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Math Help - multinominal expansion

  1. #1
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    multinominal expansion

    Hello, I have this multinominal and I also have the expanded version of it.
    The problem is I cannot understand how the expanded version was achieved.
    Can someone please explain it to me. Thanks kindly for any help.

    2x^3 + 9x^2 +12x +4

    expands to:

    (2x+1) (x^2 + 4x +4)

    but what is the method ?
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  2. #2
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    Quote Originally Posted by fran1942 View Post
    (2x+1) (x^2 + 4x +4) but what is the method ?
    The bad news is that it is tedious. Do it term by term.
    (2x)(x^2)+(2x)(4x)+(2x)(4)+(1)(x^2)+(1)(4x)+(1)(4)
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    thanks kindly for the help.
    What I really meant was what method would you follow if you did not know the answer that I provided ?
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    Quote Originally Posted by fran1942 View Post
    thanks kindly for the help.
    What I really meant was what method would you follow if you did not know the answer that I provided ?
    You follow the method that I gave you.
    (a+b+c+d)(e+f+g) is found term by term.
    ae+af+ag+be+bf+bg+ce+cf+cg+de+df+dg
    That is the way it is done.

    (x+1)(x^5+x^3+1)=x^6+x^4+x+x^5+x^3+1
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  5. #5
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    thanks, and how would go in the opposite direction ?

    e.g. expanding:

    2x^3 + 9x^2 +12x +4
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  6. #6
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    Quote Originally Posted by fran1942 View Post
    thanks, and how would go in the opposite direction ?
    e.g. expanding: 2x^3 + 9x^2 +12x +4
    You would factor the expression not expand it.
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  7. #7
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    Hello, fran1942!

    Your language is incorrect and confusing . . .


    I have this multinominal and I also have the expanded version of it.

    The problem is I cannot understand how the expanded version was achieved.

    Can someone please explain it to me. Thanks kindly for any help.


    2x^3 + 9x^2 +12x +4 . This is the expanded version!

    . . \rlap{//////////}\text{expands to: }\;\text{factors to:}

    (2x+1) (x^2 + 4x +4) . This is the factored form!

    You are asking how the polynomial was factored, right?

    Apply the Rational Roots Theorem and the Remainder Theorem.

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  8. #8
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    If this could be factored as " (ax+ b)(cx^2+ dx+ e)" then the leading coefficient would be ac and the constant term would be bd. That is, a must be a factor of the leading coefficient, 2, and b must be a factor of the constant term, 4. Further, if ax+ b is a factor, then x= -b/a (so that ax+ b= 0) would be a zero of the polynomial.

    The only possible factors of 2, the only possible values for a, are 2 and 1. The only possible factors of 4, the only possible values of b, are 4, 2, and 1. So the only possible values for x= b/a are 4/2= 2, 2/2= 1, 1/2. Further, since a sum of positive numbers cannot be 0, any possible rational zero of the polynomial must be one of -2, -1, -1/2 and -4.

    For
    x= -2, the polynomial is 2(-8)+ 9(4)+ 12(-2)+ 4= -16+ 36- 24+ 4= 0.
    So we get an immediate zero, x= -2, and a factor: x-(-2)= x+ 2.

    Now, divide 2x^3 + 9x^2 +12x +4 by x+ 2:
    x+ 2 divides into 2x^3+ 9x^2+ 12x+ 4 exactly 2x^2+ 5x+ 2 so we must have
    2x^3 + 9x^2 +12x +4= (x+ 2)(2x^2+ 5x+ 2). Repeating that argument on the last factor, or because it is quadratic, using the quadratic formula to find integer zeros, we find that it cannot be factored further in terms of integer coefficients.

    (The quadratic formula gives the roots of 2x^2+ 5x+ 2= 0 as \frac{-5\pm\sqrt{25- 8}{4}= \frac{-5}{4}\pm\sqrt{17}}{4} so that we can factor into linear terms as 2x^3 + 9x^2 +12x +4= \left(x+ 2\right)\left(x+ \frac{5}{4}+ \frac{\sqrt{17}}{4}\right)\left(x+ \frac{5}{4}- \frac{\sqrt{17}}{4}\right).
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