Question

how many x intercept does the fuction f\left(x\right)=3\left(2x-1\right)\left(x+2\right)\left(8x-5\right)^2

Answer 1

To find the \( x \)-intercepts of the function \( f(x) = 3(2x - 1)(x + 2)(8x - 5)^2 \), we need to set the function equal to zero and solve for \( x \):

\[
f(x) = 0
\]

This implies that:

\[
3(2x - 1)(x + 2)(8x - 5)^2 = 0
\]

Since \( 3 \) is a constant multiplier, we can ignore it when finding the \( x \)-intercepts. Therefore, we need to solve:

\[
(2x - 1)(x + 2)(8x - 5)^2 = 0
\]

Now we can find the \( x \)-intercepts by setting each factor equal to zero:

1. \( 2x - 1 = 0 \)
\[
2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}
\]

2. \( x + 2 = 0 \)
\[
x = -2
\]

3. \( (8x - 5)^2 = 0 \)
Since this equation is squared, it will only provide one solution, which is:
\[
8x - 5 = 0 \quad \Rightarrow \quad 8x = 5 \quad \Rightarrow \quad x = \frac{5}{8}
\]

Now we summarize the \( x \)-intercepts:
- From \( 2x - 1 = 0 \): \( x = \frac{1}{2} \)
- From \( x + 2 = 0 \): \( x = -2 \)
- From \( (8x - 5)^2 = 0 \): \( x = \frac{5}{8} \) (counted once)

Thus, the distinct \( x \)-intercepts are:
1. \( x = \frac{1}{2} \)
2. \( x = -2 \)
3. \( x = \frac{5}{8} \)

In total, there are **three distinct \( x \)-intercepts** for the function \( f(x) \).