A basic fact of algebra states that c is a root of a polynomial f(x) if and only if f(x) = (x-c)g(x) for some polynomial g(x). We say that c is a multiple root if f(x) = [(x-c)^2](h(x)) where h(x) is a polynomial.
Show that c is a multiple root of f(x) if and only if c is a root of both f(x) and f'(x)
only if:
f(x) = (x-c)^2 h(x)
f'(x) = 2(x-c) h(x) + (x-c)^2 * h'(x)
= (x-c) * (2h + (x-c)*h^2)
This should get you going. The "if" logic kind of works backwards from here.
To show that c is a multiple root of f(x) if and only if c is a root of both f(x) and f'(x), we need to prove two implications:
1. If c is a multiple root of f(x), then c is a root of both f(x) and f'(x).
2. If c is a root of both f(x) and f'(x), then c is a multiple root of f(x).
Let's start with the first implication:
1. If c is a multiple root of f(x), then c is a root of both f(x) and f'(x).
To prove this, we need to show that if c is a multiple root of f(x), then f(c) = 0 and f'(c) = 0.
Since c is a multiple root of f(x), we know that f(x) = [(x-c)^2] * h(x), where h(x) is a polynomial.
Taking the derivative of both sides of the equation, we get:
f'(x) = 2(x-c)*h(x) + [(x-c)^2]*h'(x)
Now, substitute x = c into the above equation:
f'(c) = 2(c-c)*h(c) + [(c-c)^2]*h'(c)
= 0 + 0 * h'(c)
= 0
Therefore, c is a root of f'(x), i.e., f'(c) = 0.
Similarly, substitute x = c into the original equation f(x) = [(x-c)^2] * h(x):
f(c) = [(c-c)^2] * h(c)
= 0 * h(c)
= 0
Hence, c is a root of f(x), i.e., f(c) = 0.
Therefore, if c is a multiple root of f(x), then c is a root of both f(x) and f'(x).
Now, let's move on to the second implication:
2. If c is a root of both f(x) and f'(x), then c is a multiple root of f(x).
To prove this, we need to show that if c is a root of both f(x) and f'(x), then f(x) = [(x-c)^2] * g(x), where g(x) is a polynomial.
Since c is a root of f(x), we know that f(c) = 0. This implies that [(c-c)^2] * g(c) = 0. Since (c-c)^2 = 0, we have 0 * g(c) = 0, which implies that g(c) = 0.
Therefore, c is a root of g(x).
Now, let's find the derivative of the equation f(x) = [(x-c)^2] * g(x):
f'(x) = 2(x-c) * g(x) + [(x-c)^2] * g'(x)
Substituting x = c into the above equation:
f'(c) = 2(c-c) * g(c) + [(c-c)^2] * g'(c)
= 0 + 0 * g'(c)
= 0
Hence, c is a root of f'(x), i.e., f'(c) = 0.
Therefore, if c is a root of both f(x) and f'(x), then c is a multiple root of f(x).
In conclusion, we have proved both implications, and thus, c is a multiple root of f(x) if and only if c is a root of both f(x) and f'(x).