How do you find all zeros of the function f(x)=x3−x2−4x−6 given 3 as a zero?

1 Answer
Shwetank Mauria
Feb 24, 2016

Other two zeros of f(x) are (−1+i) and (−1−i).

Explanation:

Zeros of the function f(x)=x3−x2−4x−6 are those values of x for which f(x)=0. As the term independent of x is −6, factors of −6 i.e. {1,−1,2,−2,3,−3,6,−6} could be among zeros of f(x).

It is seen that putting x=3 makes f(x)=0 and hence (x−3) isne such factor. Dividing f(x) by (x−3) we get

x3−x2−4x−6

= x2(x−3)+2x(x−3)+2(x−3)

= (x2+2x+2)(x−3)

As the determinant (b2−4ac in the function ax2+bx+c) is equai to 22−4⋅1⋅2=−4, is a negative number, no further real zeros are there.

Other complex zeros are given by −b±√b2−4ac2a

i.e. −2±√22−4⋅1⋅22⋅1 or −2±√−42

which simplifies to (−1+i) and (−1−i), which are other two zeros of f(x).

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