Posted by Sandra on Thursday, May 24, 2007 at 6:08pm.
I can't for the life of me solve this!
6x^2-23x+15 = (a x + b) (c x + d) -->
a c = 6 (1)
a d + b c = -23 (2)
b d = 15 (3)
You could write down all possibilities for a and c (Eq. 1) and all possibilites for b and d (Eq. 3) and see which one satisfies Eq. 2
But there are 16 possibilities, so let's not do this. Instead let's take another look at Eq. 3. This equation is just saying that the factorization of the quadratic must be valid at x = 0!
So, perhaps we can simplify things by just demanding that the factorization must also be valid at another point, say, x = 1. For x = 1 the quadratic 6x^2-23x+15 equals -2. If we substitute x = 1 in (a x + b) (c x + d) we get
(a + b)(c + d). This means that:
(a + b)(c + d) = -2 (4)
a + b = -2 and c + d = 1 (5)
a + b = 2 and c + d = -1 (6)
Now a and c are factors of the number 6 as you can see from Eq. 1 while b and d are factors of 15 (Eq. 3).
You now only need a little trial and error to find the solution:
a= -1; b = 3; c= -6; d = 5, because that satisfies Eq. 6:
-1 + 3 = 2
-6 + 5 = -1
Note that if you expand out Eq. 4 and use Eq. 1 and Eq. 2, you get Eq. 2 so you don't have to check if Eq. 2 is satisfied. So you have obtained the factorization:
6x^2-23x+15 = (-x+3)(-6x+5).
it doesn't specify a method, so why not just use the formula
x=(23 ± √169)/12
x=(23 ± 13)/12
x = 3 or x= 5/6
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