Given f(x)=2x3−5x2−68x+35

Question

, 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 $c=-\frac{1}{2}$: $$f(-\frac{1}{2})=2{(-\frac{1}{2})}^3-5{(-\frac{1}{2})}^2-68(-\frac{1}{2})+35$$ $$=2(-\frac{1}{8})-5\left(\frac{1}{4}\right)+34+35$$ $$=-\frac{1}{4}-\frac{5}{4}+34+35$$ $$=-\frac{6}{4}+69=-1.5+69=67.5\ne 0$$

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

3. Testing $(x-3)$:

   • Set $c=3$: $$f(3)=2(3{)}^3-5(3{)}^2-68(3)+35$$ $$=2(27)-5(9)-204+35$$ $$=54-45-204+35$$ $$=54-45-204+35=-160\ne 0$$

   So $(x-3)$ is not a factor.

4. Testing $(x+7)$:

   • Set $c=-7$: $$f(-7)=2(-7{)}^3-5(-7{)}^2-68(-7)+35$$ $$=2(-343)-5(49)+476+35$$ $$=-686-245+476+35$$ $$=-686-245+511=-420\ne 0$$

   So $(x+7)$ is not a factor.

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

$$\mathbf{\text{(x + 5)}}$$