If it is anything else this wont work and we really

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Unformatted text preview: rs. ( -1) (15 ) (1) ( -15 ) ( -3 ) ( 5 ) ( 3 ) ( -5 ) Now, we can just plug these in one after another and multiply out until we get the correct pair. However, there is another trick that we can use here to help us out. The correct pair of numbers must add to get the coefficient of the x term. So, in this case the third pair of factors will add to “+2” and so that is the pair we are after. Here is the factored form of the polynomial. x 2 + 2 x - 15 = ( x - 3) ( x + 5 ) Again, we can always check that we got the correct answer my doing a quick multiplication. Note that the method we used here will only work if the coefficient of the x2 term is one. If it is anything else this won’t work and we really will be back to trial and error to get the correct factoring form. [Return to Problems] (b) x 2 - 10 x + 24 Let’s write down the initial form again, ( x 2 - 10 x + 24 = x + )( x + ) Now, we need two numbers that multiply to get 24 and add to get -10. It looks like -6 and -4 will do the trick and so the factored form of this p...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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