Algebra.Com's Answer #54628 by bucky(2189)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "Cody, you have to look for combinations of numbers that involve all the factors of 6 and all
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document.write( "the factors of 2, and you have to also consider the signs of those factors. The factors
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document.write( "of the -2 are easy ... just 2 and 1. And you can tell that one of those numbers has to be
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document.write( "positive and one has to be negative ... so that when you multiply them together you get
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document.write( "-2, not +2. So now we can tell that our factors must be of the form:
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document.write( "(____ __ 2)*(____ __ 1)
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document.write( "and at this point we don't know which one is positive and which one is negative.
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document.write( "Now consider the possible factors of 6. Not only are there 6 and 1 which you used, but
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document.write( "there are also 3 and 2. You convinced yourself that 6x and 1x would not work. So now
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document.write( "you have to try 3x and 2x. Let's guess that the factors are:
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document.write( "(2x ___ 2)*(3x ___ 1)
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document.write( "The underlines are where we need to put the two signs ... one of them a + sign and the
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document.write( "other a - sign.
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document.write( "If you multiply 3x times 2 you get 6x and if you multiply 2x times 1 you get 2x. Is there
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document.write( "anyway that these two products can be added or subtracted to give the +x in the middle of
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document.write( "the original polynomial? Don't think so. So how about if we switch them around to get:
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document.write( "(3x ___ 2)*(2x ___ 1)
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document.write( "Then we multiply the 3x times the 1 and get 3x. And we multiply the 2x times the 2 and
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document.write( "get 4x. Is there anyway that you can add or subtract 3x and 4x and get +x? Sure is.
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document.write( "Suppose we had +4x - 3x. Those would add to give +x. So we need 2x times 2 to be positive.
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document.write( "That means that the 2 in the first set of parentheses needs to be positive, and if it is,
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document.write( "then the 1 in the second set of parentheses needs to be negative. So we can tell that
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document.write( "the factors need to be:
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document.write( "(3x + 2)*(2x - 1)
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document.write( "If you multiply these two factors out, you should find the product to be 
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document.write( "as it was given in the original statement of the problem.
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document.write( "Use these two factors as a replacement for the left side of the equation in the original
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document.write( "problem and you get:
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document.write( "Notice that this equation will be true if either of the factors on the left side equals zero.
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document.write( "Why? Because multiplying by zero on the left side makes the entire left side equal to
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document.write( "zero which is equal to the right side.
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document.write( "Set the 3x + 2 on the left side equal to zero and solve for x:
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document.write( "3x + 2 = 0 <--- subtract 2 from both sides so you have only 3x on the left side.
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document.write( "3x = -2
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document.write( "Now divide both sides by 3, the multiplier of x, and you find that:
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document.write( "When then the factor 3x + 2 will be zero. That means the entire left side
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document.write( "of the equation will be zero, and will equal the right side of the equation.
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document.write( "You can, Cody, find another value for x that will make the left side of the equation
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document.write( "equal to zero. Find it by setting the other factor 2x -1 equal to zero.
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document.write( "2x - 1 = 0 <--- add + 1 to both sides to get rid of the -1 on the left side
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document.write( "2x = +1
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document.write( "Solve for x by dividing both sides by 2 to get:
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document.write( "So the two values for x that will satisfy the original equation are  and  .
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document.write( "Factoring is a useful way of solving the equation you get when you set a quadratic
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document.write( "polynomial equal to zero. There are a couple of more ways that you can find the values
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document.write( "for x that will satisfy the equation that results from setting a quadratic polynomial
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document.write( "equal to zero. One of those ways involves graphing the polynomial and finding where the
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document.write( "graph crosses the x-axis (if it does). Another way is called completing the square
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document.write( "and a third was is a variation of completing the square. It is involves using a special
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document.write( "equation that you will learn from your teacher ... called the quadratic formula.
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document.write( "Completing the square or using the quadratic formula involve less guess work than factoring
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document.write( "might, but if you can find the factors easily, it is generally a faster method than the other
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document.write( "processes. Keep after it, Cody. You seem to have the right ideas about factoring,
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document.write( "but you just needed to know that you might have to try all the factors of the first term
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document.write( "(the term that involves the squared variable).\r
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