F(x)=x^4-12x^2-64; x-4

find the remainder when f(x) is divided by x-c & determine whether x-c is a factor of f(x)

Division Algorithm: Let f(x) and g(x) be polynomials. Then there exists unique polynomials q(x) and r(x) such that f(x) = q(x) * g(x) + r(x) and deg r(x) < deg g(x). We call r(x) the remainder.

Let f(x) = x^4 - 12x^2 - 64 and g(x) = x - 4. The by the Division Algorithm we know there exists unique polynomials q(x) and r(x) such that

x^4 - 12x^2 - 64 = q(x) * (x - 4) + r(x)

Note that deg r(x) < deg g(x) = 1 so this implies that deg r(x) = 0 so r(x) must be a constant. Then r(x) = C.

x^4 - 12x^2 - 64 = q(x) * (x - 4) + C

Now let x = 4 to solve for C.

4^4 - 12*4^2 - 64 = q(4) * (4 - 4) + C
16*16 - 12*16 - 4*16 = q(4) * 0 + C
16(16 - 12 - 4) = 0 + C
16 * 0 = C
C = 0

Thus, the remainder is zero.

We say x - 4 is a factor of f(x) if and only if f(4) = 0
f(4) = 4^4 - 12*4^2 - 64 = 0 as shown above.

Therefore, x - 4 is a factor of f(x).

remainder/factor theorem: if (x-c) a factor of f(x) then remainder is zero - check using synthetic division of 'c' i.e. f(c) = 0:

(x-4): c = 4
4 | 1, 0, -12, 0, -64
.. |_-, 4, 16, 16, 64 (add this row to top coef's of f(x) include all x^n terms)
.... 1, 4, 4, 16, 0 <-- this last value is nil so remainder =0

since rem = 0 then 4 is a root of f(x) => (x-4) is a factor.

Note: f(x) = (x^3 +4x^2 + 4x +16)(x-4)

This is easily factored to
f(x)=(x+4)(x-4)(x^2+4).
Hence x-4 is a factor.
Note f(4)=0.
If x-c is a factor of f then f can be written as f(x)=(x-c)*g(x) where g(x) is a polynomial
of degree 1 less than f.

Not sure this helps but I hope so.