Full solution

f(x)=(10x-3)(4x+1)(5x-2)

. What is the sum of all of the zeros of function

f

Given Function: We are given the function $f(x) = (10x - 3)(4x + 1)(5x - 2)$ . To find the sum of all the zeros of the function, we need to find the values of $x$ that make $f(x) = 0$ . These values are the solutions to the equations $10x - 3 = 0$ , $4x + 1 = 0$ , and $5x - 2 = 0$ .
Solve $10x - 3 = 0$ : Solve the equation $10x - 3 = 0$ for $x$ . $\newline$ $10x - 3 = 0$ $\newline$ $10x = 3$ $\newline$ $x = \frac{3}{10}$ $\newline$ The zero for this part of the function is $x = \frac{3}{10}$ .
Solve $4x + 1 = 0$ : Solve the equation $4x + 1 = 0$ for $x$ .
$4x + 1 = 0$
$4x = -1$
$x = -\frac{1}{4}$
The zero for this part of the function is $x = -\frac{1}{4}$ .
Solve $5x - 2 = 0$ : Solve the equation $5x - 2 = 0$ for $x$ . $\newline$ $5x - 2 = 0$ $\newline$ $5x = 2$ $\newline$ $x = \frac{2}{5}$ $\newline$ The zero for this part of the function is $x = \frac{2}{5}$ .
Find Sum of Zeros: Now that we have all the zeros of the function, we can find their sum. $\newline$ Sum of zeros = $(\frac{3}{10}) + (\frac{-1}{4}) + (\frac{2}{5})$ $\newline$ To add these fractions, find a common denominator, which is $20$ . $\newline$ Sum of zeros = $(\frac{6}{20}) + (\frac{-5}{20}) + (\frac{8}{20})$ $\newline$ Sum of zeros = $(6 - 5 + 8) / 20$ $\newline$ Sum of zeros = $\frac{9}{20}$