DeadShot
Determine the zeros of f(x) = x3 - 12x2 + 28x - 9.
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RONNCC
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try factoring :)
DeadShot
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so, I'd remove the x^2 from the first two terms, making it x^2 (x - 12 + 28x - 9)?
DeadShot
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and then combine like terms, making it x^2 (29x - 21)
RONNCC
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... you can't just remove x^2 from the front. what about the back? you get x^2(x - 12) + 28x - 9
ZakaullahUET
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start by hit and tril, put x=1, -1, 2, -2 till u get f(x)=0
DeadShot
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ok
ZakaullahUET
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ZakaullahUET
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|dw:1359226195996:dw|
ZakaullahUET
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try ur self till u get f(x)=0
after that v proceed next
DeadShot
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ok
ZakaullahUET
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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,
ZakaullahUET
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my laptop battery is low , m about to offline if i didnt reach power on time, good luck for the question
DeadShot
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I couldn't find any numbers that made f(x)=0
RONNCC
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ok sigh i don't know what he was doing
RONNCC
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but via factoring you get (x-9)(x^2-3x+1)
RONNCC
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therefore x = 9 and you use the quadratic formula to find the roots for x^2 - 3x + 1
DeadShot
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I'm having trouble using the quadratic formula with (x^2 - 3x + 1)
anonymous
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\[a=1,b=-3,c=1\]
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
DeadShot
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so, a=1, b=-3, and c=1?