Find the zeros of f(x)=x^3-5x^2-x+5 by factoring

-1, 1, 5

Step-by-step explanation:

Factor by grouping:

Factor out x² from the first two terms:

f(x) = x³ - 5x² - x + 5

f(x) = x²(x - 5) - x + 5

Factor out -1 from the last two terms:

f(x) = x²(x - 5) -1(x - 5)

Then, put the two terms together:

f(x) = (x² - 1)(x - 5)

(x² - 1) is a difference of squares that can be factored further:

f(x) = (x + 1)(x - 1)(x - 5)

Set this equal to 0, and set each factor equal to zero. Then, solve for x:

x + 1 = 0

x = -1

x - 1 = 0

x = 1

x - 5 = 0

x = 5

So, the zeroes are -1, 1, and 5

sine

step-by-step explanation: