please help with f(x) = (x-4)^3-4(x-4)
- Thread starter Jodene222
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D
Deleted member 4993
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What do you need to do??
pka
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\(\displaystyle \begin{array}{rcl}
\left( {x - 4} \right)^3 - 4\left( {x - 4} \right) & = & \left( {x - 4} \right)\left[ {\left( {x - 4} \right)^2 - 4} \right] \\
& = & \left( {x - 4} \right)\left( {x^2 - 8x + 12} \right) \\
\end{array}\)
Can you finish?
\left( {x - 4} \right)^3 - 4\left( {x - 4} \right) & = & \left( {x - 4} \right)\left[ {\left( {x - 4} \right)^2 - 4} \right] \\
& = & \left( {x - 4} \right)\left( {x^2 - 8x + 12} \right) \\
\end{array}\)
Can you finish?
D
Deleted member 4993
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I'll do a similar poblem for you.
Find all the zero's of
f(x) = A*(x-b)^3 - C*(x-b)
= (x-b) * [A*(x-b)^2 - C]
Now further factorize the expression within [] - it is a quadraticand easy to factorize.
Then use the definition of "zeros of function" to indicate the zeros.
Find all the zero's of
f(x) = A*(x-b)^3 - C*(x-b)
= (x-b) * [A*(x-b)^2 - C]
Now further factorize the expression within [] - it is a quadraticand easy to factorize.
Then use the definition of "zeros of function" to indicate the zeros.
D
Deleted member 4993
Guest
You dont need to expand the cubic equation!!!
Look at the solutions and hints provided above - carefully
Look at the solutions and hints provided above - carefully
please HELP
I don't know how you did this. However, I was able to solve the problem and get the right answers, but do not fully understand this formula. It seems like I am missing 1 (x-b)
f(x) = A*(x-b)^3 - C*(x-b)
= (x-b) * [A*(x-b)^2 - C]
I don't know how you did this. However, I was able to solve the problem and get the right answers, but do not fully understand this formula. It seems like I am missing 1 (x-b)
f(x) = A*(x-b)^3 - C*(x-b)
= (x-b) * [A*(x-b)^2 - C]
D
Deleted member 4993
Guest
Re: thanks
Jodene222 said: