Question

if $f(x) = x^3 + 12x^2 + 23x - 36$ and $f(-4) = 0$, then find all of the zeros of $f(x)$ algebraically.

answer attempt 1 out of 2

+ add a zero - no zeros

$x = \square$ submit answer

Explanation:

Step1: Factor using root

Since \( f(-4) = 0 \), \( (x + 4) \) is a factor of \( f(x)=x^{3}+12x^{2}+23x - 36 \). We perform polynomial division or use synthetic division. Let's use synthetic division with root \(-4\):

$$
LATEXBLOCK0
$$

So, \( f(x)=(x + 4)(x^{2}+8x - 9) \).

Step2: Factor quadratic

Factor \( x^{2}+8x - 9 \). We need two numbers that multiply to \(-9\) and add to \(8\). Those numbers are \(9\) and \(-1\). So, \( x^{2}+8x - 9=(x + 9)(x - 1) \).

Step3: Find zeros

Set each factor equal to zero:

• \( x + 4 = 0 \Rightarrow x=-4 \)
• \( x + 9 = 0 \Rightarrow x=-9 \)
• \( x - 1 = 0 \Rightarrow x=1 \)

Answer:

The zeros of \( f(x) \) are \( x = -9 \), \( x = -4 \), and \( x = 1 \).