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

User Kviktor
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1 Answer

3 votes

Answer:

x=3 and x=-1

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

User Chikitin
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7.0k points