How to find all the zeros of f(x)=x^4+4x^3-6x^2-36x-27?

amamnebrajqx

amamnebrajqx

Answered question

2023-01-31

How to find all the zeros of f ( x ) = x 4 + 4 x 3 - 6 x 2 - 36 x - 27 ?

Answer & Explanation

Natalia Arroyo

Natalia Arroyo

Beginner2023-02-01Added 4 answers

f ( x ) = x 4 + 4 x 3 - 6 x 2 - 36 x - 27
By the rational roots theorem, any rational zeros of f ( x ) are expressible in the form p q for integers p , q with p a divisor of the constant term - 27 and q a divisor of the coefficient 1 of the leading term.
Thus, the following are the only rational zeros that could exist:
± 1 , ± 3 , ± 9 , ± 27
We find:
f ( - 1 ) = 1 - 4 - 6 + 36 - 27 = 0
So x = - 1 is a zero and ( x + 1 ) a factor:
x 4 + 4 x 3 - 6 x 2 - 36 x - 27 = ( x + 1 ) ( x 3 + 3 x 2 - 9 x - 27 )
The ratio between the first and second terms and the third and fourth terms is the same in the remaining cubic. Due of the grouping, this cubic will factor:
x 3 + 3 x 2 - 9 x - 27
= ( x 3 + 3 x 2 ) - ( 9 x + 27 )
= x 2 ( x + 3 ) - 9 ( x + 3 )
= ( x 2 - 9 ) ( x + 3 )
= ( x 2 - 3 2 ) ( x + 3 )
= ( x - 3 ) ( x + 3 ) ( x + 3 )
So the remaining zeros are x=3 (with multiplicity 1) and x=-3 with multiplicity 2

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