Alg_Complete

# Mathlamaredutermsaspx college algebra initial form 5

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Unformatted text preview: rect this isn’t the correct factoring of the polynomial. That doesn’t mean that we guessed wrong however. With the previous parts of this example it didn’t matter which blank got which number. This time it does. Let’s flip the order and see what we get. ( 3x - 4 ) ( x + 2 ) = 3 x 2 + 2 x - 8 So, we got it. We did guess correctly the first time we just put them into the wrong spot. So, in these problems don’t forget to check both places for each pair to see if either will work. [Return to Problems] (f) 5 x 2 - 17 x + 6 Again the coefficient of the x2 term has only two positive factors so we’ve only got one possible © 2007 Paul Dawkins 36 http://tutorial.math.lamar.edu/terms.aspx College Algebra initial form. ( 5 x 2 - 17 x + 6 = 5 x + Next we need all the factors of 6. Here they are. (1) ( 6 ) ( -1) ( -6 ) )( x + ) ( 2 ) ( 3) ( -2 ) ( -3) Don’t forget the negative factors. They are often the ones that we want. In fact, upon noticing that the coefficient of the x is negative we can be assured that we will need one of the two pairs of negative factors since that will be the only way we will get negative coefficient there. With some trial and error we can get that the factoring of this polynomial is, 5 x 2 - 17 x + 6 = ( 5 x - 2 ) ( x - 3) [Return to Problems] (g) 4 x 2 + 10 x - 6 In this final step we’ve got a harder problem here. The coefficient of the x2 term now...
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