Quick Way to Expand Large Polynomial ExpressionsDate: 05/27/2008 at 17:17:10 From: Layla Subject: Finding terms and expanding of an equation Find the first three terms of the expansion (x-2)^4(x+1)^8. I tried expanding it but it was too difficult. I needed to have one binomial because to use the equation t(r+1)=(a choose b)a^n - r b^r where a is the first term in the binomial and b is the second, you can only use this equation with one binomial. Date: 05/28/2008 at 00:20:31 From: Doctor Greenie Subject: Re: Finding terms and expanding of an equation Hi, Layla -- It is possible, although in some cases very tedious, to determine any particular coefficient(s) in the product of any number of polynomials. The general problem can quickly get ugly; but in the case where the factors are a "relatively small" number of all linear binomials, the process is manageable. The process is based on considering all the possible "partial products"--all the products you can get by choosing one of the two terms from each binomial. Here is a simple example to demonstrate the process. Suppose we want to determine the expanded form of (x+3)(x+5)(x+7) (1) Our x^3 term comes from choosing the "x" term from each binomial; the coefficient is clearly 1. So the first term of our product is x^3. (2) Our x^2 term comes from all the partial products in which we choose the "x" term from 2 of the 3 binomials and the constant term from the other binomial. So the coefficient of the x^2 term in this example will be 3+5+7 = 15. The second term in our product is 15x^2. (3) Our x term comes from all the partial products in which we choose the "x" term from 1 of the 3 binomials and the constant term from the other 2 binomials. So the coefficient of the x term in this example will be (3*5)+(3*7)+(5*7) = 15+21+35 = 71. The third term of our product is 71x. (4) Our constant term comes from choosing the constant term from all three factors. The constant term in our example is 3*5*7 = 105. (x+3)(x+5)(x+7) = x^3+15x^2+71x+105 The binomial theorem can be viewed as resulting from this type of multiplication. If our product is (x+1)^n then (1) The x^n term comes from choosing the "x" term in all n binomials. (2) The x^(n-1) term comes from choosing the "x" term in n-1 of the n binomials and the "1" in the other binomial. The x^(n-1) term is C(n,n-1)x^(n-1) (3) The x^(n-2) term comes from choosing the "x" term in n-2 of the n binomials and the "1" in the other 2 binomials. The x^(n-2) term is C(n,n-2)x^(n-2) And so on.... Now let's look at your example. (x-2)^4 * (x+1)^8 (1) The x^12 term comes from choosing the "x" term in all 12 binomial factors. The first term in the product is x^12. (2) The x^11 term comes from choosing the "x" term in 11 of the 12 binomial factors and the constant term in the other factor. The number of partial products of this type is C(12,1)=12. Of those 12, C(8,1)=8 have a constant term of 1 and C(4,1)=4 have a constant term of -2. The coefficient of the x^11 term is then C(8,1)(1) + C(4,1)(-2) = 8(1)+4(-2) = 8-8 = 0 (3) The x^10 term comes from choosing the "x" term in 10 of the 12 binomial factors and the constant term in the other 2 factors. The number of partial products of this type is C(12,2)=66. Of those 66, C(8,2)=28 have a constant term of (1)(1)=1, C(4,2)=6 have a constant term of (-2)(-2)=4, and C(8,1)*C(4,1)=8*4=32 have a constant term of (1)(-2). The coefficient of the x^10 term is then C(8,2)(1) + C(4,2)(4) + C(8,1)*C(4,1)(-2) = 28(1)+6(4)+32(-2) = -12 The first three terms of the expansion are x^12 + 0x^11 - 12x^10 If, by "the first three terms", they mean the first three terms with nonzero coefficients, then you still have more work to do to figure out, in a similar manner, the coefficient of the x^9 term.... I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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