Is (x-2) a factor of f(x)=x^3-8x^2+14x-4?

Question

Yes, (x-2) is a factor. There is a reminder
Yes, (x-2) is a factor. The remainder is zero
No, (x-2) is not a factor. The remainder is zero
No, (x-2) is not a factor. There is a remainder

Answer 1

To determine whether \((x - 2)\) is a factor of the polynomial \(f(x) = x^3 - 8x^2 + 14x - 4\), we can use the Remainder Theorem. According to the theorem, if \(f(c) = 0\) for some \(c\), then \((x - c)\) is a factor of the polynomial \(f(x)\).

In this case, we will evaluate \(f(2)\):

\[ f(2) = (2)^3 - 8(2)^2 + 14(2) - 4 \]

Calculating each term:

1. \((2)^3 = 8\)
2. \(-8(2)^2 = -8(4) = -32\)
3. \(14(2) = 28\)
4. The constant term is \(-4\)

Now, substituting these values into the function:

\[ f(2) = 8 - 32 + 28 - 4 \]

Now simplifying:

\[ f(2) = 8 - 32 + 28 - 4 = 8 + 28 - 32 - 4 = 36 - 36 = 0 \]

Since \(f(2) = 0\), it follows that \((x - 2)\) is a factor of \(f(x)\).

Thus, the correct answer is:

Yes, (x-2) is a factor. The remainder is zero.