How to find all roots for x^3 - x^2 - 14x + 24 = 0?

Nathalie Dixon

Nathalie Dixon

Answered question

2023-02-01

How to find all roots for x 3 - x 2 - 14 x + 24 = 0 ?

Answer & Explanation

Clarence Bright

Clarence Bright

Beginner2023-02-02Added 1 answers

f ( x ) = x 3 - x 2 - 14 x + 24
By the rational root theorem, any rational zeros of f ( x ) must be expressible in the form p q for integers p , q with p a divisor of the constant term 24 and q a divisor of the coefficient 1 of the leading term.
Thus, the following are the only rational zeros that could exist:
± 1 , ± 2 , ± 3 , ± 4 , ± 6 , ± 8 , ± 12 , ± 24
Trying each in turn, we find:
f ( 1 ) = 1 - 1 - 14 + 24 = 10
f ( - 1 ) = - 1 - 1 + 14 + 24 = 36
f ( 2 ) = 8 - 4 - 28 + 24 = 0
So x = 2 is a zero and ( x - 2 ) a factor:
x 3 - x 2 - 14 x + 24 = ( x - 2 ) ( x 2 + x - 12 )
We can factor the remaining quadratic by finding a pair of factors of the absolute value 12 of the constant term which differ by the coefficient 1 of the middle term. The pair 4 , 3 works.
Thus, we discover:
x 2 + x - 12 = ( x + 4 ) ( x - 3 )
Putting it together:
x 3 - x 2 - 14 x + 24 = ( x - 2 ) ( x + 4 ) ( x - 3 )
which has zeros x = 2 , x = - 4 and x = 3

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