How to find the zeros of f(x) = 2x^3 − 5x^2 − 28x + 15?

Nathanial Frost

Nathanial Frost

Answered question

2023-02-18

How to find the zeros of f ( x ) = 2 x 3 5 x 2 28 x + 15 ?

Answer & Explanation

Cheyenne Lynn

Cheyenne Lynn

Beginner2023-02-19Added 10 answers

f ( x ) = 2 x 3 - 5 x 2 - 28 x + 15
By the rational root theorem, since f ( x ) has integer coefficients, any rational zeros must be expressible in the form p q for integers p , q with p a divisor of the constant term 15 and q a divisor of the coefficient 2 of the leading term.
That means that the only possible rational zeros of f ( x ) are:
± 1 2 , ± 1 , ± 3 2 , ± 5 2 , ± 3 , ± 5 , ± 15 2 , ± 15
Attempting the first one we locate
f ( 1 2 ) = 2 8 - 5 4 - 28 2 + 15 = 1 - 5 - 56 + 60 4 = 0
So x = 1 2 is a zero and ( 2 x - 1 ) a factor:
2 x 3 - 5 x 2 - 28 x + 15
= ( 2 x - 1 ) ( x 2 - 2 x - 15 )
To factor the remaining quadratic, find a pair of factors of 15 which differ by 2 . The pair 5 , 3 works, so:
x 2 - 2 x - 15 = ( x - 5 ) ( x + 3 )
So the two remaining zeros are x = 5 and x = - 3

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