How do you find all zeros with multiplicities of f(x)=x3−7x2+x−7?

1 Answer
George C.
Sep 17, 2017

The zeros of f(x) are 7, i, −i, all with multiplicity 1.

Explanation:

Given:

f(x)=x3−7x2+x−7

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:

x3−7x2+x−7=(x3−7x2)+(x−7)

x3−7x2+x−7=x2(x−7)+1(x−7)

x3−7x2+x−7=(x2+1)(x−7)

Note that x2+1 has no linear factors with real coefficients since x2+1>0 for all real values of x. We can factor it as a difference of squares using complex coefficients:

x2+1=x2−i2=(x−i)(x+i)

So the zeros of f(x) are 7, i, −i, all with multiplicity 1.

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