Determine the zeros of f(x) = x3 – 12x2 + 28x – 9 help please:)
1 -12 28 -9 0 <-- add these ------------- x ) 1 <- multiply these your "x" will be whatever you wanna test for a zero
1,3,9 loos to be our options
1 -12 28 -9 0 3 -27 3 ------------- 3 ) 1 -9 1 -6: -6 is NOT a zero so "3" is out
1 -12 28 -9 0 -3 45 ------------- -3 ) 1 -15 ...... this aint gonna zero out either
1 -12 28 -9 0 9 -27 9 ------------- 9 ) 1 -3 1 0 : x=9 works as a zero
x^2 -3x +1 is a quadratic form now that can be done with the quad formula if need be
thank you!
If guess that one of the roots is rational, the possible roots from the rational root theorem are ±1 , ±3 , ±9. If I try x = 9, I find that f(9) = 0. x^3 - 12x^2 + 28x - 9 = (x-9)(x^2 - 3x + 1) Using the quadratic formula, the other 2 solutions are (3 ± sqrt(9-4))/2 = (3 ± sqrt(5))/2 Solutions to f(x) = 0 : 9 , (3 + sqrt(5))/2 , (3 - sqrt(5))/2
youre welcome :)
understand?
yes thank you:)
Join our real-time social learning platform and learn together with your friends!