document.write( "Question 754231: Use the Binomial Theorem to find the binomial exspansion of the expression. (x - 3)^5 \n" ); document.write( "
Algebra.Com's Answer #458903 by Edwin McCravy(20056)\"\" \"About 
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document.write( "(x - 3)5\r\n" );
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document.write( "The exponent is 5, so there will be 1 more term than the exponent.  So there\r\n" );
document.write( "will be 6 terms.  The two terms in the binomial are x and -3. so write\r\n" );
document.write( "\" (x) (-3) \" 6 times with + signs sandwiched in between, and leaving room for\r\n" );
document.write( "an exponent and a coefficient for each term, like this:\r\n" );
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document.write( " (x) (-3)  +  (x) (-3)  +   (x) (-3)  +   (x) (-3)  +  (x) (-3)  +  (x) (-3) \r\n" );
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document.write( "Next fill in the exponents of the first term (x) descending from 5 to 0.\r\n" );
document.write( "That is, write in exponents of (x) that start with 5 and go down to 0, i.e.,\r\n" );
document.write( "5,4,3,2,1,0, like this:  \r\n" );
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document.write( "  (x)5(-3)  +  (x)4(-3)  +   (x)3(-3)  +   (x)2(-3)  +  (x)1(-3)  +  (x)0(-3) \r\n" );
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document.write( "Next fill in the exponents of the second term (-3) ascending from 0 to 5.\r\n" );
document.write( "That is, write in exponents of (-3) that start with 0 and go up to 5, i.e.,\r\n" );
document.write( "0,1,2,3,4,5, like this:\r\n" );
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document.write( "  (x)5(-3)0 +  (x)4(-3)1 +   (x)3(-3)2 +   (x)2(-3)3 +  (x)1(-3)4 +  (x)0(-3)5\r\n" );
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document.write( "Now we have everything except the coefficients and the simplification.\r\n" );
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document.write( "There are three methods for getting the 6 coefficients:\r\n" );
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document.write( "Method 1:  Pascal's triangle:          \r\n" );
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document.write( "          1\r\n" );
document.write( "        1   1\r\n" );
document.write( "      1   2   1\r\n" );
document.write( "    1   3   3   1\r\n" );
document.write( "  1   4   6   4   1\r\n" );
document.write( "1   5  10  10   5   1\r\n" );
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document.write( "So the coefficients of the 6 terms are 1, 5, 10, 10, 5, 1\r\n" );
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document.write( "Method 2:  Calculate them by formula for combinations:\r\n" );
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document.write( "C(5,0) = \"%28matrix%282%2C1%2C5%2C0%29%29\" = \"5%21%2F%280%21%2A%285-0%29%21%29\" = \"5%21%2F%280%21%2A5%21%29\" = \"5%2A4%2A3%2A2%2A1%2F%28%281%29%2A%285%2A4%2A3%2A2%2A1%29%29\" = \"cross%285%2A4%2A3%2A2%2A1%29%2F%281%2Across%28%285%2A4%2A3%2A2%2A1%29%29%29\" = 1  \r\n" );
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document.write( "C(5,1) = \"%28matrix%282%2C1%2C5%2C1%29%29\" = \"5%21%2F%281%21%2A%285-1%29%21%29\" = \"5%21%2F%281%21%2A4%21%29\" = \"5%2A4%2A3%2A2%2A1%2F%28%281%29%2A%284%2A3%2A2%2A1%29%29\" = \"5%2Across%284%2A3%2A2%2A1%29%2F%28%281%29%2Across%28%284%2A3%2A2%2A1%29%29%29\" = 5 \r\n" );
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document.write( "C(5,2) = \"%28matrix%282%2C1%2C5%2C2%29%29\" = \"5%21%2F%282%21%2A%285-2%29%21%29\" = \"5%21%2F%282%21%2A3%21%29\" = \"5%2A4%2A3%2A2%2A1%2F%28%282%2A1%29%2A%283%2A2%2A1%29%29\" = \"5%2A4%2Across%283%2A2%2A1%29%2F%28%282%2A1%29%2Across%28%283%2A2%2A1%29%29%29\" = 10\r\n" );
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document.write( "C(5,3) = \"%28matrix%282%2C1%2C5%2C3%29%29\" = \"5%21%2F%283%21%2A%285-3%29%21%29\" = \"5%21%2F%283%21%2A2%21%29\" = \"5%2A4%2A3%2A2%2A1%2F%28%283%2A2%2A1%29%2A%282%2A1%29%29\" =  = 10\r\n" );
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document.write( "C(5,4) = \"%28matrix%282%2C1%2C5%2C4%29%29\" = \"5%21%2F%284%21%2A%285-4%29%21%29\" = \"5%21%2F%284%21%2A1%21%29\" = \"5%2A4%2A3%2A2%2A1%2F%28%284%2A3%2A2%2A1%29%2A%281%29%29\" = \"5%2Across%284%2A3%2A2%2A1%29%2F%28cross%284%2A3%2A2%2A1%29%2A1%29\" = 5\r\n" );
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document.write( "C(5,5) = \"%28matrix%282%2C1%2C5%2C5%29%29\" = \"5%21%2F%285%21%2A%285-5%29%21%29\" = \"5%21%2F%285%21%2A0%21%29\" = \"5%2A4%2A3%2A2%2A1%2F%28%285%2A4%2A3%2A2%2A1%29%281%29%29\" = \"cross%285%2A4%2A3%2A2%2A1%29%2F%28cross%28%285%2A4%2A3%2A2%2A1%29%29%2A%281%29%29\" = 1\r\n" );
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document.write( "Methgod 3: Calculate them recursively:\r\n" );
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document.write( "Coefficient 1:  Always 1, so we have \r\n" );
document.write( "1(x)5(-3)0\r\n" );
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document.write( "Coefficient 2:  \r\n" );
document.write( "Take the previous coefficient, 1, multiply by the exponent of \r\n" );
document.write( "the first term, x, get 1*5 or 5. divide by the number of term \r\n" );
document.write( "which the previous term is, which is the 1st term so we divide \r\n" );
document.write( "5 by 1 and get 5. So we have\r\n" );
document.write( "1(x)5(-3)0 + 5(x)4(-3)1     \r\n" );
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document.write( "Coefficient 3:  \r\n" );
document.write( "Take the previous coefficient, 5, multiply by the exponent of \r\n" );
document.write( "the first term, x, get 5*4 or 20. divide by the number of term \r\n" );
document.write( "which the previous term is, which is the 2nd term so we divide \r\n" );
document.write( "20 by 2 and get 10. So we have\r\n" );
document.write( "1(x)5(-3)0 + 5(x)4(-3)1 + 10(x)3(-3)2\r\n" );
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document.write( "Coefficient 4:  \r\n" );
document.write( "Take the previous coefficient, 10, multiply by the exponent of \r\n" );
document.write( "the first term, x, get 10*3 or 30. divide by the number of term \r\n" );
document.write( "which the previous term is, which is the 3rd term so we divide \r\n" );
document.write( "30 by 3 and get 10. So we have\r\n" );
document.write( "1(x)5(-3)0 + 5(x)4(-3)1 + 10(x)3(-3)2 + 10(x)2(-3)3\r\n" );
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document.write( "Coefficient 5:  \r\n" );
document.write( "Take the previous coefficient, 10, multiply by the exponent of \r\n" );
document.write( "the first term, x, get 10*2 or 20. divide by the number of term \r\n" );
document.write( "which the previous term is, which is the 4th term so we divide \r\n" );
document.write( "20 by 4 and get 5. So we have\r\n" );
document.write( "1(x)5(-3)0 + 5(x)4(-3)1 + 10(x)3(-3)2 + 10(x)2(-3)3 + 5(x)1(-3)4\r\n" );
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document.write( "Coefficient 6:  \r\n" );
document.write( "Take the previous coefficient, 5, multiply by the exponent of \r\n" );
document.write( "the first term, x, get 5*1 or 5. divide by the number of term \r\n" );
document.write( "which the previous term is, which is the 5th term so we divide \r\n" );
document.write( "5 by 5 and get 1. So we have\r\n" );
document.write( "1(x)5(-3)0 + 5(x)4(-3)1 + 10(x)3(-3)2 + 10(x)2(-3)3 + 5(x)1(-3)4 + 1(x)0(-3)5\r\n" );
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document.write( "So coefficients 1,5,10,10,5,1, are what you get regardless of which \r\n" );
document.write( "of the three methods you use to get the coefficients.  There is the \r\n" );
document.write( "shortcut that the coefficients are symmetrical, so when you get \r\n" );
document.write( "half-way through, you know what the rest are because they are just \r\n" );
document.write( "like the first half.\r\n" );
document.write( "Anyway,\r\n" );
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document.write( "1(x)5(-3)0 + 5(x)4(-3)1 + 10(x)3(-3)2 + 10(x)2(-3)3 + 5(x)1(-3)4 + 1(x)0(-3)5\r\n" );
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document.write( "simplifies to:\r\n" );
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document.write( "x5 - 15x4 + 903 - 2703 + 405x - 243\r\n" );
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document.write( "I assume you can do that.\r\n" );
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document.write( "Edwin
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