Question

Show how to find the roots (zeros) of the function in the picture below.

Answer 1

Answer:

The roots (zeros) of the function are:

$x=5,\:x=-8$

Step-by-step explanation:

Given the function

$f\left(x\right)=x^2+3x-40$

substitute f(x) = 0 to determine the zeros of the function

$0=x^2+3x-40$

First break the expression x² + 3x - 40  into groups

x² + 3x - 40 = (x² - 5x) + (8x - 40)

Factor out x from x² - 5x:  x(x - 5)

Factor out 8 from 8x - 40:  8(x - 5)

Thus, the expression becomes

$0=x\left(x-5\right)+8\left(x-5\right)$

switch the sides

$x\left(x-5\right)+8\left(x-5\right)=0$

Factor out common term x - 5

$(x - 5) (x + 8) = 0$

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

$x-5=0\quad \mathrm{or}\quad \:x+8=0$

Solve x - 5 = 0

x - 5 = 0

adding 5 to both sides

x - 5 + 5 = 0 + 5

x = 5

solve x + 8 = 0

x + 8 = 0

subtracting 8 from both sides

x + 8 - 8 = 0 - 8

x = -8

Therefore, the roots (zeros) of the function are:

$x=5,\:x=-8$