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July 30, 2014

July 30, 2014

Posted by **Algebra Help** on Monday, May 7, 2007 at 8:28pm.

40x^2+2x-65

Notice that for x = 1 the value is -23. The number 23 is a prime number and the only factors are thus 23 and 1 up to signs. So, you can simply the search for the factors of the polynomial a great deal by substituting:

x = 1 + t which yields

40 t^2 + 82 t - 23 =

(10 t + 23) (4 t - 1)

You then substitute back x by putting

t = x - 1:

(10 t + 23) (4 t - 1) =

(10 x + 13) (4 x - 5)

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