Multiplying Binomials and Other PolynomialsDate: 04/16/2004 at 11:00:38 From: shirley Subject: binomials I have a hard time multiplying binomials properly, such as (2x + 5)(x + 7). I do not know what order the problem should be solved in. Date: 04/18/2004 at 15:48:19 From: Doctor Floor Subject: Re: binomials Hi, Shirley, Thanks for your question. To find the right answers you could divide 2x + 5 into the terms 2x and +5, and x + 7 into x and +7, and put them into a table: | 2x | +5 | --+----+----+ x | | | --+----+----+ +7| | | --+----+----+ Now each of the open boxes in the table we fill with the product of the terms on the left hand side of it and above it. So in the box with ... | 2x | +5 | --+----+----+ x | ...| | --+----+----+ +7| | | --+----+----+ we write the product of x and 2x, which is 2x^2: | 2x | +5 | --+----+----+ x |2x^2| | --+----+----+ +7| | | --+----+----+ This we do for all four boxes, resulting in: | 2x | +5 | --+----+----+ x |2x^2|+5x | --+----+----+ +7|+14x|+35 | --+----+----+ The results of the four multiplications we put together, and we have: 2x^2 + 5x + 14x + 35 which we can simplify to 2x^2 + 19x + 35 and we are done. Another way to think about a problem like this is to multiply each term of the first binomial times each term of the second binomial, and again add all the results. So: (2x + 5)(x + 7) Multiply the 2x times the x and the +7, then multiply the +5 times the x and the +7: 2x(x) + 2x(7) + 5(x) + 5(+7) 2x^2 + 14x + 5x + 35 2x^2 + 19x + 35 Note that with either method you can do the same thing when multiplying any two polynomials. Just remember to multiply each term of the first polynomial times each term of the second polynomial, or if you use the chart, make a row or column for each term. So, if you have (4x - 5)(x^2 - 6x + 3), you can do the same thing. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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