How do you find all the zeros of f(x)=x3+13x2+57x+85?

1 Answer
George C.
May 5, 2016

x=5 or x=−4±i

Explanation:

f(x)=x3+13x2+57x+85

By the rational roots theorem, any rational zeros of f(x) must be expressible in the form pq for integers p, q with p a divisor of the constant term 85 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational zeros are:

±1, ±5, ±17, ±85

In addition, since all of the coefficients of f(x) are positive, it has no positive zeros, so that leaves:

−1, −5, −17, −85

Trying each of these in turn we find:

f(−5)=−125+325−285+85=0

So x=−5 is a zero and (x+5) a factor:

x3+13x2+57x+85=(x+5)(x2+8x+17)

We can find the remaining two zeros by completing the square:

0=x2+8x+17

=(x+4)2−16+17

=(x+4)2+1

=(x+4)2−i2

=((x+4)−i)((x+4)+i)

=(x+4−i)(x+4+i)

So x=−4±i

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