Question

Given f(x)=2x3−5x2−68x+35
, apply the Factor Theorem and determine which of the following is a factor of f(x).(1 point)
Responses

(x+5)
(x+5)

(2x+1)
(2x+1)

(x-3)
(x-3)

(x+7)

Answer 1

To determine which of the given options is a factor of the polynomial \( f(x) = 2x^3 - 5x^2 - 68x + 35 \), we can apply the Factor Theorem. According to the Factor Theorem, \( (x - c) \) is a factor of \( f(x) \) if and only if \( f(c) = 0 \).

We will test each option by substituting the corresponding value of \( x \) into \( f(x) \).

1. **Testing \( (x + 5) \)**:
- Set \( c = -5 \):
\[
f(-5) = 2(-5)^3 - 5(-5)^2 - 68(-5) + 35
\]
\[
= 2(-125) - 5(25) + 340 + 35
\]
\[
= -250 - 125 + 340 + 35
\]
\[
= -250 - 125 + 375 = 0
\]

So \( (x + 5) \) is a factor of \( f(x) \).

2. **Testing \( (2x + 1) \)**:
- Set \( c = -\frac{1}{2} \):
\[
f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 - 5\left(-\frac{1}{2}\right)^2 - 68\left(-\frac{1}{2}\right) + 35
\]
\[
= 2\left(-\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) + 34 + 35
\]
\[
= -\frac{1}{4} - \frac{5}{4} + 34 + 35
\]
\[
= -\frac{6}{4} + 69 = -1.5 + 69 = 67.5 \neq 0
\]

So \( (2x + 1) \) is **not** a factor.

3. **Testing \( (x - 3) \)**:
- Set \( c = 3 \):
\[
f(3) = 2(3)^3 - 5(3)^2 - 68(3) + 35
\]
\[
= 2(27) - 5(9) - 204 + 35
\]
\[
= 54 - 45 - 204 + 35
\]
\[
= 54 - 45 - 204 + 35 = -160 \neq 0
\]

So \( (x - 3) \) is **not** a factor.

4. **Testing \( (x + 7) \)**:
- Set \( c = -7 \):
\[
f(-7) = 2(-7)^3 - 5(-7)^2 - 68(-7) + 35
\]
\[
= 2(-343) - 5(49) + 476 + 35
\]
\[
= -686 - 245 + 476 + 35
\]
\[
= -686 - 245 + 511 = -420 \neq 0
\]

So \( (x + 7) \) is **not** a factor.

In conclusion, the only factor from the provided options is:

\[
\textbf{(x + 5)}
\]