What are the roots of f(x)=x^3-x^2-5x-3?

x=3 and x=-1

Step-by-step explanation:

Given Descartes' Rule of Signs, there is only one positive root and one negative root.

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient.

So P could be +1, -1, +3, or -3, while Q could be 1 or -1.

Therefore, P/Q could be +1, -1, +3, or -3.

Then we use synthetic division to find the correct factor(s):

1 | 1 -1 -5   -3

___ 1_0_ -5

   1  0 -5 | -8

So (x-1) is not a factor

-1 | 1 -1 -5   -3

___-1_2_  3

    1 -2 -3 |  0

So (x+1) is a factor where the quotient is x^2-2x-3 or (x-3)(x+1)

3 | 1 -1 -5   -3

___ 3_6_ 3

    1  2  1 |  0

So (x-3) is a factor where the quotient is x^2+2x+1 or (x+1)^2

Given (x-3) and (x+1) are factors, we now use the Zero Product Property:

(x-3)(x+1)=0

x-3=0

x=3

x+1=0

x=-1

So the roots are x=3 and x=-1

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step-by-step explanation: