Do you need a complete factorization? If so, you are not finished. x2 - 7x + 10 is factorable.kenny1999 said:Homework Equations
I have no problem with (a) and f(x) = (x+1) (x^2 - 7x + 10)
eumyang said:Do you need a complete factorization? If so, you are not finished. x2 - 7x + 10 is factorable.
kenny1999 said:i have factored f(x) into (x+1)(x-5)(x-2)
but no idea to go further
eumyang said:Using the factored form of f(x), find f(x+k).
I'm assuming that HCF=GCF (greatest common factor). You are looking for a positive k such that there is a linear common factor between f(x) and f(x+k). So one of the three factors of f(x) should also be a factor of f(x+k). I used trial and error to find such a k. Find a k so that when you plug into the factored form of f(x+k), one of its factors is also a factor of f(x).
(Hope I'm not giving away too much here.)
kenny1999 said:hi how about LCM?
Solving for k and LCM in a function involves finding the value of k that makes the function have the least common multiple (LCM) with another function. This means finding the smallest value of k that results in the two functions having the same repeating pattern.
Finding the value of k and LCM in a function is important because it allows us to understand the relationship between two functions and how they behave. It can also help us simplify complex functions and make predictions about their behavior.
The first step is to set the two functions equal to each other and solve for k. Then, substitute the value of k into one of the functions to find the LCM. Finally, simplify the LCM and the original functions to find the final values of k and LCM.
Yes, the value of k and LCM can be negative. The value of k can be negative if it results in the two functions having the same repeating pattern, and the LCM can be negative if one of the original functions has a negative coefficient.
Solving for k and LCM in a function can be useful in fields such as engineering, physics, and economics. It can help in predicting periodic patterns in data, optimizing processes, and finding the most efficient solutions.