How do you find all the zeros of f(x)=x3+11x2+41x+51?

1 Answer
George C.
Mar 21, 2016

Use the rational root theorem to find zero x=−3, divide by (x+3) then solve the remaining quadratic to find zeros x=−4±i.

Explanation:

f(x)=x3+11x2+41x+51

by the rational root theorem, any rational zeros of f(x) must be expressible in the form pq for integers p and q where p is a divisor of the constant term 51 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational roots are:

±1, ±3, ±17, ±51

In addition, note that all of the coefficients of f(x) are positive, so it can only have zeros for negative values of x.

That leaves:

−1, −3, −17, −51

Trying each in turn we find:

f(−1)=−1+11−41+51=20

f(−3)=−27+99−123+51=0

So x=−3 is a zero and (x+3) a factor:

x3+11x2+41x+51=(x+3)(x2+8x+17)

Then solve:

0=x2+8x+17

=x2+8x+16+1

=(x+4)2−i2

=((x+4)−i)((x+4)+i)

=(x+4−i)(x+4+i)

So the remaining zeros are: x=−4±i

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