Question

question 1 · 1 point consider the graph of the function $f(x)=\frac{x^{3}+3x^{2}+2x}{x^{2}+3x + 2}$. which is a removable discontinuity for the graph? select all that apply. select all that apply: $x = 0$ $x = 1$ $x=-1$ $x = 2$ $x=-2$

Explanation:

Step1: Factor the numerator and denominator

Factor $x^{3}+3x^{2}+2x=x(x + 1)(x + 2)$ and $x^{2}+3x + 2=(x + 1)(x+2)$. So $f(x)=\frac{x(x + 1)(x + 2)}{(x + 1)(x + 2)}$, $x
eq - 1,-2$.

Step2: Simplify the function

After canceling out the common factors $(x + 1)$ and $(x + 2)$ (for $x
eq - 1,-2$), we get $f(x)=x$, $x
eq - 1,-2$.
A removable - discontinuity occurs when a factor in the denominator can be canceled out with a factor in the numerator.
The values of $x$ that make the original denominator zero are found by setting $x^{2}+3x + 2 = 0$, which gives $(x + 1)(x + 2)=0$, so $x=-1$ and $x=-2$. Since we can cancel out the factors $(x + 1)$ and $(x + 2)$ in the rational - function, $x=-1$ and $x=-2$ are removable discontinuities.

Answer:

C. $x=-1$, E. $x=-2$