How do you find all the zeros of f(x)=x3−4x2+9x−36?

1 Answer
George C.
May 23, 2018

The zeros are:

x=4 and x=±3i

Explanation:

Given:

f(x)=x3−4x2+9x−36

Note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

x3−4x2+9x−36=(x3−4x2)+(9x−36)

x3−4x2+9x−36=x2(x−4)+9(x−4)

x3−4x2+9x−36=(x2+9)(x−4)

x3−4x2+9x−36=(x2+32)(x−4)

x3−4x2+9x−36=(x2−(3i)2)(x−4)

x3−4x2+9x−36=(x−3i)(x+3i)(x−4)

So the zeros are:

x=4 and x=±3i

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