DeadShot
  • DeadShot
Determine the zeros of f(x) = x3 - 12x2 + 28x - 9.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
try factoring :)
DeadShot
  • DeadShot
so, I'd remove the x^2 from the first two terms, making it x^2 (x - 12 + 28x - 9)?
DeadShot
  • DeadShot
and then combine like terms, making it x^2 (29x - 21)

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anonymous
  • anonymous
... you can't just remove x^2 from the front. what about the back? you get x^2(x - 12) + 28x - 9
anonymous
  • anonymous
start by hit and tril, put x=1, -1, 2, -2 till u get f(x)=0
DeadShot
  • DeadShot
ok
anonymous
  • anonymous
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anonymous
  • anonymous
|dw:1359226195996:dw|
anonymous
  • anonymous
try ur self till u get f(x)=0 after that v proceed next
DeadShot
  • DeadShot
ok
anonymous
  • anonymous
let "a" is that value i-e f(a)=0 means "a" is a root and x-a is a factor, now divide the original equation by x-a to get another factor(that will b quadratic, factorize that and equate the three factors one by one to get the roots,
anonymous
  • anonymous
my laptop battery is low , m about to offline if i didnt reach power on time, good luck for the question
DeadShot
  • DeadShot
I couldn't find any numbers that made f(x)=0
anonymous
  • anonymous
ok sigh i don't know what he was doing
anonymous
  • anonymous
but via factoring you get (x-9)(x^2-3x+1)
anonymous
  • anonymous
therefore x = 9 and you use the quadratic formula to find the roots for x^2 - 3x + 1
DeadShot
  • DeadShot
I'm having trouble using the quadratic formula with (x^2 - 3x + 1)
anonymous
  • anonymous
\[a=1,b=-3,c=1\] \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
DeadShot
  • DeadShot
so, a=1, b=-3, and c=1?

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