ac (the discriminant) is a perfect square it can be factored rationally
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  3. #3
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    Thanks for replying. I see how practice will make this easier, but time constraints won't allow me much practice.
    There is something I'm trying to do. So the question is:

    Simplify x^2+8x+15

    x*x+8*x+3*5=
    x(x+8)+3*5=

    Am I correct so far? How do I continue in order to make "(x+3)(x+5)" ?
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  4. #4
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    Talking

    Quote Originally Posted by kappa View Post
    Simplify x^2+8x+15
    There are no "like" terms in this polynomial, so it is already fully simplified.

    Quote Originally Posted by kappa View Post
    x*x+8*x+3*5=
    x(x+8)+3*5=

    Am I correct so far? How do I continue in order to make "(x+3)(x+5)" ?
    Ah. So the instructions should have told you to "factor". That's quite a different topic.

    To learn how to factor quadratics (which is very different from the common-factor factoring with which you appear to be familiar), try some online lessons. (The topic usually comprises much or all of a textbook chapter, so obviously it's a bit much to present here.)

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