document.write( "Question 181388This question is from textbook
\n" ); document.write( ": Explain how to factor a polynomial of the form ax^2+bx+c when a is not equal
\n" ); document.write( "to 1. Describe both the grouping as well as reversing FOIL. Contrast the two
\n" ); document.write( "methods by means of an example. Discuss which is the best approach and why.
\n" ); document.write( "I am sooo lost. Thank You\r
\n" ); document.write( "\n" ); document.write( "Denise \n" ); document.write( "

Algebra.Com's Answer #135994 by Mathtut(3670) $\"\"$ $\"About$

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grouping is a method by which you are trying to see a pattern so that you can factor more easily:
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\n" ); document.write( "such as $\"2x%5E2%2B8x%2B8\"$ . In this example we can factor out a 2 to get an (a) coeffcient of 1. which is easier to factor than problems where a is not 1.
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\n" ); document.write( " $\"2%28x%5E2%2B4x%2B4%29\"$ $\"x%5E2%2B4x%2B4\"$ can be further factored to (x+2)(x+2)
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\n" ); document.write( "grouping is uncommon in degrees of 2 or less but would be the easier method if you spotted a common pattern. This method is used extensively for polynomials that are 3rd degree and above.
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\n" ); document.write( "Reverse FOIL is simply looking at the coefficients of the $\"x%5E2\"$ and the constant terms in other words a and c
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\n" ); document.write( "the first part of the factor are made up of the factors of a and the 2nd part of the factor is made up from the factors of c.
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\n" ); document.write( "the combination of The OI from FOIL is what produces b the coefficient of the x term. A disadvantage of this method is if a or c is large it could take some time to figure out the right combination for the b coefficient. I believe this method is better 2nd degree poly's. example would be
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\n" ); document.write( " $\"2x%5E2-11x%2B12\"$ we know that the factors will be of the pattern (2x+?)(x+?) in order to get $\"2x%5E2\"$ .....the ? will be composed of factors of c which is 1 and 12, 2 and 6, and 3 and 4. We can begin testing to see which pair will give us the -11 part of x. Lets try 2 and 6
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\n" ); document.write( "(2x+2)(x+6) will not give us a negative 11 so we have to try both factors with minus signs (2x-2)(x-6). that will give us -14x now switch the 2 and 6
\n" ); document.write( "(2x-6)(x-2)...that gives us -10x....so lets move on to 3 and 4
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\n" ); document.write( "cant have pluses in the factors because we will never get a negative middle term
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\n" ); document.write( "so try(2x-4)(x-3) that gives us -10x. reverse them(2x-3)(x-4)= $\"2x%5E2-11x%2B12\"$ so that is it we found the factors. \n" ); document.write( "

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