Question

Using long division method, show that x-5 is a factor of x3 - 125

Answer 1

Step-by-step explanation:

If the remainder is zero when (x-5) divides (x³ - 125) then we can say that (x-5) is a factor. Let us work out the long division

x-5 ) x³ + 0.x² + 0.x -125 ( x² + 5x - 25

x³ - 5x²

⁽⁻⁾ ⁽⁺⁾

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

5x² + 0.x

5x² - 25x

⁽⁻⁾ ⁽⁺⁾

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

25x - 125

25x - 125

⁽⁻⁾ ⁽⁺⁾

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

0

Since the remainder is zero, it is shown that (x-5) is a factor of (x³- 125)