f(x) = x^2 - 2x - 8 = (x-1)^2 - 9
so, f(x) = 0 at x = 1±√9 = 4,-2
f(x) = (x-4)(x+2)
the real question was
What multiplies to -8 and adds to -2?
That is, which two factors of 8 differ by 2?
What multiplies to 4 and adds to -2?
I'm trying to find the zeroes of the function f(x)=x^2-2x-8. I did the AC/B tree (1*-8/-2) and from there I can't figure it out.
2 answers
Y = x^2 - 2x - 8.
-8 = -4*2. Sum = -4 + 2 = -2 = B.
Y = (x-4)(x+2) = 0.
x-4 = 0, X = 4.
x + 2 = 0, X = - 2.
Solution set: (4,0), (-2, 0).
-8 = -4*2. Sum = -4 + 2 = -2 = B.
Y = (x-4)(x+2) = 0.
x-4 = 0, X = 4.
x + 2 = 0, X = - 2.
Solution set: (4,0), (-2, 0).
1 answer